<s>[INST] Prove that there is no integer a for which $a^2 - 3a - 19$ is divisible by 289 Prove that there is no integer a for which $a^2 - 3a - 19$ is divisible by 289.
Not got any clue. Please help.
 [/INST] Inspired by mercio
let us find $x,y$ such that $x-y=3,x+y=17\implies x=10,y=7$
$$a^2-3a-19=(a-10)(a+7)+51$$
As $51$ is divisible by $17,$ so must be $(a-10)(a+7)$ to make $17|(a^2-3a-19)$
Now $17|(a-10)\iff 17|(a+7)$ as $(a+7)-(a-10)=17$
So, $17^2|(a-10)(a+7),$ but $17^2\not|51$
Generalization:
Let us find integer $x,y$ such that 
$\displaystyle x-y=-3$(the coefficient of $a$)
and $\displaystyle x+y=17c$ so that either both or none $(a+x),(a-y)$ will be divisible by $17$
$\implies 2x=17c-3\implies c$  must be odd $=2d+1$(say) for some integer $d$
$\implies x=17d+7$ and $y=x+3=17d+10$
Now, $\displaystyle (a+x)(a+y)=\{a+(17d+7)\}\{a-(17d+10)\}=a^2-3a-(17d+7)(17d+10)$
$\displaystyle\implies a^2-3a-19=\{a+(17d+7)\}\{a-(17d+10)\}+(17d+7)(17d+10)-19$
$\displaystyle\implies a^2-3a-19=\underbrace{\{a+(17d+7)\}\{a-(17d+10)\}}_{\text{terms with difference }=17}+17^2(d^2+d)+51$
Now the logic is exactly same as the one above
In the above method $d=0$
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