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Prévia do material em texto

ROMANIAN MATHEMATICAL SOCIETYROMANIAN MATHEMATICAL SOCIETY
Mehedinți BranchMehedinți Branch
 SUMMER EDITION 2021 SUMMER EDITION 2021 
R. M. M. - 29 R. M. M. - 29 
ROMANIAN MATHEMATICAL ROMANIAN MATHEMATICAL 
MAGAZINEMAGAZINE
ISSN 2501-0099ISSN 2501-0099
Romanian Mathematical Society-Mehedinți Branch 2021 
 
1 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
 
 
ROMANIAN MATHEMATICAL 
SOCIETY 
 Mehedinți Branch 
 
 
 
 
ROMANIAN MATHEMATICAL MAGAZINE 
R.M.M. 
 
 
 
Nr.29-SUMMER EDITION 2021 
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
2 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
 
ROMANIAN MATHEMATICAL 
SOCIETY 
 Mehedinți Branch 
DANIEL SITARU-ROMANIA EDITOR IN CHIEF 
ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT 
ISSN 1584-4897 
GHEORGHE CĂINICEANU-ROMANIA 
 
 
 
 
EDITORIAL BOARD 
 
D.M.BĂTINEȚU-GIURGIU-ROMANIA 
CLAUDIA NĂNUȚI-ROMANIA 
FLORICĂ ANASTASE-ROMANIA 
NECULAI STANCIU-ROMANIA 
MARIAN URSĂRESCU-ROMANIA 
MARIN CHIRCIU-ROMANIA 
DAN NĂNUȚI-ROMANIA 
IULIANA TRAȘCĂ-ROMANIA 
EMILIA RĂDUCAN-ROMANIA 
DRAGA TĂTUCU MARIANA-ROMANIA 
DANA PAPONIU-ROMANIA 
GIMOIU IULIANA-ROMANIA 
DAN NEDEIANU-ROMANIA 
OVIDIU TICUȘI-ROMANIA 
LAVINIU BEJENARU-ROMANIA 
ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL 
ISSN 2501-0099 WWW.SSMRMH.RO 
DANIEL WISNIEWSKI-USA 
EDITORIAL BOARD VALMIR KRASNICI-KOSOVO 
 
 
 
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
3 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
CONTENT 
 
INEQUALITIES WITH CONVEX FUNCTIONS-Daniel Sitaru,Claudia Nănuți……………….……………..4 
ABOUT RMM INEQUALITY IN TRIANGLE 899 - Marin Chirciu …….………………………..……………..7 
AN EXTENSION OF A TRIANGLE IDENTITY TO THE EUCLIDEAN N-SIMPLEX- Vasile Jiglău ......10 
NEW INEQUALITIES WITH FIBONACCI AND LUCAS’ NUMBERS- Florică Anastase......................12 
120 YEARS OF LALESCU SEQUENCES (I)- Bătineţu-Giurgiu, Neculai Stanciu................................. 16 
SOME APPLICATIONS OF THE ROLLE SEQUENCES - Marian Ursărescu.......................................... 26 
THREE FUNDAMENTAL THEOREMS FROM NUMBERS THEORY - Angela Niţoiu.....................29 
ABOUT NAGEL AND GERGONNE CEVIANS (III)- Bogdan Fuştei …………………………..…………….32 
ABOUT IONESCU-NESBITT INEQUALITY – D.M.Bătinețu-Giurgiu and Daniel Sitaru-……………40 
GAKOPOULOS’ LEMMA (II) - Thanasis Gakopoulos …………………………………………………….......42 
STRUCTURI ALGEBRICE(VI) – Vasile Buruiană…………………………………………..…………................45 
ABOUT A TRIANGLE INEQUALITY - D.M.Bătineţu-Giurgiu, Claudia Nănuţi, Daniel 
Sitaru…………………………………………………………………………………………………………………….….......49 
A BEAUTIFUL EQUIVALENCE-Daniel Sitaru………………………………………………………………………..50 
ABOUT AN INEQUALITY BY GEORGE APOSTOLOPOULOS-Marin Chirciu…………………………….52 
REVERSE HUYGENS’ INEQUALITY-Dorin Mărghidanu…………………………………………………..……53 
HOLDER’S AND BERGSTROM’S INEQUALITIES-NEW APPLICATIONS- Florică 
Anastase....................................................................................................................................................................58 
PROPOSED PROBLEMS…………………………………….………………………………………………………………...63 
 
INDEX OF PROPOSERS AND SOLVERS RMM-29 PAPER MAGAZINE.……………………….……………119 
 
 
 
 
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
4 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
INEQUALITIES WITH CONVEX FUNCTIONS 
By Daniel Sitaru, Claudia Nănuți – Romania 
Abstract: 
In this paper it is proved a property of convex functions which can generate beautiful 
inequalities. 
Keywords: convex functions, inequalities 
Main result: 
Theorem 1(Hardy-Polya-Szego-Littlewood) 
If �, �, �, �, � ∈ ℝ; � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; �: [�, �] → ℝ; � convexe, then: 
�(�) + �(�) ≤ �(�) + �(� + � − �) (1) 
Proof: 
� ≤ � ⇒ 0 ≤ � − � ⇒ � ≤ � + � − � 
� ≤ � ⇒ 0 ≤ � − � ⇒ � ≤ � + � − � 
� ∈ [�; � + � − �] ⇒ (∃)� ∈ [0,1] 
� = �� + (1 − �)(� + � − �) (2) 
� ∈ �� + (1 − �)(� + � − �) (3) 
By adding (2); (3): � + � = (� + �)� + (� + � − �)(2 − � − �) (4) 
� + � = � + (� + � − �).Replace in (4): 
� + (� + � − �) = (� + �)� + (� + � − �)(2 − � − �) 
(� + � − 1)� + (� + � − �)(1 − � − �) = 0, (� + � − 1)(� − � − � + �) = 0 
(� + � − 1)(2� − � − �) = 0 
Case I: 2� − � − � = 0 ⇒ � + � = 2� (5) 
But � ≤ �; � ≤ � ⇒ 2� ≤ � + � (6) 
By (5); (6) ⇒ � = � = �. Inequality (1) becomes: 
�(�) + �(�) ≤ �(�) + �(� + � − �) (true) 
Case II: � + � − 1 = 0 ⇒ � + � = 1 ⇒ 2 − � − � = 1 
� convexe; �, � ∈ [�, � + � − �] ⇒ (∃)�, � ∈ [0,1] such that: 
�(�) ≤ ��(�) + (1 − �)�(� + � − �) (7) 
�(�) ≤ ��(�) + (1 − �)�(� + � − �) (8) 
By adding (7); (8): 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
5 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
�(�) + �(�) ≤ (� + �)�(�) + (2 − � − �)�(� + � − �) 
�(�) + �(�) ≤ �(�) + �(� + � − �) 
Theorem 2 
If �, �, �, �, � ∈ ℝ; � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; �: [�, �] → ℝ; � concave, then: 
�(�) + �(�) ≥ �(�) + �(� + � − �) (9) 
Application 1:If �, �, �, �, � ∈ ℝ; 0 < � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then 
�
��
+
�
��
≤
�
��
+
�
(�����)�
 (10) 
Solution: We take in (1): �(�) =
�
��
; ��(�) =
������
���
 
��(�) =
−�
����
; ���(�) =
�(� + 1)��
�����
=
�(� + 1)
����
> 0 
� convexe 
�(�) + �(�) ≤ �(�) + �(� + � − �) 
1
��
+
1
��
≤
1
��
+
1
(� + � − �)�
 
Equality holds for � = � = �. 
Application 2:If �, �, �, �, � ∈ ℝ; 0 < � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then: 
��
� + √�
�
≤ √�
�
+ �� + � − �
� 
Solution: We take in (9): �(�) = √�
�
; ��(�) =
�
�
�
�
�
�� 
���(�) =
�
�
�
�
�
− 1� �
�
�
�� < 0; � concave 
�(�) + �(�) ≥ �(�) + �(� + � − �); �: (0, ∞) → ℝ 
��
� + √�
�
≥ √�
�
+ �� + �
� − � (11) 
Equality holds for � = � = �. 
Application 3:If �, �, �, �, � ∈ ℝ; 0 < � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then: 
arccot � + arccot � ≤ arccot � + arccot(� + � − �) (12) 
Solution:We take in (1): �(�) = arccot � ; 
��(�) =
��
����
; ���(�) =
��
(����)
> 0; � convexe 
�(�) + �(�) ≤ �(�) + �(� + � − �); �: (0, ∞) → ℝ 
arccot � + arccot � ≤ arccot � + arccot(� + � − �) 
Equality holds for: � = � = �. 
Application 4:If �, �, �, �, � ∈ ℝ; � ≤ � ≤ � ≤ � ≤ �; � + � ≤ � + �; � ∈ ℕ ∖ {0} then: 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
6 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
��
�
+ ��
�
≤ ��
�
+ �(�����)
�
 (13) 
Solution: We take in (1): �(�) = ��
�
; ��(�) = 2���
�
 
���(�) = 2��
�
+ 4����
�
= 2��
�
(1 + 2��) > 0; � convexe 
�(�) + �(�) ≤ �(�) + �(� + � − �) 
��
�
+ ��
�
≤ ��
�
+ �(�����)
�
 
Equality holds for: � = � = �. 
Application 5: If 0 < � ≤ �; � ∈ ℕ ∖ {0} then: 
1
�√���
� +
1
�
���
�
�
� ≤
1
�
���
���
�
� +
1
�√�� +
���
�
−
���
���
�
� 
Solution: We take in (10): 
� =
2��
� + �
; � = √��; � =
� + �
2
 
By AM-GM-HM: 
0 < � ≤
2��
� + �
≤ √�� ≤
� + �
2
≤ � 
1
��
+
1
��
≤
1
��
+
1
(� + � − �)�
 
1
�√���
� +
1
�
���
�
�
� ≤
1
�
���
���
�
� +
1
�√�� +
���
�
−
���
���
�
� 
Equality holds for � = �. 
Application 6: If 0 < � ≤ �; � ∈ ℕ ∖ {0} then: 
√��
��
+ �
� + �
2
�
≥ �
2��
� + �
�
+ �√�� +
� + �
2
−
2��
� + �
�
 
Solution: 
We take in (11): � =
���
���
; � = √��; � =
���
�
 
By AM-GM-HM: 
0 < � ≤
2��
� + �
≤ √�� ≤
� + �
2
≤ � 
��
� + √�
�
≥ √�
�
+ �� + � − �
� 
�√��
�
+ �
� + �
2
�
≥ �
2��
� + �
�
+ �√�� +
� + �
2
−
2��
� + �
�
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
7 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
√��
�
+ �
� + �
2
�
≥ �
2��
� + �
�
+ �√�� +
� + �
2
−
2��
� + �
�
 
Equality holds for � = �. 
Application 7: If 0 < � ≤ � then: 
arccot(√��) + arccot �
� + �
2
� ≤ arccot �
2��
� + �
� + arccot �√�� +
� + �
2
−
2��
� + �
� 
Solution: We take in (12): 
� = √��; � =
� + �
2
; � =
2��
� + �
 
By AM-GM-HM: 0 < � ≤
���
���
≤ √�� ≤
���
�
≤ � 
arccot � + arccot � ≤ arccot � + arccot(� + � − �) 
arccot(√��) + arccot �
� + �
2
� ≤ arccot �
2��
� + �
� + arccot�√�� +
� + �
2
−
2��
� + �
� 
Equality holds for � = �. 
Application 8: If 0 < � ≤ � then: 
��� + �
�
���
�
�
�
≤ �
�
���
���
�
�
+ �
�√���
���
�
�
���
���
�
�
 
Solution: We take in (13): 
� =
2��
� + �
; � = √��; � =
� + �
2
 
��
�
+ ��
�
≤ ��
�
+ �(�����)
�
, ��√���
�
+ �
�
���
�
�
�
≤ �
�
���
���
�
�
+ �
�√���
���
�
�
���
���
�
�
 
��� + �
�
���
�
�
�
≤ �
�
���
���
�
�
+ �
�√���
���
�
�
���
���
�
�
 
Equality holds for � = �. 
Reference: 
ROMANIAN MATHEMATICAL MAGAZINE – www.ssmrmh.ro 
ABOUT RMM-INEQUALITY IN TRIANGLE 899 
By Marin Chirciu – Romania 
1) In ���� the following relationship holds: 
� �
�
��
+
�
��
� �� ≥ ��� − �� 
Proposed by Mehmet Șahin – Ankara – Turkey 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
8 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Solution: We prove the following lemma 
Lemma. 
2) In ���� the following relationship holds: 
� �
�
��
+
�
��
� �� =
�� + �� − ���
�
 
