New puzzle about prime

wpolly · 2003-07-29, 12:21

Assume that p>=7 is a prime and p=1(mod 6).
Let S=(1^2+1+1)(2^2+2+1)(3^2+3+1)......(p^2+p+1).
Prove that p|S.

maxal · 2009-03-16, 19:31

p divides a^2 + a + 1 = (a^3 - 1)/(a - 1), where a is a cube root of 1 modulo p, different from 1. Since p=1(mod 6) such a cube root exists. Namely, if x is a primitive root modulo p, then we can take a = x^((p-1)/3) mod p.

davar55 · 2009-07-02, 20:27

Does this tie arithmetic to algebra?

2003-07-29, 12:21	#1
wpolly Sep 2002 Vienna, Austria 11011011₂ Posts	New puzzle about prime Assume that p>=7 is a prime and p=1(mod 6). Let S=(1^2+1+1)(2^2+2+1)(3^2+3+1)......(p^2+p+1). Prove that p\|S.

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2009-03-16, 19:31	#2
maxal Feb 2005 2²×3²×7 Posts	p divides a^2 + a + 1 = (a^3 - 1)/(a - 1), where a is a cube root of 1 modulo p, different from 1. Since p=1(mod 6) such a cube root exists. Namely, if x is a primitive root modulo p, then we can take a = x^((p-1)/3) mod p.

2009-07-02, 20:27	#3
davar55 May 2004 New York City 4234₁₀ Posts	Does this tie arithmetic to algebra?