Question

Show that there is no integer a such that $a^2-3a-19$ is divisible by $289$.

Answer 1

Let $\frac{a^2-3a-19}{289}=k$

We need to show that $k$ is not an integer.

Assume, for contradiction, that it is an integer.

$a^2-3a-19=289k$

$\Rightarrow a^2-3a-(19+289k)=0$

For $a{x}^2+bx+c=0$ to have a rational solution, the discriminant $b^2-4ac$ must be a perfect square.

Here $a=1,b=-3,c=-19-288k$

Discriminant$={(-3)}^2-4\times 1\times (-19-288k)$

$=9+76+1156k=17(5+68k)$

In order for this discriminant, $17(5+68k)$, to be a perfect square, since it has one factor of $17$, the other factor $5+68k$ must also have a factor of $17$. So there

must be an integer $n$ such that

$5+68k=17n$

$\Rightarrow 5=17n-68k$

$\Rightarrow n-4k=\frac{5}{17}$

But the right side is an integer but the left side is not.This contradicts our assumption that the discriminant is a perfect square. $\therefore a$ cannot be a rational number. And since $a$ is not a rational number, it certainly cannot be an integer