Atualização matemática_ (a b)p ap bp (mod p)

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25/01/2023 12:16 Atualização matemática: (a + b)p ≡ ap + bp (mod p)
mathrefresher.blogspot.com/2008/03/b-p-p-b-p-mod-p.html 1/3
Revisão de conceitos matemáticos fundamentais de forma direta e acessível.
Atualização matemáticaAtualização matemática
sábado, 01 de março de 2008
(a + b) 
p
 ≡ a 
p
 + b 
p
 (mod p)
Lema 1:
se p é primo, então:
(a + b) 
p 
≡ a 
p 
+ b 
p 
(mod p)
Prova:
(1) Usando o Teorema Binomial (ver Teorema aqui ), sabemos que: (2) Agora, como p é primo,
fica claro que para cada termo p!/(m!)(pm)! , p não é divisível por nenhum termo ≤ m ou por
qualquer termo ≤ pm de modo que temos: p!/(m!)(pm)! = p*([(p-1)*...*p-m+1]/[m!]) (3) Isso
mostra que cada um desses termos é divisível por p e, portanto: [p!/(m !)(pm)!]a 
p-m
 b 
m
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25/01/2023 12:16 Atualização matemática: (a + b)p ≡ ap + bp (mod p)
mathrefresher.blogspot.com/2008/03/b-p-p-b-p-mod-p.html 2/3
Postado por Larry Freeman às 4h01 
≡ 0 (mod p)
(4) Então existe um inteiro n tal que:
(a + b) 
p
 = a 
p
 + np + b 
p
(5) Visto que a 
p
 + nb + b 
p
 ≡ a 
p
 + b 
p
 (mod p) , segue que:
(a + b) 
p 
≡ a 
p 
+ b 
p 
(mod p)
QED
 
2 comentários:
Desconhecido disse...
Fiquei um pouco confuso com "p!/(m!)(pm)! = p*([(p-1)*...*p-m+1]/[m!])". Como você
conseguiu isso?
10 de novembro de 2010, 12:44:00
Larry Freeman disse...
Oi David,
Isso vem da definição de um fatorial.
p! = p*(p-1)*...*2*1
(a + b)p ≡ ap +
bp (mod p)
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25/01/2023 12:16 Atualização matemática: (a + b)p ≡ ap + bp (mod p)
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Se assumirmos que m é menor que p, temos:
(pm)! = (pm)*(pm-1)*...*2*1
Então:
p!/(pm)! = p*(p-1)*...*(p-m+1)
Espero ter ajudado.
10 de novembro de 2010, 14:49:00
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