problem on divisiblity How can I show that there is no integer such that $a^2 − 3a − 19$ is divisible by $289$.
 A: We have $a^2-3a-19\equiv (a+7)^2\bmod 17$, so that $17^2\mid a^2-3a-19$ implies that $a\equiv -7 \bmod 17$. With $a=17k-7$ we obtain
$$
\frac{a^2-3a-19}{289}=\frac{17k^2 - 17k + 3}{17}=k^2-k+\frac{3}{17},
$$
which is not an integer.
A: Applying the quadratic formula, we see that $x^2 − 3x − 19=0$ has a solution mod $289$ iff the discriminant $85$ is a square mod $289$.
Now, $85=5\cdot17$ and $289=17^2$.
So $b^2 \equiv 85 \bmod 289$ implies $b^2 \equiv 0 \bmod 17$ and so $b=17c$, but then $b^2 \equiv 0 \not\equiv 85 \bmod 289$.
A: Just another thought.
Let say $(a^2-3a-19)$ is divisible by $289$, so we can write it as $(a^2-3a-19)=289k$ where $k$ is an integer. So, 
$$a(a-3)=289k+19$$ 
$$ 289k+19 \equiv 2 (\text{mod} 17)$$
$$\implies a(a-3) \equiv 2 (\text{mod} 17)\ \ \dots(1)$$
So, there is a possibility that, $a=17\lambda+2$, but then $a-3=17\lambda-1$ which will contradicts $(1)$.  ($\lambda$ is an integer.)
Another possibility, $a=17\lambda-2$, but then $a-3=17\lambda-5$ which will again contradicts $(1)$.
Another possibility, $a=17\lambda+1$, but then $a-3=17\lambda-2$ which will again contradicts $(1)$.
Another possibility, $a=17\lambda-1$, but then $a-3=17\lambda-4$ which will again contradicts $(1)$.
So, this proves that our first assumption was wrong. 
