For any prime $\displaystyle p \ge 29$. Prove that $\displaystyle 182 | p^2 - 1$

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- Mar 20th 2013, 06:49 PMKanwar245Prime number
For any prime $\displaystyle p \ge 29$. Prove that $\displaystyle 182 | p^2 - 1$

- Mar 20th 2013, 07:24 PMrgrotheeRe: Prime number
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- Mar 21st 2013, 05:11 AMHallsofIvyRe: Prime number
- Mar 21st 2013, 02:57 PMKanwar245Re: Prime number
- Mar 21st 2013, 03:59 PMHallsofIvyRe: Prime number
$\displaystyle p^{12}$! Oh, Dear. Well, $\displaystyle p^{12}- 1= (p+1)(p^2- p+ 1)(p-1)(p^2+ p+ 1)(p^6+ 1)$

$\displaystyle 182= 2(91)= 2(7)(13)$

Obviously, if p is prime, larger than or equal to 29, then both p+1, and p- 1 are even so it remains to show that one of the factors is divisible 7 and one by 13. - Mar 21st 2013, 08:27 PMNehushtanRe: Prime number
By Fermat’s little theorem, 13 divides $\displaystyle p^{12}-1$. Also by Fermat’s little theorem, 7 divides $\displaystyle p^6-1$ and hence 7 divides $\displaystyle p^{12}-1=(p^6-1)(p^6+1)$. And obviously 2 divides $\displaystyle p^2 - 1$ since $\displaystyle p$ is odd. The result follows.

NB: This works for all integers $\displaystyle p$ which are coprime with 2, 7, and 13.