Proof. 
Using �� =
�
���
 and ℎ� =
��
�
 we obtain: 
� �
1
��
+
1
ℎ�
� �� = � �
� − �
�
+
�
2�
� �� =
1
2�
�(� + �) �� =
�� + �� − 2��
�
 
which follows from ∑ ��(� + �) = 2�(�� + �� − 2��) 
Let’s get back to the main problem. 
Using the Lemma we write the inequality: 
�� + �� − 2��
�
≥ 28� − 2� ⇔ �� ≥ 27�� 
which follows from Mitrinovic’s inequality: � ≥ 3�√3. 
Equality holds if and only if the triangle is equilateral. 
Remark.The inequality can be strengthened: 
3) In ���� the following relationship holds: 
� �
�
��
+
�
��
� �� ≥ ��� − �� 
Solution: Using the Lemma the inequality can be written: 
���������
�
≥ 14� − 4� ⇔ �� ≥ 16�� − 5�� (Gerretsen’s inequality) 
Equality holds if and only if the triangle is equilateral. 
Remark. Inequality 3) is stronger than inequality 1): 
4) In ���� the following relationship holds: 
� �
�
��
+
�
��
� �� ≥ ��� − �� ≥ ��� − �� 
Solution: See inequality 3) and Euler’s inequality � ≥ 2�.Equality holds if and only if the 
triangle is equilateral. 
5) In ���� the following inequality holds: 
� �
�
��
+
�
��
� �� ≤
���
�
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
9 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Proposed by Marin Chirciu – Romania 
Solution 
�� + �� − 2��
�
≤
6��
�
⇔ �� ≤ 6�� + 2�� − �� 
which follows from Gerretsen’s inequality: �� ≤ 4�� + 4�� + 3��. 
It remains to prove that: 
4�� + 4�� + 3�� ≤ 4�� + 4�� + 3�� ⇔ �� − �� − 2�� ≥ 0 ⇔ (� − 2�)(� + �) ≥ 0 
obviously from Euler’s inequality � ≥ 2�. 
Equality holds if and only if the triangle is equilateral. 
Remark.We can write the double inequality: 
6) In ���� the following inequality holds: 
��� − �� ≤ � �
�
��
+
�
��
� �� ≤
���
�
 
Proposed by Marin Chirciu – Romania 
Solution See inequalities 3) and 5).Equality holds if and only if the triangle is equilateral. 
Remark. If we split the above sum after �� and ℎ� we propose: 
7) In ���� the following relationship holds: 
�(�� − �) ≤ �
��
��
≤
��
�
(�� − �) 
Solution We prove the following lemma: 
Lemma. 
8) In ���� the following relationship holds: 
�
��
��
=
�� + �� − ���
�
 
Proof. 
Using �� =
�
���
, we obtain: 
�
��
��
= �
� − �
�
⋅ �� =
1
�
� ��(� − �) =
1
��
⋅ �(�� + �� − ��) =
�� + �� − 8��
�
 
which follows from ∑ �� (� − �) = �(�� + �� − 8��) 
Let’s get back to the main problem. 
Using Lemma and Gerretsen’s inequality: 16�� − 5�� ≤ �� ≤ 4�� + 4�� + 3�� 
Equality holds if and only if the triangle is equilateral. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
10 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
9) In ���� the following relationship holds: 
��� ≤ �
��
��
≤ �� 
Solution 
We use ∑
��
��
= 6� and Euler’s inequality � ≥ 2� 
Equality holds if and only if the triangle is equilateral. 
Remark. 
Between the sums ∑
��
��
 and ∑
��
��
 we write the following relationship: 
10) In ���� the following relationship holds: 
�
��
��
≤ �
��
��
 
Proposed by Marin Chirciu – Romania 
Solution: Using the sums ∑
��
��
= 6� and ∑
��
��
=
���������
�
 we write the inequality: 
6� ≤
���������
�
⇔ �� ≥ 14�� − ��, which follows from Gerretsen’s inequality 
�� ≥ 16�� − 5��. Equality holds if and only if the triangle is equilateral. 
Remark. We can write the sequence of inequalities: 
1) In ���� the following relationship holds: 
��� ≤ �
��
�
≤ �� ≤ �(�� − �) ≤ �
��
�
≤
��
�
(�� − �) 
Solution: See inequalities 4) and 6) and Euler’s inequality � ≥ 2�.Equality holds if and only if 
the triangle is equilateral. 
Refference: 
ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro 
 
AN EXTENSION OF A TRIANGLE IDENTITY TO THE EUCLIDEAN N-SIMPLEX 
By Vasile Jiglău-Romania 
Abstract: In this note we’ll extended a geometric identity from triangle to an Euclidean n-
simplex. 
Let ������ be an arbitrary triangle and denote by �� the radius of the circle which is tangent 
to the circumcircle of ������ and to the side ���� of the given triangle, simultaneously 
({�, �, �} = {1,2,3}). In [1] the authors proved that: 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
11 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
1
��
+
1
��
+
1
��
=
2
�
+
1
�
 
where � and � are the circumradius and the inradius of ������, respectively. 
In the following we’ll prove an extension of this identity to the Euclidean n-simplex. 
Taking in the proposition enunciated below and in it’s proof � = 3 one obtains the 
corresponding identity for tetrahedron. 
Let ���� … �� be an Euclidean n-simplex.We’ll use the following notations: 
−�, �, �, � the centre if it’s circumscribed hypersphere, it’s volume, it’s circumradius and it’s 
inradius,respectively. 
−�′�, �′� the centre and the radius of the hypersphere tangent to the circumscribed sphere 
of ���� … �� in the vertex �� and to the hyperplane ���� … �������� … �� simultaneously. 
−�� the volume of the n-simplex ����� … �������� … �� 
−�′� the volume of the n-simplex �′����� … �������� … �� 
−�"� the volume of the (n-1)-simplex ���� … �������� … �� 
Proposition: With the above notations, the following identity holds: 
�
1
��
�
���
=
�
�
+
1
�
 
Proof: Since the hypersphere having −�′� as centre defined above is tangent to the 
circumscribed hypersphere of the given n-simplex, it results that ��, �′�, � are collinear. We 
recall that the volume of ���� … �� is given by the formula � =
�
�
ℎ��"�, where ℎ� is the 
distance from the vertex �� to the hyperplane ���� … �������� … ��.Projecting �, �′� on 
the hyperplane ���� … �������� … ��, (� ≠ �) applying the Thales’ theorem, then the 
precedent formula, we remark that 
���
��
=
��
�
. We have: 
� = � ���
�
���
���
+ ��� = �
��
�
��
�
���
���
+
���"�
�
= ��
⎝
⎜
⎛1
�
� ��
�
���
���
+
�"�
�
⎠
⎟
⎞
⇒
�
��
=
1
�
� ��
�
���
���
+
�"�
�
 
Since the sum ∑ ∑ ��
�
���
���
�
��� any �� appears n times, we have 
�
�
∑ ∑ ��
�
���
���
�
��� =
�
�
�� 
On the other hand �� = � ∑ �"�
�
��� , therefore 
∑
�
��
�
��� =
��
�
+
�
�
⇒ ∑
�
��
�
��� =
�
�
+
�
�
, q.e.d. 
Reference: 
[1] I.Isaev, Y.Maltsev, A.Monastyreva-On some geometric relations of a 
triangle, Journal of Classical Geometry, volume 4. 
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
12 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
NEW INEQUALITIES WITH FIBONACCI AND LUCAS’ NUMBERS 
By Florică Anastase-Romania 
 
Let be (��)��� −Fibonacci sequences, �� = ���� + ����; �� = 0; �� = 1, ∀� ∈ ℕ, 
(��)��� −Lucas sequences, �� = ���� + ����; �� = 2; �� = 1, ∀� ∈ ℕ 
Application 1. 
��
(�� − �)�
+
��
(�� − �)�
+ ⋯ +
��
(���� − �)�
≤
���� + ���� − ���
���
�(���� − ��)
 
Solution: 
�� = � ��
�
���
= �� + �� + ⋯ + �� = ���� − �� ⇒ 
���� − 2�� = ���� + ���� − 2�� = �� + ����, ∀� = 1, � 
�� − 2�� = �� + �� = �� + �� + �� = 2�� + ��; �� − 2�� = �� + �� = 2�� + 2�� + �� 
���� − 2�� = �� + ���� = 2�� + 2�� + ⋯ + 2���� + �� 
��
(�� − 2)�
+
��
(�� − 2)�
+ ⋯ +
��
(���� − 2)�
= 
=
��
(2�� + ��)
�
+
��
(2��+ 2�� + ��)
�
+ ⋯ +
��
(2�� + 2�� + ⋯ + 2���� + ��)
�
 
��
��(�� + ��)
+
��
(�� + ��)(�� + �� + ��)
=
�� + ��
��(�� + �� + ��)
 
Suppose: 
��
��(�� + ��)
+
��
(�� + ��)(�� + �� + ��)
+ ⋯ +
��
(�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��)
=
�� + �� + ⋯ + ��
��(�� + �� + ⋯ + ��)
⇒ 
��
��(�� + ��)
+
��
(�� + ��)(�� + �� + ��)
+ ⋯ +
��
(�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��)
+ 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
13 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
+
����
(�� + �� + ⋯ + ��)(�� + �� + ⋯ + ����)
= 
=
�� + �� + ⋯ + ��
��(�� + �� + ⋯ + ��)
+
����
(�� + �� + ⋯ + ��)(�� + �� + ⋯ + ����)
= 
=
(�� + �� + ⋯ + ��)
� + ��(�� + �� + ⋯ + ��) + ����(�� + �� + ⋯ + ��) + ������
(�� + �� + ⋯ + ��)��(�� + �� + ⋯ + ����)
= 
=
�� + �� + ⋯ + ����
��(�� + �� + ⋯ + ����)
 
From (� + �)� ≥ 4�� ⇒
�
��
≥
�
(���)�
 we have: 
��
��(�� + ��)
≥
4��
(2�� + ��)�
 
��
(�� + ��)(�� + �� + ��)
≥
4��
(2�� + 2�� + ��)�
 
��
(�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��)
≥
4��
(2�� + 2�� + ⋯ + 2���� + ��)
�
 
Adding up relationships, we have: 
4��
(2�� + ��)�
+
4��
(2�� + 2�� + ��)�
+ ⋯ +
4��
(2�� + 2�� + ⋯ + 2���� + ��)�
≤ 
≤
��
��(�� + ��)
+
��
(�� + ��)(�� + �� + ��)
+ ⋯ +
��
(�� + �� + ⋯ + ����)(�� + �� + ⋯ + ��)
=
�� + �� + ⋯ + ��
��(�� + �� + ⋯ + ��)
⇒ 
��
(2�� + ��)�
+
��
(2�� + 2�� + ��)�
+ ⋯ +
��
(2�� + 2�� + ⋯ + 2���� + ��)�
≤
�� + �� + ⋯ + ��
4��(�� + �� + ⋯ + ��)
 
��
(2�� + ��)�
+
��
(2�� + 2�� + ��)�
+ ⋯ +
��
(2�� + 2�� + ⋯ + 2���� + ��)�
≤
���� − ��
��
�(���� − ��)
 
��
(�� − 2)�
+
��
(�� − 2)�
+ ⋯ +
��
(���� − 2)�
≤
���� − ��
��
�(���� − ��)
 
From: 5�� = ���� + ���� ⇒ ���� − �� =
�������������
�
⇒
�������
��
�(�������)
=
�������������
���
�(�������)
 
��
(�� − 2)�
+
��
(�� − 2)�
+ ⋯ +
��
(���� − 2)�
≤
���� + ���� − 5��
5��
�(���� − ��)
 
Application 2. 
 If (��)��� is in arithmetic progression with �� > 0, � > 0 then prove: 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
14 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
�
�����
+
�
�����
+ ⋯ +
�
�����
≥
���
� + ������ − �(���� − �� − ��) − ����
 
Solution: From A-G-M we have: √� ≤
���
�
⇒
�
√�
≥
�
���
, ∀� ∈ ℝ�
∗ = (0, ∞) then: 
1
�����
+
1
�����
+ ⋯ +
1
�����
≥
2
1 + ����
+
2
1 + ����
+ ⋯ +
2
1 + ����
≥
��������̈�
 
≥
2��
� + ���� + ���� + ⋯ + ����
 
 From ���� = ���� + ��; �� = �� = 1 we have ���� = ���� + ��, ∀� = 1, ������ ⇒ 
���� − ���� = ��, ∀� = 1, ������ ⇒ �� ∙ �� = (�� + (� − 1)�)�� = (�� − �)�� + ���� 
But: � ∙ �� = �(���� − ����) = (� + 2)���� − (� + 1)���� − 2(���� − ����) − ���� = 
= (� + 2)���� − (� + 1)���� − 2(���� − ����) − (���� − ����) ⇒ 
� � ∙ ��
�
���
= �[(� + 2)���� − (� + 1)����]
�
���
− 2 �(���� − ����)
�
���
− �(���� − ����)
�
���
= 
= (� + 2)���� − 2�� − 2(���� − ��) − (���� − ��) = ����� − ���� + �� 
� �� ∙ ��
�
���
= (�� − �) � ��
�
���
+ � � � ∙ ��
�
���
= (�� − �)(���� − ��) + �(����� − ���� + ��)
= ������ − �(���� − �� − ��) − ����. 
So, 
�
�����
+
�
�����
+ ⋯ +
�
�����
≥
���
����������(����������)�����
 
Application 3. 
�
����
+
�
����
+ ⋯ +
�
�(� + �)��
≥
���
(� + ����)� + ��
 , ∀� ≥ � 
Solution: From A-G-M we have: √� ≤
���
�
⇒
�
√�
≥
�
���
, ∀� ∈ ℝ�
∗ = (0, ∞) then: 
1
�3��
+
1
�4��
+ ⋯ +
1
�(� + 2)��
≥
2
1 + 3��
+
2
1 + 4��
+ ⋯ +
2
1 + (� + 2)��
≥
��������̈�
 
