How many positive integers $n$ are there such that $2n$ and $2n^2+1$ are both perfect squares?

Question

How many positive integers $n$ are there such that $2n$ and $2n^2+1$ are both perfect squares? $n=2$ is the only solution I can find. Are there others?

Answer 1

Not a perfect solution... but maybe can be a hint.

$2n$ is perfect square, so let $n=2k^2$.

Therefore, $8k^4+1$ is also a perfect square.

let $8k^4+1=l^2$.

Then, $l^2-8k^4=1.$

let $k^2=m$, which leads to:

$l^2-8m^2=1.$

This is a Pell equation. solving this, we get:

$(l, m)=(3, 1), (17, 6), ..., \left( \dfrac{(3+\sqrt8)^{\alpha}+(3-\sqrt{8})^{\alpha}}{2}, \dfrac {(3+\sqrt{8})^{\alpha}-(3-\sqrt{8})^{\alpha}} {2\sqrt{8}} \right), ...$ .

So, the number of $n$ will be the number of the perfect square of $\dfrac{(3+\sqrt8)^{\alpha}-(3-\sqrt8)^{\alpha}}{2\sqrt8}$. This can be also written:

$ (3+\sqrt{8})^{\alpha-1} + (3+\sqrt{8})^{\alpha-2}(3-\sqrt{8})+\cdots+(3+\sqrt{8})(3-\sqrt{8})^{\alpha-2}+(3-\sqrt{8})^{\alpha-1} $.

Use that $(3+\sqrt{8})(3-\sqrt{8})=1.$