2010/021) Prove that there is no a for which a^2-3a -19 is divisible by 289

Proof:
As a first step as we see that 289 = 17^2.

Now a^2-3a-19 = (a-10)(a+7) + 51

The 2nd term that is 51 is divisible by 17 and for the 1st term that is product to be divisible by 17 either (a-10) or (a+7) is divisible by 17. but if one of them is divisible by 17 then the 2nd one is divisible by 17.

So 1st term is divisible by 289 and 2nd one is not divisible by 289 so sum is not divisible by 289. Or the 2nd term is divisible by 17 and 1st term is not divisible by 17 so sum is not divisible by 17.

So the expression is not divisible by 289.