≥
2��
� + 3�� + 4�� + ⋯ + (� + 2)��
 
For �� = 3, � = 1 we get: 3�� + 4�� + ⋯ + (� + 2)�� = (1 + ����)� + ��, then 
1
�3��
+
1
�4��
+ ⋯ +
1
�(� + 2)��
≥
2��
(1 + ����)� + ��
, ∀� ≥ 2 
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
15 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Application 4: 
�
�2��
+
�
����
+ ⋯ +
�
����
≥
���
(� + ����)� − �� + �
, ∀� ≥ � 
Solution: From A-G-M we have: √� ≤
���
�
⇒
�
√�
≥
�
���
, ∀� ∈ ℝ�
∗ = (0, ∞) then: 
1
�2��
+
1
�3��
+ ⋯ +
1
����
≥
2
1 + 2��
+
2
1 + 3��
+ ⋯ +
2
���
≥
��������̈�
 
≥
2��
� + 2�� + 3�� + ⋯ + ���
 
For �� = 2, � = 1 we get: 2�� + 3�� + ⋯ + ��� = ∑ (� + 1) ∙ ��
�
��� = 
= (� + 1)���� − (���� − �� − ��) − 2�� = ����� − ���� + 1 
Then 
1
�2��
+
1
�3��
+ ⋯ +
1
����
≥
2��
(1 + ����)� − �� + 1
, ∀� ≥ 2 
Application 5: 
�
��
� + ��
�
��
+ �
��
� + ��
� + ��
�
��
+ ⋯ + �
��
� + ��
� + ⋯ + ����
�
��
≥
�������
(� + �)���� − ��
 
Solution: We have: ���� = ���� + ��, ∀� ≥ 0 ⇒ ���� − �� = ���� ∀� ≥ 0 ⇒ 
���� ∙ ���� − ���� ∙ �� = ����
� ⇒ ��
� + ��
� + ⋯ + ����
� = ���� ∙ ���� − ���� ⇒ 
��
� + ��
� + ��
� + ⋯ + ����
� = ���� ∙ ���� ⇒
1
��
� + ��
� + ��
� + ⋯ + ����
� =
1
���� ∙ ����
 
⇒
��
��
� + ��
� + ��
� + ⋯ + ����
� =
���� − ����
���� ∙ ����
=
1
����
−
1
����
⇒ 
�
��
��
� + ��
� + ��
� + ⋯ + ����
�
�
���
= 1 −
1
����
 
From A-G-M we have: √� ≤
���
�
⇒
�
√�
≥
�
���
, ∀� ∈ ℝ�
∗ = (0, ∞) then: 
�
��
� + ��
�
��
+ �
��
� + ��
� + ��
�
��
+ ⋯ + �
��
� + ��
� + ⋯ + ����
�
��
≥ 
≥
2
1 +
��
��
����
�
+
2
1 +
��
��
����
����
�
+ ⋯ +
2
1 +
��
��
����
��⋯�����
�
≥
��������̈�
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
16 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
≥
2��
� +
��
��
����
� +
��
��
����
����
� + ⋯ +
��
��
����
��⋯�����
�
= 
=
2��
� + 1 −
�
����
=
2������
(� + 1)���� − 1
 
Application 6: 
(� + �)���
� + (� + �)���
� + ⋯ + (�� + �)������
� ≥
�(������ − �) − ����� + �
�
 
Solution: 
(� + 1)���
� + (� + 2)���
� + ⋯ + (3� + 1)������
� = 
=
[(� + 1)��]
�
�
���
+
[(� + 2)��]
�
�
���
+ ⋯ +
[(3� + 1)�����]
�
�
����
≥
��������̈�
 
≥
[(� + 1)�� + (� + 2)�� + ⋯ + (3� + 1)�����]
�
�
���
+
�
���
+ ⋯ +
�
����
 
But: 
�
���
+
�
���
+ ⋯ +
�
����
< 2, ∀� ∈ ℕ from mathematical induction by � ∈ ℕ and 
(� + 1)�� + (� + 2)�� + ⋯ + (3� + 1)����� = 
= �(�� + �� + ⋯ + �����) + 1 ∙ �� + 2 ∙ �� + ⋯ + (2� + 1) ∙ ����� = 
= �(����� − ��) + 2������ − ����� + 2 = �(3����� − 1) − ����� + 2 
(� + 1)���
� + (� + 2)���
� + ⋯ + (3� + 1)������
� ≥
�(3����� − 1) − ����� + 2
2
 
Refference: 
ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro 
 
120 YEARS OF LALESCU SEQUENCES (I) 
By D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania 
Abstract. In this paper we present new methods to calculate certain limits from math 
journals from all over the world . 
MathEduc Subject Classification: D55, I35 
MSC Subject Classification: 97D50, 97I30 
Key words and phrases: Limits of Lalescu type sequences; Limits of Lalescu type functions; 
Problem Solving. 
1. RESULTS – LALESCU TYPE LIMITS 
PROBLEMS FROM MATH JOURNALS FROM ALL OVER THE WORLD 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
17 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
The Stolz-Cesàro criterion (C-S). Let the sequences (��)��� and (��)��� such that(��)��� is 
strictly monotonous and boundless. If there exists the limit ���
�→�
�������
�������
 , then ���
�→�
��
��
 
The Cauchy-D’Alembert criterion (C-D’A). Let the sequence (��)���with �� > 0 If there 
exists the limit ���
�→�
����
��
= � > 0, then ���
�→�
���
� = � 
We present new solutions of certain problems of Lalescu type limits from math journals from 
all over the world. 
In the next we use the following abbreviations: Crux Mathematicorum (CM); Gazeta 
Matematică Seria B (GMB); La Gaceta de la RSME (LG); Math Problems (MP); Pi Mu 
Epsilon Journal (PME); Mathematics Magazine from Timişoara (RMT); Mathematical 
Recreation from Iaşi (RM); Revista Escolar de la Olimpiada Iberoamericana de Matematica 
(REOIM); Romanian Mathematical Magazine (RMM); School Science and Mathematics 
(SSM); The Spark of the Mind (SM); The American Mathematical Monthly (AMM); The 
College Mathematics Journal (CMJ); The Pentagon (P); The Fibonacci Quarterly (FQ). 
Problem 1. Traian Lalescu’s limit, GMB, Vol. VI, 1900-1901, problem 579, p. 148. 
���
�→�
� �(� + �)!
���
− √�!
�
� =⏞
�!≅�
�
�
�
�
⇒ √�!
�
≅
�
�
���
�→�
�
� + �
�
−
�
�
� =
�
�
 
Problem 2. D. M. Bătineţu-Giurgiu’s limit, GMB, Vol. XCIV, 1989, problem C:890,p. 139. 
���
�→�
�
(� + �)�
�(� + �)!
���
−
��
√�!
� � =⏞
�!≅�
�
�
�
�
⇒ √�!
�
≅
�
�
���
�→�
�
(� + �)�
���
�
−
��
�
�
� = ���
�→�
�(� + �)� − ��� = � 
Problem 3. D.M. Bătineţu-Giurgiu, Neculai Stanciu, GMB 5/2012. 
If (��(�))��� is defined by ��(�) = �
��� �(� + �)�� √� + �
���
�
�
− ��� √�
�
�
�
� , ∀� ∈ ℝ, then 
compute ���
�→�
��(�). 
 Solution: ��(�) = � √��
�
∙ (�� − 1) = � √�
�
�
�
∙
����
�����
∙ �����
�, ∀� ≥ 2, where 
�� = �
� + 1
�
�
�
∙ �
√� + 1
���
√�
� �
�
, ∀� ≥ 2; ���
�→�
�� = 1 and then ���
�→�
�� − 1
�����
= 1 
 ���
�→�
��
� = ���
�→�
��
� ∙ ���
�→�
�
√� + 1
���
√�
� �
��
= �� ∙ ���
�→�
�
� + 1
�
∙
1
√� + 1
��� �
�
= �� ∙ 1 = ��. 
���
�→�
��(�) = 1 ∙ 1 ∙ ��� � ���
�→�
��
�� = ����� = �. 
Observation: For � = 1 we deduce lim
�→�
�� = ���
�→�
��(1) = 1, i.e. the limit of Romeo T. 
Ianculescu. 
Problem 4. D.M. Bătineţu-Giurgiu, Neculai Stanciu, CM 7/2013. 
 Let (��)��� be a positive real sequence such that ���
�→�
�������
�
= � ∈ ℝ�
∗ .We define ��! by 
���� = ��! ∙ ����, ∀� ∈ ℕ
∗. Compute ���
�→�
�
�����!
���
���
−
���!
�
�
� . 
Solution: ���
�→�
��
��
=
���
���
�→�
�������
(���)����
= ���
�→�
�
�������
�
∙
�
����
� =
�
�
. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
18 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
���
�→�
���!
�
��
= ���
�→�
�
��!
���
�
=
�����
���
�→�
����!
(� + 1)�(���)
∙
���
��!
= ���
�→�
����
(� + 1)�
∙ ���
�→�
�
�
� + 1
�
��
=
�
2
∙
1
��
=
�
2��
 
 
So, � = ���
�→�
�
�����!
���
���
−
���!
�
�
� = ���
�→�
���!
�
�
∙ (�� − 1) = ���
�→�
�
���!
�
��
∙
����
�����
∙ �����
�� ; 
���
�→�
�� = ���
�→�
�
�����!
���
� + 1
∙
�
���!
�
� = ���
�→�
�
�����!
���
(� + 1)�
∙
��
���!
�
∙
� + 1
�
� =
�
2�
∙
2�
�
∙ 1 = 1; 
���
�→�
�� − 1
�����
= 1, 
���
�→�
��
� = ���
�→�
�
�����!
���
���!
�
∙
�
� + 1
�
�
= ���
�→�
�
����!
��!
∙ �
�
� + 1
�
�
∙
1
�����!
���
� = 
= ���
�→�
�
����
(� + 1)�
∙
(� + 1)�
�����!
���
� ∙ ���
�→�
�
�
� + 1
�
�
=
�
2
∙
2��
�
∙
1
�
= �. 
We obtain: � =
�
���
∙ 1 ∙ ���� =
�
���
. 
Problem 5. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RM 2/2012. 
If �� =
(���)���
(���)�
, ∀� ∈ ℕ∗ then compute ���
�→�
� ����� … ������
��� − ����� … ��
� � 
Solution: �� = ����� … ������
��� − ����� … ��
� = ����� … ��
� (�� − 1) = 
= ����� … ��
� ∙
����
�����
∙ �����
�, ∀� ≥ 2, where �� =
�����…������
���
�����…��
� , ∀� ≥ 2; 
���
�→�
����� … ��
�
�
= ���
�→�
�
���� … ��
��
�
= ���
�→�
�
���� … ������
(� + 1)���
∙
��
���� … ��
� = 
= ���
�→�
�
����
� + 1
∙ ��
��� = ��� ∙ ���
�→�
(� + 3)���
(� + 2)���(� + 1)
= ��� ∙ ���
�→�
���� ∙
� + 2
� + 1
= ��� ∙ � ∙ 1
= 1 
Where we denote �� = �1 +
�
�
�
�
; so ���
�→�
�� = 1, ���
�→�
����
�����
= 1. 
���
�→�
��
� = ���
�→�
�
���� … ������
���� … ��
∙
1
����� … ������
���
� = ���
�→�
�
����
� + 1
∙
� + 1
����� … ������
���
�
= � ∙ 1 = �. 
We obtain ���
�→�
�� = 1 ∙ 1 ∙ ���� = 1. 
Problem 6. D.M. Bătineţu-Giurgiu, Neculai Stanciu, AMM 9/2012. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
19 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Compute ���
�→�
�����
�� ��Г(� + �)�
�����
��� − �Г(� + �)�
�����
� �� where � ∈ ℝ and Г is gamma 
function. 
Solution: ���
�→�
�Г(���)�
�
�
�
= ���
�→�
�∈ℕ∗
�Г(���)�
�
�
�
= ���
�→�
√�!
�
�
= ���
�→�
�
�!
��
�
=
�����
���
�→�
�
(���)!
(���)���
∙
��
�!
� = 
= ���
�→�
�
�
� + 1
�
�
=
1
�
 
