Prove that there is no integer a for which $a^2 - 3a - 19$ is divisible by 289

Solution 1:

$a^2-3a-19 = (a+7)^2-17(a+4)$. For $17^2$ to divide this, $17$ has to divide both $a+7$ and $a+4$, which is impossible.

Solution 2:

Assume that $a^2 -3a -19 = 289n$ for some integers $a$, and $n$.

Then: $a^2 - 3a -19 - 289n = 0$, taking "delta"

$$\Delta = b^2 - 4ac = (-3)^2 -4(1)(-19-289n) = 9 + 76 + 1156n = 85 + 1156n = 17(5 + 4\times17n)$$

must be a perfect square.

Thus $17(5 + 4 \times 17n)$ can be written as a product of squares of primes. Since $17$ is a prime itself, the factor $5 + 4 \times 17n = 17p^2 \times q^2 \times ...t^2$ with $p, q, ..., t$ are primes.

But this means $5 + 4 \times 17n = 0 \pmod{17}$ a contradiction since it is $5 \pmod{17}$.

Solution 3:

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 $d=0$