Italian Mathematical Olympiad 1990 - Problem 5

Problem: 
Prove that, for every integer $x$, the number $x^2+5x+16$ is not divisible by $169$.

Solution:
If $x^2+5x+16$ would be divisible by $169$, then would be divisible by $13$. Then,
$$x^2+5x+16 \equiv x^2-8x+16 = (x-4)^2 \equiv 0 \pmod{13}.$$ So, if $x \neq 4 + 13k$, $k \in \mathbb{Z}$, then $x^2+5x+16$ is not divisible by $13$ and, a fortiori, by $169$. Now, let $x=4+13k$ with $k \in \mathbb{Z}$. Hence,
$$(4+13k)^2+5(4+13k)+16=169(k^2+k)+52 \equiv 52 \pmod{169}$$
then $x^2+5x+16$ is not divisible by $169$ for $x=4+13k$ and the desired conclusion follows.