�(�) = ����
�� ��Г(� + 2)�
�����
��� − �Г(� + 1)�
�����
� � = ����
���Г(� + 1)�
�����
� (�(�) − 1) 
�: ℝ�
∗ → ℝ, �(�) = �
�Г(� + 2)�
�
���
�Г(� + 1)�
�
�
�
�����
; 
���
�→�
�(�) = ���
�→�
�
�Г(� + 2)�
�
���
� + 1
∙
�
�Г(� + 1)�
�
�
∙
� + 1
�
�
�����
= �
1
�
∙ � ∙ 1�
�����
= 1 
���
�→�
�(�) − 1
����(�)
= 1. 
���
�→�
��(�)�
�
= ���
�→�
�
�Г(� + 2)�
�
���
�Г(� + 1)�
�
�
�
������
= ���
�→�
�
Г(� + 2)
Г(� + 1)
∙
1
�Г(� + 2)�
�
���
�
�����
= 
= ���
�→�
�
� + 1
�Г(� + 2)�
�
���
�
�����
= ����
��; 
���
�→�
�(�) = − ���
�→�
⎝
⎜
⎛
����
�� �
�Г(� + 1)�
�
�
�
∙ ��
�����
∙
�(�) − 1
����(�)
∙ ����(�)
⎠
⎟
⎞
= 
= − ���
�→�
⎝
⎜
⎛
�
�Г(� + 1)�
�
�
�
�
�����
∙ ����
�������� ∙
�(�) − 1
����(�)
∙ ����(�)
⎠
⎟
⎞
= 
= ����
�� ∙ lim
�→�
�(�) − 1
����(�)
∙ log ����
�→�
��(�)�
�
� = ����
�� ∙ 1 ∙ �������
�� = ����� ∙ ����
��; 
Problem 7. D.M. Bătineţu-Giurgiu, Neculai Stanciu, MP 2/2013. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
20 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
If � ∈ ℝ and (��(�))��� is defined by ��(�) = �
����� �� �(� + �)!
���
�
�����
− � √�!
�
�
�����
� 
then compute ���
�→�
��(�). 
 Solution: ��(�) = �
����� �� �(� + 1)!
���
�
�����
− � √�!
�
�
�����
� = 
= ����
��� √�!
�
�
�����
(�� − 1) = 
= �
√�!
�
�
�
�����
����
�������� ∙
�� − 1
�����
∙ �����
� = �
√�!
�
�
�
�����
� ∙
�� − 1
�����
∙ �����
�, 
where �� = �
�(���)!
���
√�!
� �
�����
 and ���
�→�
�� = ���
�→�
�
�(���)!
�
���
∙
�
√�!
� ∙
���
�
�
�����
= �
�
�
∙ � ∙ 1�
�����
= 1 
���
�→�
�� − 1
�����
= 1 
���
�→�
��
� = ���
�→�
�
�(� + 1)!
���
√�!
� �
������
= ���
�→�
�
(� + 1)!
�!
∙
1
�(� + 1)!
���
�
�����
= 
= ���
�→�
�
� + 1
�(� + 1)!
���
�
�����
= ����
��, 
We obtain ���
�→�
��(�) = �
�
�
�
�����
∙ 1 ∙ �������
�� =
�����
����
��
. 
Observation: If � =
�
�
, then ���� = 1, ���� = 0 so �� �
�
�
� = �(� + 1)!
���
− √�!
�
 ,i.e.we 
obtained the limit of Traian Lalescu: ���
�→�
�� �
�
�
� = ���
�→�
�� =
�
�
. 
Problem 8. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RMM 2020. 
Let (��)���, �� ∈ ℝ�
∗ , ∀� ∈ ℕ∗ with ���
�→�
(���� − ��) = � ∈ ℝ�
∗ . 
Compute ���
�→�
����� √� + �
���
− �� √�
�
�. 
Solution 1. 
�� = ���� √� + 1
���
− �� √�
�
= (���� − ��) √� + 1
���
+ ��� √� + 1
���
− √�
�
�; ∀�
∈ ℕ∗ − {1} 
���
�→�
�� = ���
�→�
(���� − ��) ∙ ���
�→�
√� + 1
���
+ ���
�→�
��
�
∙ ���
�→�
��� √� + 1
���
− √�
�
�� = 
= � ∙ 1 + ���
�→�
���� − ��
(� + 1) − �
∙ ���
�→�
� √�
� (�� − 1) = � + � ���
�→�
�
�� − 1
�����
∙ �����
�� 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
21 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
where �� =
√���
���
√�
� , ∀� ∈ ℕ
∗ − {1}; ���
�→�
�� = 1 then ���
�→�
����
�����
= 1; 
���
�→�
��
� = ���
�→�
�
� + 1
�
∙
1
√� + 1
��� � ; 
So, ���
�→�
�� = � + � ∙ 1 ∙ 1 ∙ log � ���
�→�
��
�� = � + � ∙ 1 ∙ 1 ∙ 0 = �. 
Solution 2. �� = ���� √� + 1
���
− �� √�
�
= �� √�
� (�� − 1), where �� =
����
��
∙
√���
���
√�
� , 
∀� ∈ ℕ∗ − {1} then ���
�→�
�� = ���
�→�
����
��
∙ ���
�→�
√���
���
√�
� = ���
�→�
���������
�������
∙
�
�
=
�
�
∙ 1 = 1 
 So, ���
�→�
�� = 1, ���
�→�
����
�����
= 1 and ���
�→�
��
� = ���
�→�
�
����
��
�
�
∙ ���
�→�
�
���
�
∙
�
√���
��� � = 
= ���
�→�
�
����
��
�
�
∙ 1 ∙ 1 = ���
�→�
��1 +
���� − ��
��
�
��
�������
�
(�������)∙
�
��
 
= ��∙
�
� = �. 
We obtain: ���
�→�
�� = ���
�→�
�� √�
�
�
∙ ���
�→�
����
�����
∙ ��� � ���
�→�
��
�� = ���
�→�
��
�
∙ ���
�→�
√�
�
∙ 1 ∙ ���� = 
= � ∙ 1 ∙ ���� = �. 
Problem 9. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RMM 1/2018. 
Let (��)��� be a positive real sequence such that ���
�→�
(���� − ��) = � > � Compute 
 ���
�→�
�
(���)����
�(����)!!
��� −
���
�(����)!!
� �. 
Solution. Let (��)���, �� =
(���)����
�(���)!!
��� −
���
�(����)!!
� =
���
�(����)!!
� (�� − 1) = 
=
���
�(����)!!
� ∙
����
�����
∙ �����
�, ∀� ∈ ℕ∗ − {1} ,where 
�� =
� + 1
�
∙
����
��
∙
�(2� − 1)‼
�
�(2� + 1)‼
���
=
����
� + 1
∙
�
��
∙
(� + 1)�
��
∙
�(2� − 1)‼
�
�(2� + 1)‼
���
 
We have ���
�→�
��
�
= ���
�→�
�������
(���)��
= � and ���
�→�
�(����)‼
�
�
= ���
�→�
�
(����)‼
��
�
= 
= ���
�→�
�
(2� + 1)‼
(� + 1)���
∙
��
(2� − 1)‼
� = ���
�→�
2� + 1
� + 1
∙ �
�
� + 1
�
�
=
2
�
. 
 So, ���
�→�
�� = � ∙
�
�
∙ 1 ∙
�
�
∙
�
�
= 1, ���
�→�
����
�����
= 1; 
���
�→���
� = ���
�→�
�
� + 1
�
�
�
∙ ���
�→�
(2� − 1)‼
(2� + 1)‼
∙ �(2� + 1)‼
���
= 
= � ∙ ���
�→�
��1 +
���� − ��
��
�
��
�������
�
(�������)∙
�
��
 
∙ ���
�→�
�(2� + 1)‼
���
2� + 1
= 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
22 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
= � ∙ � ∙ ���
�→�
�
(2� − 1)‼
(2� − 1)�
���
= �� ∙ ���
�→�
�
(2� + 1)‼
(2� + 1)���
∙
(2� − 1)�
(2� − 1)‼
� = �� ∙ ���
�→�
�
2� − 1
2� + 1
�
�
= 
= �� ∙ ��� = �. 
Then ���
�→�
�� = ���
�→�
��
�(����)‼
� ∙ ���
�→�
����
�����
∙ ��� � ���
�→�
��
�� = ���
�→�
��
�
∙ ���
�→�
�
�(����)‼
� ∙ ���� = 
= � ∙
�
2
∙ 1 =
��
2
. 
Problem 10. D.M. Bătineţu-Giurgiu, Neculai Stanciu, AMM 4/2014. 
Compute ���
�→�
� �(�� − �)‼
�
����
� �(���)!
���
� √�!
� − ���. 
Solution. ���
�→�
√�!
�
�
= �; 
���
�→�
�(2� − 1)‼
�
�
= ���
�→�
�
(2� − 1)‼
��
�
=
�����
���
�→�
(2� + 1)‼
(� + 1)���
∙
��
(2� − 1)‼
= ���
�→�
2� + 1
(� + 1)��
=
2
�
. 
where we denote �� = �1 +
�
�
�
�
, ∀� ∈ ℕ∗ , 
�� =
�(� + 1)!
���
√�!
� , ∀� ≥ 2; ����→�
�� = ���
�→�
�
�(� + 1)!
���
� + 1
∙
�
√�!
� ∙
� + 1
�
� = 1; ���
�→�
�� − 1
�����
= 1; 
���
�→�
��
� = ���
�→�
(� + 1)!
�!
∙
1
�(� + 1)!
���
= ���
�→�
� + 1
�(� + 1)!
���
= �. 
We denote �� =
�
�
��, ∀� ≥ 1 and we have ���
�→�
� �(2� − 1)‼
�
����
� �(���)!
���
� √�!
� − 1�� = 
= ���
�→�
�(2� − 1)‼
���
�
∙ ���
�→�
� ������ − ���
�
4
� =
2
�
∙ ���
�→�
sin ��� −
�
�
�
��������
�
�
∙ � = 
=
2
�
∙ ���
�→�
�
sin ��� −
�
�
�
�� −
�
�
∙ � ��� −
�
4
�� ∙
1
����
�
�
=
4
�
∙ 1 ∙ ���
�→�
� ��� −
�
4
� = 
=
4
�
∙ ���
�→�
� ∙
�
4
∙ (�� − 1) =
4
�
∙
�
4
∙ ���
�→�
� �
�(� + 1)!
���
√�!
� − 1� = 
=
�
�
∙ ���
�→�
�
√�!
� ∙ ����→�
� �(� + 1)!
���
− √�!
�
� =
�
�
∙ � ∙ ���
�→�
√�!
�
(�� − 1) = 
= � ∙ ���
�→�
√�!
�
�
∙ ���
�→�
�(�� − 1) =
�
�
∙ ���
�→�
�
�� − 1
�����
∙ �����
�� =
�
�
∙ 1 ∙ ���� =
�
�
. 
Problem 11. D.M. Bătineţu-Giurgiu, Neculai Stanciu, SSM 5/2014. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
23 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Let � ∈ ℝ�
∗ and (��)��� is defined by �� = ∑
�
�!
�
��� . Compute ���
�→�
√�!
�
�� ���
� �� − ��. 
Solution. �� = √�!
�
�� ���
� �� − 1� =
√�!
�
�
∙ � ∙ �� ���
� �� − 1� = 
=
√�!
�
�
∙ � ∙ � ���
� − 1� ∙
� ���
� �� − 1
���
� − 1
=
√�!
�
�
∙
� ���
� ��
���
� − 1
∙
���
� − 1
�
�
�����
∙ �����, ∀� ≥ 2. 
���
�→�
√�!
�
�
=
1
�
; ���
�→�
� ���
� ��
���
� − 1
= ���� ; ���
�→�
���
� − 1
�
�
�����
= 1 and ���
�→�
�� = � 
We obtain ���
�→�
�� =
�
�
∙ ���� ∙ 1 ∙ ���� =
����
�
. 
Problem 12. D.M. Bătineţu-Giurgiu, Neculai Stanciu, RMM 2020. 
Let (��)���, �� = −���� + ∑
�
�
�
��� , with ���
�→�
�� = � (�. �. � is Euler Mascheroni constant). 
Compute ���
�→�
(����� − ����) √�!
�
. 
Solution. �� = (����� − ����) √�!
�
= 2 ∙ √�!
�
∙ ���
����
�
∙ ���
����
�
= 
=
√�!
�
�
∙ ���
�� + �
2
∙
���
����
�
����
�
∙ � ∙
�� − �
2
, ∀� ≥ 2; ���
�→�
√�!
�
�
=
1
�
; ���
�→�
���
�� + �
2
= ���� 
���
�→�
���
����
�
����
�
= 1 and ���
�→�
�(�� − �) = ���
�→�
�� − �
�
�
=
���
���
�→�
���� − ��
�
���
−
�
�
= 
= ���
�→�
�� − ����
�
�
−
�
���
= ���
�→�
−
�
���
+ ���
���
�
�
�(���)
= ���
�→�
�� ����
� + 1
�
−
1
� + 1
� = 
= ���
�→�
�� �log(� + 1) − ���� −
1
� + 1
� = lim
�→�
���
log �1 +
�
�
� + ���� −
�
���
��
= 
= lim
�→�
���
log(1 + �) −
�
���
��
=
���
lim
�→�
���
�
���
−
�
(���)�
2�
= lim
�→�
���
� + 1 − 1
2�(� + 1)�
=
1
2
⇒ 
���
�→�
�� =
1
�
∙
1
2
∙ ���� =
����
2�
. 
Problem 13. D.M. Bătineţu-Giurgiu, Neculai Stanciu, REOIM 2013. 
Let �, � ∈ ℝ and (��(�, �))��� be a sequence defined by 
��(�, �) = (� + �)
� ∙ �((� + �)!)�
���
− �� ∙ �(�!)�
�
. Evaluate ���
�→�
��(�, �) = �(�, �) 
Solution. lim
�→�
√�!
�
�
= �; ��(�, �) = �
� ∙ � √�!
�
�
�
∙ (�� − 1) = �
� ∙ � √�!
�
�
�
∙
����
�����
∙ ����� = 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
24 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
= ������ ∙ �
√�!
�
�
�
�
∙
����
�����
∙ �����
�, ∀� ∈ ℕ∗ − {1} where �� = �
���
�
�
�
∙ �
�(���)!
���
√�!
� �, 
∀� ∈ ℕ∗ − {1}; ���
�→�
�� = 1; ���
�→�
�� − 1
�����
= 1; 
���
�→�
��
� = ���
�→�
�
� + 1
�
�
��
∙ ���
�→�
�
(� + 1)!
�!
∙
1
�(� + 1)!
���
�
�
= �� ∙ ���
�→�
�
� + 1
�(� + 1)!
���
�
�
= ����. 
So, �(�, �) = ���
�→�
��(�, �) = � ∙ log � ���
�→�
��
�� ∙ ���
�→�
������ = ������� ∙ ��� ∙ ���
�→�
������ = 
= (� + �) ∙
1
��
= �
0, if � + � < 1
���, if � + � = 1
∞, if � + � > 1
 
