Let $n$ be a positive integer and let $d$ be a positive divisor of $2n^2$. Prove that $n^2+d$ is not a perfect square.
My working:
$d \mid 2n^2$
Let $d \cdot k=2n^2 \implies d=\dfrac {2n^2}k$
Therefore, $n^2+d=n^2\dfrac {k+2}k$
How do I proceed further?
Since $d$ is a divisor of $2n^2$ we'll have:
$$\frac{2n^2}{d}=t\in \Bbb{Z}$$
So our expression becomes:
$$n^2+\frac{2n^2}{t}$$ $$n^2(1+\frac 2t)$$
This means that $1+\frac 2t$ have to be a square and this is banally never verified
:)