Prove that $n^2+11n+2$ is not divisible by $12769$

Question

My Attempt :

Prime factorisation of $12769$ is $113^2$

$n^2+11n+2-113^2m=0$

The conjugate of this quadratic equation becomes:

$\sqrt {113 (113m+1)} $

which can never be a rational as (113,113m+1)=1 and 113 is not a perfect square (and its prime )

I belive this method is right,but I am asking for a modular method of doing this, which I believe will be easier. Thanks

Answer 1

$113$ is prime.

$$n^2+11n+2\equiv 0\pmod{113^2}$$

$$\stackrel{\cdot 4}\iff (2n+11)^2\equiv 113\pmod{113^2}$$

$$\implies 113\mid (2n+11)^2\iff 113\mid 2n+11$$

$$\iff 113^2\mid (2n+11)^2$$

$$\iff (2n+11)^2\equiv 0\not\equiv 113\pmod{113^2}$$