Problem 14. D.M. Bătineţu-Giurgiu, Neculai Stanciu, SM 1/2015. 
If �� = ∑
�
�!
�
��� compute ���
�→�
(� − ��) ∙ (� + �)!. 
Solution. ���
�→�
(� − ��) ∙ (� + 1)! =
���
���
�→�
(������)�(����)
�
(���)!
�
�
(���)!
= ���
�→�
�������
���
(���)!
= ���
�→�
�
(���)!
�
(���)!
= 
= ���
�→�
� + 2
� + 1
= 1 
Problem 15. D.M. Bătineţu-Giurgiu, Neculai Stanciu, P 2/2014. 
Compute ���
�→�
√� ∙ � �(� + �)!
�(���)
− √�!
��
� 
Solution 1. ���
�→�
�
√�!
� = ���
�→�
�
��
�!
�
=
�����
���
�→�
�
(���)���
(���)!
∙
�!
��
� = �. 
 �� = √� ∙ � �(� + 1)!
�(���)
− √�!
��
� = √� ∙ √�!
��
∙ (�� − 1) =
√�!
��
√�
∙ (�� − 1) ∙ � = 
= �
√�!
�
�
∙
�� − 1
�����
∙ �����
�, ∀� ≥ 2, �ℎ��� 
 �� =
�(� + 1)!
�(���)
√�!
�� =
�(� + 1)!
�(���)
√� + 1
∙
√�
√�!
�� ∙ �
� + 1
�
 
So, ���
�→�
�� = √� ∙ �
�
�
∙ 1 = 1; ���
�→�
����
�����
= 1. 
���
�→�
��
� = ���
�→�
��
�(� + 1)!
���
√�!
� �
�
= � ���
�→�
(� + 1)!
�!
∙
1
�(� + 1)!
���
∙ � ���
�→�
� + 1
�(� + 1)!
���
= √�. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
25 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
We obtain ���
�→�
�� = �
�
�
∙ 1 ∙ log � ���
�→�
��
�� =
�
√�
∙ ���√� =
�
�√�
. 
Solution 2. 
�� = √� ∙ � �(� + 1)!
�(���)
− √�!
��
� = √� ∙
�(���)!
���
� √�!
�
�(���)!
�(��!)
� √�!
��
=
�(���)!
���
� √�!
�
� �(���)!
���
���
∙
���
�
��
√�!
�
�
, 
So, ���
�→�
�� = ���
�→�
�(���)!
���
� √�!
�
� �(���)!
���
���
∙
���
�
��
√�!
�
�
=
�
�
∙
�
�
�
�
��
�
�
=
�
�√�
. 
Solution 3. �� = √� ∙ � �(� + 1)!
�(���)
− √�!
��
� = √� ∙
�(���)!
�(���)
� √�!
��
�����
=
√�!
��
�
∙
����
����
∙ √� = 
= �
√�!
�
�
∙
�� − 1
�� − 1
= �
√�!
�
�
∙
�� − 1
�����
∙
�����
�� − 1
∙
�����
�����
= �
√�!
�
�
∙
�� − 1
�����
∙
�����
�� − 1
∙
�����
�
�����
�, 
where �� =
�(���)!
�(���)
√�!
�� , �� = 1 +
�
�
, so ���
�→�
�� = ���
�→�
�� = 1, then ���
�→�
����
�����
= ���
�→�
����
�����
= 1. 
We have ���
�→�
��
� = √�, ���
�→�
��
� = ���
�→�
�1 +
�
�
�
�
= �, �� ���
�→�
�� =
�
√�
∙ 1 ∙ 1 ∙
���√�
����
=
�
�√�
. 
REFERENCES 
[1] D. M. Bătineţu-Giurgiu, Sequences, Albatros Publishing House, Bucharest, 1979. 
[2] D. M. Bătineţu-Giurgiu, Lalescu sequences, RMT, Vol. 20, No. 1-2, 1989, pp. 33-36. 
[3] D. M. Bătineţu-Giurgiu, Lalescu sequences and gamma function. Euler-Lalescu function, 
Romanian Mathematical Gazette - A, Vol. 11, No. 1, 1990, pp. 21-26. 
[4] D.M. Bătineţu-Giurgiu, The generalizations of Traian Lalescu sequence, Romanian 
Mathematical Gazette, No. 8-9/1990, pp. 219-224. 
[5] D.M. Bătineţu-Giurgiu, D. Sitaru, N. Stanciu, New classes of sequences-functions and 
famous limits, Octogon Mathematical Magazine, Vol. 27, No.2, October, 2019, 784-797. 
[6] D.M. Bătineţu-Giurgiu, D. Sitaru, N. Stanciu, Two classes of Lalescu’s sequences, Octogon 
Mathematical Magazine, Vol. 27, No. 2, October, 2019, 805-813. 
[7] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of Traian Lalescu 
type with Fibonacci and Lucas numbers (II), Octogon Mathematical Magazine, Vol. 27, No. 1, 
April, 2019, 101-111. 
[8] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of Traian Lalescu 
type (II), Octogon Mathematical Magazine, Vol. 27, No. 1, April, 2019, 184-209. 
[9] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of definite integrals 
of Lalescu’s type, Octogon MathematicalMagazine, Vol. 27, No. 1, April, 2019, 312-328. 
[10] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Some limits of Traian Lalescu 
type (II), Octogon Mathematical Magazine, Vol. 26, No. 2, October, 2018, 664-690. 
[11] D.M. Bătineţu-Giurgiu, M. Bencze, N. Stanciu, Some limits of Traian Lalescu type with 
Fibonacci and Lucas numbers, Octogon Mathematical Magazine, Vol. 26, No. 1, April, 2018, 
54-87. 
[12] D.M. Bătineţu-Giurgiu, M. Bencze, N. Stanciu, Some limits of Traian Lalescu type (I), 
Octogon Mathematical Magazine, Vol. 26, No. 1, April, 2018, 116-136. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
26 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
[13] D.M. Bătineţu-Giurgiu, M. Bencze, D. Sitaru, N. Stanciu, Certain classes of Lalescu 
sequence, Octogon Mathematical Magazine, Vol. 26, No. 1, April, 2018, 243-251. 
[14] D.M. Bătineţu-Giurgiu, N. Stanciu, Several results of some classes of sequences, The 
Pentagon , Volume 73, Number 2, Spring 2014, 10-24. 
[15] D.M. Bătineţu-Giurgiu, A. Kotronis, N. Stanciu, Calculating the limits of some real 
sequences, Math Problems, Issue 1, 2014, 252-257. 
[16] D.M. Bătineţu-Giurgiu, N. Stanciu, New methods for calculations of some limits, The 
Teaching of Mathematics, Vol. XVI, Nr. 2, 2013, 82-88. 
 
SOME APPLICATIONS OF THE ROLLE SEQUENCES 
By Marian Ursărescu-Romania 
In this article we want to highlight some types of problems that are elegantly solved with to 
the Rolle sequences. 
I.1.a) For any � ∈ ℕ∗, let the equation �� + ���� = 0. Prove that the equation admits a 
single root �� ∈ ℝ. 
b) Study the convergence of the sequences (��)���. 
Proof: 
a) Let be the function ��: (0, +∞) → ℝ, ��(�) = �
� + ���� derivable with 
��
�(�) = ����� +
1
�
> 0, ∀� > 0 
� 0 + ∞ 
��′(�) | + + + + + + + + + + + + + + + + 
��(�) |−∞ ↗ ↗ ↗ 
 
lim�→� ��(�) = +∞; lim�→�� ��(�) = −∞, from the Rolle sequences ∃! �� > 0. 
b) ��(1) = 1 > 0 ⇒ ∃ �� ∈ (0,1), ∀� ∈ ℕ
∗ ⇒ (��)��� is limited. 
��(��) = 0 ⇒ ��
� + ����� = 0 (1) 
����(����) = 0 ⇒ ����
��� + ������� = 0 (2) 
��(����) = ����
� − ������� =
(�)
����
� − ����
��� = ����
� (1 − ����) > 0 
�
��(����) > 0
��(��) = 0
⇒ ��(����) > ��(��) ⇒ ���� > �� ⇒ (��)��� − increasing. 
So, (��)��� is convergent. 
Proposed problems: 
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27 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
 1.a) For any � ∈ ℕ∗, let the equation �� + ������ = 1. Prove that the equation admits a 
single root �� ∈ ℝ. 
b) Study the convergence of the sequences (��)���. 
2.a) For any � ∈ ℕ∗, let the equation �� − ���� − ⋯ − � − 1 = 0. Prove that the equation 
admits a single root �� ∈ ℝ. 
b) Study the convergence of the sequences (��)���. 
c) Find: lim�→� �� 
II.1.a) Show that it exists only one function �: ℝ → ℝ such that 
��(�) + 3�(�) + � = 0, ∀� ∈ ℝ 
b) Prove to the function � is derivable on ℝ. 
Proof: a) Let �(�) = � ⇒ �� + 3� + � = 0 and let be the function �: ℝ → ℝ, 
�(�) = �� + 3� + � continuous and derivable with ��(�) = 3�� + 3 > 0, ∀� ∈ ℝ 
� −∞ + ∞ 
��(�) + + + + + + + + + + + + + + + + 
�(�) −∞ ↗ ↗ ↗ +∞ 
 
lim
�→�
�(�) = ∞, lim
�→��
�(�) = −∞ ⇒ ∃! � = �(�). 
b) To begin with, let’s show that the function � is continuous on ℝ. ⇔ 
∀�� ∈ ℝ, lim
�→��
�(�) = �(��) 
�
��(�) + 3�(�) + � = 0
��(��) + 3�(��) + �� = 0
⇒ ��(�) − ��(��) + 3��(�) − �(��)� + � − �� = 0 
��(�) − �(��)�(�
�(�) + �(�)�(��) + �
�(��) + 3) = −(� − ��) ⇒ 
�(�) − �(��) =
−(� − ��)
��(�) + �(�)�(��) + ��(��) + 3
 (1) 
��(�) + �(�)�(��) + �
�(��) + 3 = ��(�) +
�
�
�(��)�
�
+
�
�
��(��) + 3 ≥
�
�
��(��) + 3 (2) 
From (1),(2) we get: |�(�) − �(��)| ≤
|����|
�
�
��(��)��
→ 0 by the criterion of the edge, 
lim�→�� �(�) = �(��) then � is continuous in any point �� ∈ ℝ. 
��(��) = lim
�→��
�(�) − �(��)
� − ��
 (3) 
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28 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
From (1) we get: 
�(�)��(��)
����
= −
�
��(�)��(�)�(��)���(��)��
 then 
lim�→��
�(�)��(��)
����
= − lim�→��
�
��(�)��(�)�(��)���(��)��
, � −continuous in any point �� 
��(��) =
��
�(��(��)��)
∈ ℝ then � −is derivable in any point �� ∈ ℝ. 
Proposed problems: 
1.a) Show that it exists only one function �: ℝ → ℝ such that 
�� − ��(�) + ��(�) = 0, ∀� ∈ ℝ 
b) Prove to the function � is derivable on ℝ∗. 
2.a) Show that it exists only one function �: ℝ → ℝ such that 
��(�) + ���(�) − �� = 0, ∀� ∈ ℝ 
b) Prove to the function � is continuous and derivable on ℝ. 
III. Let be the equation: �� − �� − � − 1 = 0. 
a) Prove to the equation admits only real root, noted ��. 
b) Find: lim�→�(��
� + ��
�), where ��, �� −are the complex roots of the equation. 
Proof:a) Let be the function �: ℝ → ℝ, �(�) = �� − �� − � − 1 continuous and derivable 
with 
��(�) = 3�� − 2� − 1 
��(�) = 0 ⇔ �� = −
1
3
, �� = 1 
� 
−∞ −
1
3
 1 + ∞ 
��(�) 0 0 
�(�) −∞ − −2 + ∞ 
 
lim�→� �(�) = ∞, lim�→�� �(�) = −∞ by the Rolle sequences ∃! �� > 1. 
b) If ��, �� ∈ ℂ ⇒ �� = �����. 
Let �� = �(���� + �����) ⇒ �� = ����� = �(���� − �����), where � = |��| = |��|. 
lim
�→�
(��
� + ��
�) = lim
�→�
����cos(��) + ����(��)� + ���cos(��) − ����(��)��
= lim
�→�
2�� cos(��) (1) 
By Vieta’s relations ������ = 1 ⇒ |��||��||��| = 1 ⇒ �
� =
�
��
< 1 ⇒ � < 1 (2) 
|�� cos(��)| = ��|cos(��)| ≤ �� → 0 (3) 
From (1),(2),(3) we have: lim�→�(��
� + ��
�) = 0 
Proposed problem: 
Let be the equation: �� − � − 1 = 0. 
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29 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
a) Prove to the equation admits only real root, noted ��. 
b) Find: lim�→�(��
� + ��
�), where ��, �� −are the complex roots of the equation. 
IV. If �, �, � ∈ ℝ are such that � + � + � = 2 and �� + �� + �� = 1. Prove that ��� ∈ �0,
�
��
� 
Proof: 
The equation of degree (III) who admits the roots �, �, � is �� − ���
� + ��� − �� = 0, where 
�
�� = � + � + � = 2
�� = �� + �� + �� = 1
�� = ���
⇒ �� − 2�� + � − �� = 0. 
Because �, �, � ∈ ℝ we must show that all roots are real numbers. 
Let be the function �: ℝ → ℝ, �(�) = �� − 2�� + � − �� continuous and derivable on ℝ with 
��(�) = 3�� − 4� + 1 = 0 ⇔ �� =
1
3
, �� = 1 
� 
−∞ 
1
3
 1 + ∞ 
��(�) 0 0 
�(�) 
−∞ 
4
27
− �� − �� + ∞ 
 
lim�→� �(�) = +∞; lim�→�� �(�) = −∞ by the Rolle sequences the equation admits 3 real 
roots, then 
�
��
− �� ≥ 0 and −�� ≤ 0 ⇒ �� ∈ �0,
�
��
� ⇒ ��� ∈ �0,
�
��
� 
 
