Exercise 3.2.20* ($(x^2 - 2)/(2y^2 + 3)$ is never an integer)

Proof. Let $x,y$ be integers. Assume, for the sake of contradiction, that $(x^2-2)∕(2y^2+3)$ is an integer, so that

Let $p$ be any prime factor of $2y^2+3$. By transitivity, $p$ is also a prime factor of $x^2-2$.

Thus $x^2\equiv 2(\mathit{mod}p)$, and $2y^2\equiv -3(\mathit{mod}p)$. This gives ${(\mathit{xy})}^2\equiv 2y^2\equiv -3(\mathit{mod}p)$.

Moreover, $p\ne 2$, because $2y^2+3$ is odd, and $p\ne 3$, otherwise $x^2\equiv 2(\mathit{mod}3)$, which is impossible since $2$ is not a quadratic residue modulo $3$.

Therefore $-3$ and $2$ are quadratic residues modulo $p$, so

By the law of quadratic reciprocity,

Therefore, (1) gives

So $p$ is a solution of one of the two systems

which give

In both cases, $p\equiv 1(\mathit{mod}6)$. Under the hypothesis $2y^2+3\mid x^2-2$, this proves that every prime factor of $2y^2+3$ is of the form $6k+1$. Since $2y^2+3$ is a product of such primes, we obtain

Therefore $2y^2\equiv -2(\mathit{mod}6)$, and

This is a contradiction, because $-1$ is not a quadratic residue modulo $3$.

Hence $(x^2-2)∕(2y^2+3)$ is never an integer. □