# Perfect squares and divisor

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

:)

• 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? – Tapi Apr 20 at 18:46
• @LenaDas exactly – Eureka Apr 20 at 23:51