THREE FUNDAMENTAL THEOREMS FROM NUMBERS THEORY 
By Angela Niţoiu-Romania 
Euler’s theorem: 
If � ∈ �, � ∈ �, � ≥ 2 and (�, �) = 1 then ��(�) ≡ 1(mod �), 
where �: �∗ → �∗, �(�) = ����({� ∈ �∗|1 ≤ � ≤ � and (�, �) = 1 }) (that is,the number 
of natural numbers smaller or equal with � and prime with �) is the arithmetic function by 
Euler. 
Fermat Little theorem: 
If � ∈ �∗is an prime number and � ∈ � an integer number, then �� ≡ �(mod �). 
Wilson theorem: 
If � ≥ 2 it is an natural number,then the following affirmations are equivalent: 
a) p it is an prime number; 
b) (� − 1)! + 1 ≡ 0(mod �). 
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30 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Aplications: 
Problem 1: Let p be an prime number. Let it look like the equation �� ≡ −1(mod �) it has a 
solution if and only if � = 2 or � ≡ 1(mod 4). 
Solution: From Wilson theorem, � prime number⇒ (� − 1)! ≡ −1(mod �). If � > 2 then p 
it is odd, 
���
�
 it is integer and we have: 
(� − 1)! = �1 ∙ 2 ∙ … ∙
� − 1
2
� �
� + 1
2
… (� − 1)� = � �(� − �)
���
����
≡ � �(−�)
���
�
���
(mod �). 
So, (� − 1)! ≡ (−1)
���
� �� and there it is the relationship (−1)
���
� �� ≡ −1(mod �). 
For � ≡ 1(mod 4),
���
�
 even number, so the relationship becomes: �� ≡ −1(mod �). 
If � ≡ 3(mod 4) and suppose it exists � with �� ≡ −1(mod �), we have: 
���� ≡ (��)
���
� ≡ (−1)
���
� ≡ −1(mod �), 
But, how v���� ≡ 0(mod �) ⇒ � do not divide �, so,from Fermat little theorem, we get 
���� ≡ 1(mod �) and we come to a contradiction: 1 ≡ (−1)(mod �), because p > 2. 
For � = 2 we have 1� ≡ 1 ≡ (−1)(mod 2), so the equation �� = −1 it has a solution. 
Problem 2: Let p be an prime number and k an natural number with the condition 
1 ≤ � ≤ �. Prove that number (� − �)! (� − 1)! + (−1)��� it is divisible by p. ( A. 
Simionov) 
Solution: We have the congruences modulo p: 1 ≡ −(� − 1) 
 2 ≡ −(� − 2) 
 …………………….. 
 � − 1 ≡ −(� − � + 1) 
Which multiplies it gives: (� − 1)! ≡ (−1)���(� − 1)(� − 2) … (� − � + 1) . 
So, (� − �)! (� − 1)! ≡ (−1)���(� − 1)! ≡ (−1)� (from Wilson theorem). 
We get, obviously, to (� − �)! (� − 1)!−(−1)� ≡ 0(mod �), where, obviously, (� −
�)! (� − 1)! + (−1)��� it is divisible by p. 
Problem 3: Prove that if p is an prime number and a is an integer numbers, then p it is 
divisible by (� − 1)! �� + �. 
Solution: From Fermat and Wilson theorems we can written: � ∣ �(� − 1)! + 1�, 
� ∣ (�� − �) , where � ∣ ����(� − 1)! + 1� − (�� − �)� or � ∣ (��(� − 1)! + �). 
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31 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Problem 4: Let p be an prime number and a, b integer numbers. 
Prove that if �|(�� − ��) then ��|(�� − ��). 
Solution: From Fermat theorem ⇒ �� ≡ �(mod �), �� ≡ �(mod �), so � ≡ �(mod �). 
We can written � = � + �� and we get: 
�� − �� = (� + �� )� − �� = ��
������� + ��
��������� + ⋯ + ��
�
���� = 
�������� + ��
������� + ⋯ + ������� where � ≥ 2 . So, �� − �� it is divisible with ��. 
Problem 5: Prove that for any natural numbers a,b relative prime, the number ���(�) +
��(�) − 1� it is divisible with the product ab. 
Solution: From Euler theorem,we get: ��(�) ≡ 1(mod �) and ��(�) ≡ 1(mod �). 
So, ��(�) = 1 + �� and ��(�) = 1 + ��, with �, � ∈ �. 
Then ���(�) − 1����(�) − 1� = (��)(��), so ���(�)��(�) − ��(�) − ��(�) + 1� it is 
divisible with ab and how ��(�)��(�) it is divisible with ab,result the conclusion. 
Problem 6: Prove the following characterization of the prime numbers: 
An natural number � ≥ 2 it is prime if and only if �(�)|(� − 1) and (� + 1)|�(�), where � 
is the Euler function and � is the function sum of divisors. 
Solution: " ⇒ " If p is an prime number, then �(�) = � − 1 and �(�) = � + 1. 
" ⇐ " Let be � ∈ �∗, with the property � ≥ 2, �(�)|(� − 1) and (� + 1)|�(�). 
We must to abserve that for any number � > 2, the number �(�) it is even number: from 
the expression by �(�), if n contains in the prime factorization ,prime numbers p different 
two by two 2 (so odd), then �(�) = (�� − ����)� = ����(� − 1)� , which is even 
number; if � = 2�, � > 2 then �(�) = 2� − 2��� = 2��� which it is even number. 
 From the relationship �(�)|(� − 1) we get that n must be even number(contains in the 
prime factorization only powers of odd numbers). 
 We show that all the factors from the factorization it is prime number (without 
exponent). 
If, through the absurd, it would exists ��
�� therms by n, with �� ≥ 2 then 
���
�� − ��
�����|�(�)|(� − 1) so, ��
����|(� − 1) = ��
����
�� … ��
�� − 1, which it is false. 
So, � = ���� … ��. Then, �(�) = (�� − 1)(�� − 1) … (�� − 1) and 
�(�) = (�� + 1)(�� + 1) … (�� + 1), with ��, ��, … , �� odd numbers and then result 
2�|�(�) and 2�|�(�). 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
32 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
 If � ≥ 2 then 4|�(�)|(� − 1), so 4�|(� + 1). From hypothesis, the numbers 
�(�)
���
 it is 
integer, n+1 it is even numbers, not divisible with 4, and �(�) it is divisible with 2�. Then 
2��� � 
�(�)
���
 and result 
2��� <
�(�)
�
=
(�� + 1)(�� + 1) … (�� + 1)
���� … ��
= �1 +
1
��
� … �1 +
1
��
� < �1 +
1
3
�
�
= �
4
3
�
�
, 
And the relationship it is false; so � = 1, therefore � = �� it is prime numbers. 
References: 
1. Romanian Mathematical Magazine-www.ssmrmh.ro 
2. Vasile Pop, Viorel Lupşor- Matematică pentru grupele de performanţă, clasa a XII-a, Ed. 
Dacia Educaţional, Cluj- Napoca, 2004 
3. P. Radovici- Probleme de teoria elementară a numerelor, Ed. Tehnică, Bucureşti, 1986 
ABOUT NAGEL AND GERGONNE’S CEVIANS (III) 
By Bogdan Fuştei-Romania 
 
Let’s consider the triangle ABC with standard notations; From previous parts: 
���
� = ��
� + ��
� + �����(��� �������); 
���
� = ����� + (� − �)
�(��� �������); 
���� = �(� − �)(��� �������); 
�� + �� = ��
� + ��
� + ���� (��� �������); 
��
� = ���� +
(� − �)�
�
� (��� �������); 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
33 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
�� = ���� + ��� (��� �������); 
����� + (� − �)
� = ���� +
(� − �)�
�
� + ��
� + ����(��� �������); 
 It’s easy to prove that: 
��
� = �(� − �) +
(� − �)�
�
(� − �) = (� − �) �� −
(� − �)�
�
� (��� �������); 
We was show it that : ��
� = � �� − � +
(���)�
�
� (and analogs); 
��
� ��
� = �(� − �) �� −
(� − �)�
�
� �� − � +
(� − �)�
�
� ≥ �(� − �)�(� − �) 
�(� − �) − (� − �)
(� − �)�
�
+
(� − �)�
�
� −
(� − �)�
�
(� − �)�
�
≥ �(� − �) 
(� − �)�
�
(� − � + �) ≥ 
(� − �)�
�
(� − �)�
�
 
(� − �)�
�
� ≥ 
(� − �)�
�
(� − �)�
�
 
(� − �)� ≥
(� − �)�
�
(� − �)�
�
 
If b=c then we get equality in up relationship; 
If b≠ � ⇒ 1 >
(���)�
��
⇒ �� > (� − �)� ⇒ � > � − � or � > � − � ;which means the triangle 
inequality; Finally,we have the inequality: 
���� ≥ ����(��� �������)(�);⇒����� + ��
� + ��
� ≥ ���
� 
So, we get: �� + �� ≥ ���(��� �������);(2) 
From up relationships, it’s easy to see that: 
�� + �� = ��
� + ��
� + ���� ≥ ����� + ���� ≥ ����� + ����; 
�� + �� ≥ �(���� + ���) ≥ ���(��� �������);(3) 
����� = ��(�� + ��)(��� �������); 
 ����� ≥ ��(�� + ��) 
����
��
≥
�� + ��
�
(��� �������); (�) 
Adding up relationships, we get: 
∑
����
��
≥ ��+��+�� = �� + �;(5) and 
��
��
≥
�����
���
(��� �������);(6) 
Again, adding (5),(6) we get: ∑
��
��
≥
�
�
∑
�����
��
(7); ���
�
�
=�
�(���)
��
(and analogs); 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
34 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
�� =
2��
� + �
 ���
�
2
(and analogs); 
From: ���� ≥ �(� − �) = ���� =
���
�
 ���
�
�
 �� ⇒ 
 ����
��
≥
���
�
 ���
�
�
(��� �������);(8) 
We know that: �� ≥
���
�
 ���
�
�
(and analogs) then, we get ∶ ���� ≥ �(� − �) = ����; 
So, we can write : ��������� ≥ ����(and analogs)(9); ∑ ���� = �
� ; 
So, we have: 
∑ ��������� ≥ �
� = ��
� + �����(��� �������);(10) 
From the abrove relationships, we have: 
���������
��
≥
�����
�
(��� �������);(11) 
Adding up relationships, we have: 
∑
���������
��
≥ ��+��+�� = �� + � ; (12) 
���������
�����
≥
��
�
 (and analogs); again, adding, we have: 
∑
���������
�����
≥
�
�
(�� + �� + ��);(13) 
From: ������ ≥ ������ = �
��; (and analogs) ⇒ 
������� ≥ �√�(��� �������);(14) 
∑ ������� ≥ ��√�;(15) 
From: �� ≥
���
�
 ���
�
�
 and 
 ����
��
≥
���
�
 ���
�
�
 we get: 
�
������
��
≥
� + �
�
���
�
�
 (��� �������)(��); 
Adding, we get: 
� �
������
��
≥ �
� + �
�
���
�
�
(��) 
������
�
≥ ��(��� �������) 
��
�
=
��
�����
(and analogs)so, we get:
������
�����
≥ ��=>
����
�����
≥
��
��
(and analogs) 
�
��
+
�
��
+
�
��
=
�
�
 ; 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
35 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
So, we will get a newralationship: 
�
����
�� − ��
≥
��
�
(��) 
�� = ��
� + �����(��� �������) 
 So, we will get a new relationship: 
����
�����
≥
��
�
��
+ ���(19) 
Adding up relationships, we get: 
�
����
�� − ��
≥ �(�� + �� + ��) + �
��
�
��
= �(�� + �) + �
��
�
��
(��) 
�� �
��
�� − ��
−
��
��
� ≥ ��� =>
��
�� − ��
−
��
��
≥
���
��
(��� �������)(��) 
Adding, we get: 
 �
��
�� − ��
≥ �
��
��
+ � �
��
��
(��) 
������
(�� − ��)(�� − ��)(�� − ��)
≥ � �
��
��
+
���
��
� (��) 
We know that: AI=�2�(ℎ� − 2�)=
�
���
�
�
(and analogs); sin
�
�
=�
���
��
(and analogs) 
������� = �
�; ��� = 4�� 
So, we have: 
(ℎ� − 2�)(ℎ� − 2�)(ℎ� − 2�) =
2��
�
 
