1. ## Prime

Hi,

p is a prime number such as p $\ge$ 5, how to show that $p^2+11 \equiv 0[12]$

2. Because is...

$p^{2} -1 = (p+1)\cdot (p-1)$ (1)

... and p is prime, p+1 and p-1 are both even numers and one of them is a multiple of 3, so that 12 devides $p^{2}-1$ and then 12 devides also $p^{2} + 11$...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by bhitroofen01
Hi,

p is a prime number such as p $\ge$ 5, how to show that $p^2+11 \equiv 0[12]$

Any prime number $p\geq 5$ equals $1,5,7\,\,\,or\,\,\,11\!\!\!\pmod {12}$ (why?) . Now just check that each of these ones satisfy the given congruence.

Tonio