<s>[INST] Number of solutions to Pell-type equation $k(x^2-2y^2) = d$ Given equation $$k(x^2-2y^2) = d$$
Where d is a constant. k,x,y are variables. All are positive integers.
Is there some characterization for the values of d for which there are unique solutions, multiple solutions or no solutions? If this is unsolved, then any tips towards studying this topic would be appreciated. Thanks.
EDIT:
So the original equation multiplied by k is:
$$k^2x^2-2k^2y^2=kd$$
and substituting $X=kx$ and $Y=ky$ and $Z=kd$
we have
$X^2-2Y^2=Z$
Does this help?
 [/INST] If there is a solution then there are infinitely many solutions. If $a$ and $b$ are so that $a^2-2b^2=1$ (of which there are infinitely many), and $x,y,k$ are solutions to $k(x^2-2y^2)=d$,  then $n=ax+2by$ and $m=ay+bx$ will also satisfy $k(n^2-2m^2)=d$. 
This is because $1=a^2-2b^2=(a+\sqrt{2}b)(a-\sqrt{2}b)$ and $(a+\sqrt{2}b)(x+\sqrt{2}y)=n+\sqrt{2}m$. You can use this to shows show that
$$
(n+\sqrt{2}m)(n-\sqrt{2}m)= (a^2-2b^2)(x^2-2y^2)\\
$$
Which is equal to $\frac{d}{k}$. But this also equals $n^2-2m^2$. So if there is a solutions, there are infinitely many.
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