Then, the inequality (23) becomes: 
�
��
 
������
��
≥ ∏ �
��
��
+
���
��
� (��) 
From 
��
�����
−
��
��
≥
���
��
(and analogs)we get the relationships: 
� �
��
�� − ��
−
���
��
� ≥
������
������
(��) 
� �
��
�� − ��
−
��
��
� ≥
�������
������
(��) 
From ���� ≥ ����(and analogs) =>
����
��
� ≥
����
��
� (and analogs); 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
36 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
��
� =
4��
(� + �)�
���� ; so, we will get the next relationships:
����
���
=
(� + �)�
4��
=
1
2
+
1
4
�
�
�
+
�
�
� ; 
Adding up relationships, we get: ∑
����
��
� =
�
�
+
�
�
∑
���
�
, so, we will write that: 
�
����
���
≥
�
�
+
�
�
�
� + �
�
(��) 
But: 
���
�
=
����
����
(and analogs); so, we get: 
 �
����
���
≥
�
�
+
�
�
�
�� + �
�� − �
(��) 
2S=ℎ�� = ℎ�� = ℎ�� =>
�
�
=
��
��
(and analogs) ⇒
�
�
+
�
�
=
��
��
+
��
��
⇒ ∑ �
�
�
+
�
�
� =
∑ �
��
��
+
��
��
� 
�
� + �
�
= �
ℎ� + ℎ�
ℎ�
, so, we get a new relationship: 
∑
����
��
� ≥
�
�
+
�
�
∑
�����
��
 (29), but 
��
�
= 1 +
���
�
(and analogs), adding, we get: 
ℎ� + ℎ� + ℎ�
�
= 3 + �
� + �
�
, following simple calculations, we have: 
�
����
���
≥
�� + �� + �� + ��
��
(��) 
We know that : 
�� =
2��
�� + ��
��(and analogs) 
⇒
��
��
=
1
2
�
�
�
+
�
�
� ⇒
����
���
=
1
2
(1 +
��
��
)(and analogs) 
�
����
���
=
1
2
(3 + �
��
��
) 
 So, we get a new relationship: 
�
����
���
≥
�
�
(� + �
��
��
) (��) 
We know that: 
 
���
�
=
�����
��
= 1 +
��
��
(and analogs) ⇒ ∑
���
�
= 3 + ∑
��
��
 
From simple calculations, we have: 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
37 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
 �
����
���
=
�
�
+
�
�
�
��
��
 
�
����
���
≥
�
�
+
�
�
�
��
��
 
From (� + � + �) �
�
�
+
�
�
+
�
�
� = 3 +
���
�
+
���
�
+
���
�
 we have: 
 �
����
���
=
1
4
�3 + (� + � + �) �
1
�
+
1
�
+
1
�
�� 
So, we get a new inequality: 
�
����
���
≥ �
����
���
=
�
�
�� + (� + � + �) �
�
�
+
�
�
+
�
�
�� (��) 
(� + � + �) �
1
�
+
1
�
+
1
�
� = 3 + �
� + �
�
; 
(ℎ� + ℎ� + ℎ�) �
1
ℎ�
+
1
ℎ�
+
1
ℎ�
� = 3 + �
ℎ� + ℎ�
ℎ�
 
But: ∑
���
�
= ∑
�����
��
 then, we get a new inequality: 
�
����
���
≥
�
�
�� + (�� + �� + ��) �
�
��
+
�
��
+
�
��
�� ; (��) 
From: �� = 2�ℎ�(and analogs); �� + �� + �� = �
� + �� + 4��, we have: 
ℎ� + ℎ� + ℎ� =
�� + �� + 4��
2�
, then 3 +
ℎ� + ℎ� + ℎ�
�
=
�� + �� + 10��
2��
 
So, we have a new inequality: 
�
����
���
≥
�� + �� + ����
���
=
�
�
+
�� + ��
���
(��) 
If, we applying Gerretsen inequality: 
�� ≥ 16�� − 5��, we get: 
�
����
���
≥
�� + �� + 10��
2��
=
5
4
+
�� + ��
8��
≥
5
4
+
16�� − 5�� + ��
8��
=
5
4
+ 2 −
4��
8��
 
Finally, we get a new inequality: 
�
����
���
≥
��
�
−
�
��
(��) 
We will improve this relationship. Let’s consider Yang inequality: 
�� ≥ 16�� − 5�� +
��(� − 2�)
� − �
(��� ������ ������������ �������� � − ���) 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
38 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Following simple calculations, we get a new ralationship: 
�
����
���
≥
��
�
−
�(�� − ��)
��(� − �)
 ; (��) 
We know the following relationship: 
�� = �2�(ℎ� − 2�) ; (and analogs) 
�� =
�
sin
�
�
(and analogs); 
����
�
2
=
�
2�
��
ℎ�
=
�� − �
4�
(and analogs) 
���
����
�
�
=
2�(ℎ� − 2�)4�
�� − �
= 8��
ℎ� − 2�
�� − �
; (and analogs) 
1
����
�
�
= 8 �
�
�
�
� ℎ� − 2�
�� − �
(and analogs); 
1
����
�
�
= 4 �
�
�
�
�
�
ℎ�
��
�
�
(and analogs); 
So, we have the following relationships: 
��
��
= √2�
�����
����
(and analogs); 
�
����
���
≥
9
4
+
1
4
�
ℎ�
��
; 
And from up relationships, we get: 
 �
����
���
≥
�
�
+
�
�√�
 � �
�� − ��
�� − �
 (��) 
 (� − �)� = ��
� + ��
� + 2��� − 2��� − 2���� 
 (� − �)� = ��
� + ��
� − 2����(and analogs); 
(�� − ��)
� = ��
� + ��
� − 2���� (and analogs); 
From up relationships, we get: 
 |� − �| ≥ �� − ��(��� �������)(��) 
Adding up relationships, we get: 
|� − �| + |� − �|+|� − �| ≥ (�� + �� + ��) − (�� + �� + ��)(��) 
 �� + �� ≥ 2�� ⇒ �� − �� ≥ 2(�� − ��), 
So, we have the following relationships: 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
39 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
�
�
|� − �| ≥ �� − ��(��)(��� �������) 
But: �� ≥ ��(and analogs) we have: 
�
�
|� − �| ≥ �� − ��(��)(��� �������); 
So, we have: 
| (� − �)(� − �)(� − �) | ≥ (�� − ��)(�� − ��)(�� − ��)(��) 
From the above relationships: 
�
�
| (� − �)(� − �)(� − �) | ≥ (�� − ��)(�� − ��)(�� − ��)(43) 
From the above relationships: 
�
�
| (� − �)(� − �)(� − �) | ≥ (�� − ��)(�� − ��)(�� − ��)(��) 
Let’s consider the triangle OIH by area : 
���� =
 | (���)(���)(���) |
��
 . 
So, we have the following relationships: 
���� ≥
(�� − ��)(�� − ��)(�� − ��)
��
(��) 
���� ≥
(�� − ��)(�� − ��)(�� − ��)
�
(��) 
���� ≥
(�� − ��)(�� − ��)(�� − ��)
�
(��) 
 �� + �� ≥ ��� => �� − �� ≥ �� − ��(��� �������)(��) 
(�� − ��)(�� − ��)(�� − ��) ≥ (�� − ��)(�� − ��)(�� − ��)(��) 
�� ≤ �� + � ≤ �� 
So, we get: 
(�� − ��)(�� − ��)(�� − ��) ≥ (�� − ��)(�� − ��)(�� − ��)(��) 
We know that: ∑ ���� = �
� then, we have: 
� ���� ≥ �
�(��) 
 �� + �� ≥ 2�� then adding, we get: 
∑(�� + ��) ≥ �(�� + �� + ��)(��) 
References: 
ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
40 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
ABOUT IONESCU-NESBITT INEQUALITY 
By D.M.Bătinețu-Giurgiu and Daniel Sitaru-Romania 
 
Let be �, � ∈ ℝ� = [�, ∞); �, � ∈ ℕ, � ≥ � and �, �, �, � ∈ ℝ�
∗ = (�, ∞), � = �, ������, 
�� = ∑ ��
�
��� such that � ∙ �� > � ∙ ���
�����
{��}. We want to proof the following 
inequalities: 
� ∙ � + � �
��� + ���
��� − ��� + �
�
����
���
≥
(� + �)�(�� + �)��
(�� − �)�� + ��
 ; (�) 
� + �� ∙ � �
��� + ���
��� − ��� + �
�
����
���
≥
(� + �)�(�� + �)��
(�� − �)�� + ��
; (�) 
Proof: 
� ∙ � + � �
��� + ���
��� − ��� + �
�
����
���
≥
�����
� ∙ � +
1
��
∙ ��
��� + ���
��� − ��� + �
�
���
�
���
≥
�����
 
≥ (� + 1) �� ∙ � ∙ … ∙ ��������
�� �������
∙
1
��
∙ ��
��� + ���
��� − ��� + �
�
���
�
���
���
= 
= (� + 1) ∙ �
��� + ���
��� − ��� + �
�
���
=
(∗)
 
= (� + 1) ∙ �
��� + ���
��� − ��� + �
�
���
+
(� + 1)��
�
−
(� + 1)��
�
= 
= (� + 1) ∙ � �
��� + ���
��� − ��� + �
+
�
�
�
�
���
−
(� + 1)��
�
= 
= (� + 1) ∙ ��
���� + ���� + ���� − ���� + ��
�(��� − ��� + �)
�
���
� −
(� + 1)��
�
= 
=
� + 1
�
��
(�� + ��)�� + ��
��� − ��� + �
�
���
− ��� = 
=
� + 1
�
��(�� + ��)�� + ��� ∙ �
1
��� − ��� + �
�
���
− ��� ≥
���������
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
41 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
≥
� + 1
�
�
�(�� + ��)�� + ����
�
∑ (��� − ��� + �)
�
���
− ��� =
� + 1
�
�
�(�� + ��)�� + ����
�
���� − ��� + ��
− ��� = 
=
� + 1
�
∙
������ + ���
��� + ���
� − ������ + ����� − ��
�
(�� − �)�� + ��
= 
=
� + 1
�
∙
������ + ����
(�� − �)�� + ��
=
� + 1
�
∙
(�� + �)��
(�� − �)�� + ��
=
(� + 1)(�� + �)���
(�� − �)�� + ��
 
namely the inequality (1) is proved. 
� + �� ∙ � �
��� + ���
��� − ��� + �
�
����
���
≥
�����
� + �� ∙
1
��
∙��
��� + ���
��� − ��� + �
�
���
�
���
= 
= � + ��
��� + ���
��� − ��� + �
�
���
�
���
≥
�����
 
≥ (� + 1) �1 ∙ 1 ∙ … ∙ 1�������
�� �������
∙ ��
��� + ���
��� − ��� + �
�
���
�
���
���
= (� + 1) ∙ �
��� + ���
��� − ��� + �
�
���
(∗)
�� 
� + �� ��
��� + ���
��� − ��� + �
�
���
�
���
≥
(� + 1)�(�� + �)��
(�� − �)�� + ��
 
namely the inequality (2) is proof. 
If in (1)or (2) we take � = 0, � = 0, � = � = � = 1, � = 0 we get: 
�
��
�� − ��
�
���
≥
�
� − 1
; (� − �) 
For � = 3, �� = �, �� = �, �� = � we get: 
�
� + �
+
�
� + �
+
�
� + �
≥
3
2
 ; (� − �) 
Namely Ionescu-Nesbitt inequality. 
If adding relations (1), (2) we get: 
�(� + 1) + (1 + ��) ∙ ��
��� + ���
��� − ��� + �
�
���
�
���
≥
2(� + 1)�(�� + �)��
(�� − �)�� + ��
; (3) 
References: 
ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro 
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
42 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
GAKOPOULOS’ LEMMA (II) 
By Thanasis Gakopoulos – Larisa – Greece 
(A) Lemma (I): (I) 
��
��
��
��
=
��
��
��
��
 
 
Proof. Let �� ∥ ��. Is: 
��
��
=
��
��
 (1) 
��
��
=
��
��
 (Thales’s theorem) (2) 
Is: (Menelaus’s theorem Δ���, ���): 
��
��
⋅
��
��
⋅
��
��
= 1 →
(�) ��
��
⋅
��
��
⋅
��
��
= 1 →
(�) ��
��
⋅
��
��
⋅
1
��
⋅
��
��
⋅ �� = 1 → 
→
��
��
��
��
=
��
��
��
��
 
Note: the lemma (I) has been proved (with PLAGIOGONAL system) and so is known in my 
previous paper, which has been in RMM group. 
(B) Lemma (II) Let: 
��
��
= �,
��
��
= �,
��
��
= �,
��
��
= �,
��
��
= �,
��
��
= �,
��
��
= �. 
�
�
=
�⋅�
�⋅�
 (II) 
 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
43 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Proof. 
��� (����� �):
��
��
��
��
=
��
��
��
��
 
��� (����� �):
��
��
��
��
=
��
��
��
�� ⎭
⎪⎪
⎬
⎪⎪
⎫
→
(×)
��
��
⋅
��
��
��
��
⋅
��
��
=
��
��
⋅
��
��
��
��
⋅
��
��
→
��
��
��
��
=
��
��
⋅
��
��
��
��
⋅
��
��
→
�
�
=
� ⋅ �
� ⋅ �
 
(C) APPLICATIONS. Relationships that holds to any COMPLETE 
QUADRILATERAL. 
��⋅��
��⋅��
=
��⋅��
��⋅��
 (C1) 
 
Proof.From lemma (I), if � ≡ � then we have: 
��
��
��
��
=
��
��
��
��
→
��⋅��
��⋅��
=
��⋅��
��⋅��
,
[���]
[���]
=
[���]
[���]
 (C2) 
 
There is a harmonious relationship between the triangle’s areas. 
Proof. 
From (C1): 
��⋅��
��⋅��
=
��⋅��
��⋅��
→
��⋅��⋅��� ��
��⋅��⋅��� �
=
��⋅��⋅��� ��
��⋅��⋅��� �
→
[���]
[���]
=
[���]
[���]
 
�� ⋅ �� = �� ⋅ �� (C3) 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
44 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
There is a harmonious relationship between the segment’s lengths. 
 
