Prove that there is no integer a for which $a^2 – 3a – 19$ is divisible by 289.
Not got any clue. Please help.
Number Theory – Prove No Integer a Makes a^2 – 3a – 19 Divisible by 289
elementary-number-theory
elementary-number-theory
Prove that there is no integer a for which $a^2 – 3a – 19$ is divisible by 289.
Not got any clue. Please help.
Best Answer
$a^2-3a-19 = (a+7)^2-17(a+4)$. For $17^2$ to divide this, $17$ has to divide both $a+7$ and $a+4$, which is impossible.