Math Help - Prime number

1. Prime number

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

0

3. Re: Prime number

Originally Posted by Kanwar245
For any prime $p \ge 29$. Prove that $182 | p^2 - 1$
If p= 31, $p^2- 1= 960$ which is NOT divisible by 182: 960/182= 5.274725...

Further p- 1= 30 and p+ 1= 32 and neither divides 182 so I have no idea what you or rgrothee are asserting.

4. Re: Prime number

Originally Posted by HallsofIvy
If p= 31, [tex]p^2- 1= 960 which is NOT divisible by 182: 960/182= 5.274725...

Further p- 1= 30 and p+ 1= 32 and neither divides 182 so I have no idea what you or rgrothee are asserting.
My bad, you're right. There's a typo. We need to show that $182|p^{12}-1$

5. Re: Prime number

$p^{12}$! Oh, Dear. Well, $p^{12}- 1= (p+1)(p^2- p+ 1)(p-1)(p^2+ p+ 1)(p^6+ 1)$
$182= 2(91)= 2(7)(13)$

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.

6. Re: Prime number

Originally Posted by Kanwar245
For any prime $p \ge 29$. Prove that $182 | p^2 - 1$
By Fermat’s little theorem, 13 divides $p^{12}-1$. Also by Fermat’s little theorem, 7 divides $p^6-1$ and hence 7 divides $p^{12}-1=(p^6-1)(p^6+1)$. And obviously 2 divides $p^2 - 1$ since $p$ is odd. The result follows.

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