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May divide ?p3 (a + b − −)p ap bp
 Asked 6 years, 1 month ago Modified 6 years, 1 month ago 1k timesViewed
9
Do there exist positive integers and a prime such that divides ?a, b p > max(a, b) p3 (a + b − −)p ap bp
The reader of Kvant magazine A. T. Kurgansky asked to prove that such do not exist, see . But discussion herea, b, p here
On the exact reference of a cute Diophantine problem
suggests that it should be very hard to prove. Maybe, a counterexample may be bound? Roughly speaking, a probability of this event is about
, for each we have about events (even for , I was previously wrong that it may be supposed without loss of generality, thanks for
Noam Elkies for noting this) and so as , we may expect them (and even infinitely many!) But this series converges very slowly, so
the minimal example may be large.
1/p2 p p b = 1
∑ 1/p = ∞
nt.number-theory prime-numbers experimental-mathematics
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25/01/2023 12:33 nt.number theory - May $p^3$ divide $(a+b)^p-a^p-b^p$? - MathOverflow
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edited Apr 13, 2017 at 12:58
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asked Nov 29, 2016 at 14:43
Fedor Petrov
 How far has it been checked numerically? Computing power is much more plentiful now than it was in 1984. – Noam D. Elkies Nov 29, 2016 at 14:54
1 That does not preserve congruences mod .p3 – Noam D. Elkies Nov 29, 2016 at 14:57
1 "We may suppose that " -- I don't see it: replacing by won't usually preserve the region .b = 1 (a, b) (a/b, 1) mod p2 a, b ≤ p – Noam D. Elkies Nov 29, 2016 at
15:00
 
 
@NoamD.Elkies you are correct: of course, we can not assume this. Moreover, finding a root of modulo , if it exists, we may choose
a pair for appropriate , and something like that works or almost works by Thue's lemma.
x ((x + 1 − − 1)/p)p xp p2
(a, b) = (a mod , ax mod )p2 p2 a –  Fedor Petrov Nov 29, 2016
at 15:09
 @AlexeyUstinov what exactly is the question? Of course we may always add to .p3 a, b –  Fedor Petrov Nov 30, 2016 at 15:01
3 Answers Sorted by: Highest score (default)
13
There are apparently lots of examples. The smallest is , , and .a = 1 b = 2 p = 7
Share Cite Improve this answer Follow answered Nov 29, 2016 at 14:57
Pace Nielsen
6 The small example was , , and .cutest a = 5 b = 6 p = 7 – Pace Nielsen Nov 29, 2016 at 14:58
1 Great. I did not expect that they missed such a small example. –  Fedor Petrov Nov 29, 2016 at 15:02
3
 
 , arguably is even cuter, besides being the first example where is divisible not just by but even by
.
Pace Pace Nielsen (a, b, p) = (3, 5, 7) (a + b − −)p ap bp p3
p5 – Noam D. Elkies Nov 29, 2016 at 22:23
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25/01/2023 12:33 nt.number theory - May $p^3$ divide $(a+b)^p-a^p-b^p$? - MathOverflow
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 Wow -- computers must have been pretty rubbish in 1984... – wrigley Nov 30, 2016 at 9:26
20
We explain the pattern observed by , deducing the existence of infinitely many solutions, some of which even have
.
Joe Silverman
|(a + b − −p5 )p ap bp
 Lemma. If then the homogeneous polynomial has a factor .n ≡ 1 mod 3 (a, b) := (a + b − − ∈ Z[a, b]Pn )n an bn ( + ab +a2 b2)2
 : Either
i) evaluate and its derivative at a cube root of unity , or
ii) use the symmetry of where : double roots at and are
the only ways to get the total number of roots to be . 
Proof
(x, 1)Pn ρ
S3 − (a, b) = + +Pn a
n bn cn a + b + c = 0 (a : b : c) = (1 : ρ : )ρ2 (1 : : ρ)ρ2
1 mod 3 □
[These -rational points on the -th Fermat curve are well known.]Q(ρ) n
 Corollary. If is a prime congruent to , and are integers such that , then .p 1 mod 3 a, b | + ab +pk a2 b2 | (a, b)p2k+1 Pn
 : Observe that and use the Lemma.Proof ∈ pZ[a, b]Pp
Now , and thus , is easy to obtain: choose arbitrarily, and let where is a cube root of unity mod . We can
even get a factor of by choosing such that (that is, so that are sides of a triangle with a angle); for
example,
k = 1 2k + 1 = 3 a b ≡ ra mod p r p
p5 a, b + ab + =a2 b2 p2 a, b, p 120∘
(3 + 5 − − = 120 ⋅ .)7 37 57 75
Share Cite Improve this answer Follow answered Nov 29, 2016 at 17:42
Noam D. Elkies
1
 
Yes of course (or as we'd do it nowadays, by applying Hensel's Lemma), but once we can no longer find positive solutions with as the Kvant
problem asked.
k > 2 a, b < p
– Noam D. Elkies Nov 29, 2016 at 18:30
 For the first 200 primes I find by exhaustive search that there is no solution mod with .p6 0 < a, b < p – Neil Strickland Nov 29, 2016 at 18:53
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25/01/2023 12:33 nt.number theory - May $p^3$ divide $(a+b)^p-a^p-b^p$? - MathOverflow
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 Motivated by Neil's computation, I've now checked all primes up to , and there are still no solutions modulo .8803 p6 – Pace Nielsen Dec 2, 2016 at 14:50
10
For what it's worth, consider . It seems (experimentally at least, I haven't tried to prove it) that for every there exists
an in the range such that
I checked for , and in this range, for each there is a unique such . Further, again just for , if , then there are
no solutions to .
b = p − 1 p ≡ 1 (mod 6)
a 2 ≤ a ≤ p1
2
(a + p − 1 − − (p − 1 ≡ 0 (mod ). (∗))p ap )p p3
p < 500 p a p < 500 p ≢ 1 (mod 6)
(∗)
Share Cite Improve this answer Follow answered Nov 29, 2016 at 15:31
Joe Silverman
2
 
The unique solution is always a cube root of unity mod (verified up to now); presumably it's easy to prove in general that this works,and intractable to
show that there are no other solutions.
p 1000
– Noam D. Elkies Nov 29, 2016 at 15:59
 (Actually that was for . For it's a sixth root of unity.)b = 1 b = p − 1 – Noam D. Elkies Nov 29, 2016 at 16:02
 
Interesting. This congruence is equivalent to , or . But I do not see why it
has roots modulo so often and why modulo 3 matters.
(a − 1 − + 1 ≡ 0 (mod p)p ap )3 a + /2 + ⋯ + /(p − 1) ≡ 0 (mod pa2 ap−1 )2
p2 p –  Fedor Petrov Nov 29, 2016 at 16:05
 Well, why the cubic root of unity works is seen from this formula. –  Fedor Petrov Nov 29, 2016 at 16:07
 @NoamD.Elkies Thanks. I was going to try to figure out what the was when I got a chance. I'm not actually convinced of the uniqueness.a – Joe Silverman Nov
29, 2016 at 16:07
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