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

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 $0^2+1=1^2=(-1)^2$.