Proof. From (C2): 
[���]
[���]
=
[���]
[���]
→
��⋅��
��⋅��
=
��⋅��
��⋅��
→ ℎ� ⋅ ℎ� = ℎ� ⋅ ℎ� 
��⋅��
��⋅��
=
��⋅��⋅��
��⋅��⋅��
 (C4) 
 
Proof. 
From lemma (II), we have: �
�
�
=
�⋅�
�⋅�
� →
��
��
��
��
=
�
��
��
�⋅�
��
��
�
�
��
��
�⋅�
��
��
�
→
��⋅��
��⋅��
=
��⋅��⋅��
��⋅��⋅��
 
��
��
��
��
⋅
��
��
=
��
��
��
��
⋅
��
��
= � (C5.1) 
��⋅��
��⋅��
=
��⋅��⋅��⋅��
��⋅��⋅��⋅��
 (C5.2) 
Proof. 
���� ����� (�): ���:
��
��
��
��
=
��
��
��
��
→
��
��
��
��
⋅
��
��
= 1
���� ����� (�): ���:
��
��
��
��
=
��
��
��
��
→
��
��
��
��
⋅
��
��
= 1
⎭
⎪⎪
⎬
⎪⎪
⎫
→
��
��
��
��
⋅
��
��
=
��
��
��
��
⋅
��
��
= 1 
Proof. 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
45 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
From Lemma (II): �
�
�
=
�⋅�
�⋅�
� →
��
��
��
��
=
�
��
��
�⋅�
��
��
� 
�
��
��
��
��
��
�
→
��⋅��
��⋅��
=
��⋅��⋅��⋅��
��⋅��⋅��⋅��
 
(D) Complete Quadrilateral ��� − ��� 
Cyclic Quadrilateral ���� 
��⋅��
��⋅��
=
��⋅��
��⋅�� 
 (D1) 
��⋅��
��⋅��
=
��⋅��
��⋅��
 (D2) 
 
Proof D1. Is �� = �� = �� = �� = � (1) 
From lemma (II) we have: �
�
�
=
�⋅�
�⋅�
� →
��
��
��
��
=
�
��
��
��
��
��
�
�
��
��
��
��
��
�
→
(�) ��⋅��
��⋅��
=
��⋅��
��⋅��
 
Proof D2. From lemma (II) we have: �
�
�
=
�⋅�
�⋅�
� →
��
��
��
��
=
�
��
��
��
��
��
�
�
��
��
��
��
��
�
→
(�) ��⋅��
��⋅��
=
��⋅��
��⋅��
 
References: 
ROMANIAN MATHEMATICAL MAGAZINE-www.ssmrmh.ro 
STRUCTURI ALGEBRICE(VI) 
By Vasile Buruiană-Romania 
7. Să se rezolve în ℤ ecuația următoare: 
�� + ��� = ���� 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
46 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Proprietatea 5|�, � ⇒ 5|� ± � și generalizarea ei s-au demonstrat întâi, apoi s-a menționat 
teorema: �|��, (�, �) = 1 ⇒ �|� ⇒ 5|7��, (5,7) = 1 ⇒ 5|�� ⇒ 5|� deci � = 5��. Înlocuind 
⇒ � + 35��� = 320 și apoi ⇒ 5|� deci � = 5��. Înlocuind cercul �� + 7��
�
= 64 și prin 
încercări se găsesc (57; 1), (36; 2), (1; 3) pentru (��, ��), de unde soluția 
{(285,5), (180,10), (5,15)} 
S-a atras atenția asupra noțiunii de rădăcină a ecuației cu două necunoscute. S-a schițat 
geometric soluția {(�, �)|�, � ∈ ℝ} și ce înseamnă că (�, �) este formată doar din întregi 
pentru verificarea ecuației date. 
8. (clasa a X a, după construcția lui �, cerc de matematică) 
Cum găsim numere prime < 20 în (T) 
Cum putem găsi, cu un “ciur” al lui Erotostene, numerele prime din ℤ[�] cu norma < � dat? 
După ce s-a reamintit cum se procedează în primul caz, s-a introdus noțiunea de normă ca 
generalizare a modului, noțiunile de ireductibil, prim și s-a demonstrat că 
�(��) = �(�)�(�) cu �|� ⇒ �(�)|�(�). S-a definit unitatea și s-au găsit unitățile ±1, ±� 
arătând că � e unitate ⇔ �(�) = 1. S-a demonstrat că 2 este reductibil și factorii săi 1 ± � 
sunt ireductibili și asociați. 
Ca în cazul ciurului lui Eratostene (numai că acum se lucrează în plan) s-au șters unitățile și 
multiplii lui 1 + � dintr-un cerc dat (de ex. cercul cu centrul în origine și de rază 4). Cum 
�(�) = 4 ⇔ � = ±2 sau � = ±2� și cum deja au fost șterse ca multipli ai lui 1 + �, se caută 
cele de normă 5 (găsiți: �(1 + 2�), �(2 ± �) cu � ∈ {±1, ±�}. S-au scos cele ireductibile, s-au 
șters multiplii lor aflați în cerc ș.a.m.d. 
S-a insistat asupra aspectului geometric, s-a aplicat teorema împărțirii cu rest în acest caz, s-
a arătat ce este aceea o generalizare. 
 
9. (clasa a XI a, definiții, lecție). Fie �, � ∈ ℕ∗ astfel încât ����, �����, �����, �����, … . Să se 
arate că � = �. 
S-a insistat asupra scrierii unice � = ��
�� … ��
��, � = ��
�� … ��
��, a noțiunii de șir și faptului că 
�����|����� și �����|����� ⇒ (2� + 1)�� ≤ (2� + 2)�� și (2� + 2)�� ≤ (2� + 3)��; 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
47 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
� = 1, … , � ⇒ �� ≤
����
����
�� și �� ≤
����
����
��. Cu definiția limitei avem că 
����
����
→ 1,
����
����
→ 1 și 
deci trecând la inegalități obținem �� ≤ �� respectiv �� ≤ ��; ∀� = 1, … �. 
Deci �� = ��, � = 1 ⇒ � = �. 
10. (clasa a XII a, după teorema Bezout, lecție). Să se descompună în factori în ℝ[�, �, �] 
polinomul (� + � + �)� − �� − �� − �� = �. 
Privind ca polinom în ℝ[�, �][�] avem întâi că: 
� ⋮ � + � fiindcă �(−�) = (−� + � + �)� − (−�)� − �� − �� = �� + �� − �� − �� = 0. 
Analog �: � + �, � ⋮ (� + �) ⇒ (factori comuni) 
� = �(� + �)(� + �)(� + �). Cum gradul în fiecare variabilă este 2 ⇒ � este o constantă și 
găsirea lui se face dând lui �, �, � valoarea 1 ⇒ 27 − 3 = � ⋅ 8 ⇔ � = 3. Deci 
� = 3(� + �)(� + �)(� + �). 
11. (clasa a XII a, cap. Polinom, cerc). Criterii de ireductibilitate pentru polinoame din ℤ[�]. 
S-au reamintit noțiunile de prim, divizor, polinom ireductibil, primitiv, s-a enunțat și 
demonstrat criteriul Ersenstein și s-a aplicat pe cazuri: �� − 2 (pentru a dovedi că în ℚ[�] 
există polinoame ireductibile de orice grad) apoi �� − �� − �� − � ca polinom din ℚ[�][�]. 
S-a formulat teorema descompunerii unice pentru polinoame. 
12. (clasa a XII a, cap. Polinoame, cerc). Să se descompună în fracții simple fracția 
�
(���)������
. 
S-a definit noțiunea de polinom ireductibil în ℝ[�]și s-a arătat care sunt toate elementele 
ireductibile în acest inel, s-a format teorema de descompunere în factori primi în ℝ[�], s-a 
definit noțiunea de fracție simplă și s-a exemplificat pentru fracții ordinare. 
S-a formulat și demonstrat teorema în general.S-au făcut exemple și în particular problema 
de mai sus. S-au revăzut și fixat: noțiunile de polinoame identice, rădăcină, rădăcină 
multiplă. Teorema împărțirii cu rest în �[�] rădăcini conjugate, c.m.m.d.c, ideal, elemente 
compuse. Apoi cum din scrierea unei fracții 
�(�)
�(�)
 și a �(�) = ��
�� … ��
����
�� … ��
�� în factori 
ireductibili peste ℝ (�� liniari, �� de gradul �� cu rădăcini complexe), deoarece polinoamele 
următoare sunt coprime: 
��
�� … ��
����
�� … ��
�� = � 
��
����
�� … ��
����
�� … ��
�� = � 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
48 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
⋮ 
��
�� … ��
����
�� … ��
�� = � 
⋮ 
��
�� … ��
����
�� … ����
���� = � 
rezultă că există ��, … ��, �, … , �� astfel încât. 
� = ��� + ��� + ⋯ ������� + ����� deci 
�
�
=
��
��
�� + ⋯ +
��
��
�� +
��
��
��
+ ⋯
��
��
��
. Cum �� se dezvoltă după puterile lui �� și �� după cele ale 
lui ��, ajungem la scrierea cunoscută și ultilizată la integrarea funcțiilor raționale. Fiecare pas 
a fost exemplificat și pe fracții ordinare. 
13. (clasa a X a, după definirea numerlor complexe, cerc). Să se găsească numerele 
pitagorice (care verifică ecuația �� + �� = ��) și sunt în ℤ. 
Mai întâi căutăm soluțiile cu (�, �) = 1, deci �, � nu sunt ambii pari și obligatoriu � este 
impar. În �[�], factorial, avem (� + ��)(� − ��) = ��. Fie � ireductibil astfel ca 
 �|� + ��, �|� + �� ⇒ �|2� și �|2�; � fiind impar ⇒ (�) ≠ (1 + �) căci dacă sunt egale 
idealele ⇒ ��� = 2|�� ceea ce este imposibil căci ℤ este impar, deci � ⊺ 2 ⇒ �|� și 
�|� ⇒ �(�) = �|�, �|� ceea ce arată că nu avem că �, � sunt compuse, absurd. ⇒ � + ��, 
� − �� sunt coprime și dacă � = ���
��, … , ��
��, � fiind unitate, datorită factorialității lui ℤ[�] 
prin scrierea unică ⇒ � + �� = ���. Dacă � = � + ��, luând � = 1 obținem 
� + �� = �� + 2��� + �� ⇒ �
� = �� − ��
� = 2��
� = −
�
�
. Variantele pentru celelalte unități sau situații 
(� − �� = ��� sau (�, �) ≠ 1) se reduc la acestea, având pentru �, � semnele eventual 
schimbate. Deci orice soluție este de această forma. Orice soluție de forma respectivă, 
verifică ecuația �� + �� = ��, deci toate soluțiile sunt cu �
� = �� − ��
� = 2��
� = �� + ��
; �, � ∈ ℤ 
Mai întâi s-au precizat noțiunile de prim, ireductibil, inel ℤ[�], imagine geometrică, norma 
proprietății, unități, factorialitate, apoi s-a rezolvat. (Se poate compara eventual această 
soluție cu una care nu folosește numerele complexe) 
14. (clasa XII, după cl de resturi mod �, cerc) 
Romanian Mathematical Society-Mehedinți Branch 2021 
 
49 ROMANIAN MATHEMATICAL MAGAZINE NR. 29 
 
Să se construiască în ℤ[�] corpul claselor de resturi modulo 1 + � = � (sau mai general 
pentru � ireductibil). S-a arătat că în ℤ[�] este adevărată teorema împărțirii cu rest, că 
numărul elementelor în corpul respectiv este �(�), s-au figurat resturile posibile și s-a 
construit corpul finit corespunzător, apoi s-au făcut calcule în el. S-a remarcat numărul de 
elemente. S-a făcut paralela cu clasele de resturi mod �. Principial, relația de echivalență nu 
diferă. 
15. (clasa X,lecție) 
Care este exponentul lui � în �!? 
S-a redefinit [�] și datorită factorialității, exponentul căutat este finit. Dacă ��(�) este 
exponentul căutat este clar că � > � ⇒ �� = 0, iar � = � ⇒ �� = 1. Deci e interesant 
pentru � < �. Atunci în �! multiplii lui � sunt 1 ⋅ �, 2�, 3�, … , �
�
�
� �. 
Fie �� = �
�
�
� ⇒ înmulțind acești multipli avem 1 ⋅ 2 … �
�
�
� ⋅ ��� 
La fel în ��! ⇒ � întâi ca factor de �
��
�
� ori și cum �
��
�
� = �
�
�
�
�
�
� = �
�
��
� ⇒ 
��(�) = �� + ��(��), apoi ��(��) = �� + ��(��) etc., unde �� = �
�
�
� 
și în final α�(����) = �� 
ABOUT A TRIANGLE INEQUALITY 
By D.M.Bătineţu-Giurgiu, Claudia Nănuţi, Daniel Sitaru-Romania 
In ∆��� we’ll note with �, �, � −the lengths of the sides, � −semiperimeter and 
� −area.We will consider �, � ∈ ℝ� = [�, ∞); � + �, �, �, � ∈ ℝ�
∗ = (�, ∞), then: 
 
(� + �)� + �� + ��
�� + ��
∙ �� +
(� + �)� + �� + ��
�� + ��
∙ �� +
(� + �)� + �� + ��
�� + ��
∙ �� ≥ �√��, (�) ��� 
 
�� + ��
�� + �� + (� + �)�
∙ �� +
�� + ��
�� + �� + (� + �)�
∙ �� +
�� + ��
�� + �� + (� + �)�
∙ �� ≥ �√��; (�) 
 
Proof: We have: 
�
�� + ��
�� + �� + (� + �)�
∙ ��
���
= �
�� + ��
�� + �� + (� + �)�
∙ ��
���
+ � ��
���
− � ��
���
=

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