Why is $(x^2-2)/(2y^2+3)$ never an integer for any integers $x$ and $y$?

Solution 1:

Try to show that $2y^2 + 3$ must have at least one prime divisor which is not of the form $8k+1$ or $8k+7$.

Let's suppose it is an integer;

$x^2 - 2 = k(2y^2+3)$

$x^2 = (2k)y^2 + 3k + 2;$

since $x$ is an integer; l.h.s is also a perfect square, which is a quadratic equation in $y$;

that means roots of the equation are equal, and discriminant = 0;

$b^2-4ac = 0$ $\ \Rightarrow\ $ $0-(3k + 2)(2K) = 0\ $; $\ \Rightarrow\ $ $k = -3/2$ or $k = 0$;

and $k = 0$ only for $x^2 = 2$;

which is a contradiction...