Hi,
p is a prime number such as p $\displaystyle \ge$ 5, how to show that $\displaystyle p^2+11 \equiv 0[12]$
Because is...
$\displaystyle p^{2} -1 = (p+1)\cdot (p-1)$ (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 $\displaystyle p^{2}-1$ and then 12 devides also $\displaystyle p^{2} + 11$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$