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


  • $\begingroup$ 1+2/t can be either 2 or 3, therefore, this necessarily means that 2n^2 or 3n^2 can never be a perfect square, right? $\endgroup$ – Tapi Apr 20 at 18:46
  • $\begingroup$ @LenaDas exactly $\endgroup$ – Eureka Apr 20 at 23:51

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