Integer solutions to $x^2 - 15 y^2 = 15$

Question

How would one show that $x^2 - 15 y^2 = 15$ has no integer solutions ?

I got that $x = \pm \sqrt{15 (y^2 +1) }$ and then WLOG I assume that $x \in \mathbb{N}$ and $x = \sqrt{15 (y^2 +1) }$. From there, I want to show that for every $y \in \mathbb{Z}$, $15 (y^2 +1)$ is not a square.

Clearly, $y^2 +1$ is not a square but how do I know that $15(y^2 +1)$ is not one ?

Any help would be appreciated !

Answer 1

If $15(y^2+1)$ is a square then $y^2+1$ is divisible by $15$, so in particular it is divisible by $3$, a contradiction.

Note that your remark

Clearly, $y^2+1$ is not a square...

is false because $0^2+1=1^2=(-1)^2$.

Answer 2

Hint: $x$ must be divisible by $3$, hence $x^2$ is divisible by $9$. Putting $x=3z$, we have in fact that $5(y^2+1)$ is divisible by $3$, which you can reject by a case distinction on the remainder of $y$ modulo $3$.