Exercise 5.4.12 (Equation $x^2 = y^3 + 23$)

Proof. Assume for the sake of contradiction that the pair $(x,y)\in {\mathbb{Z}}^2$ satisfies

If $y$ is even, then $x^2=y^3+23\equiv -1(\mathit{mod}8)$. This is impossible, so $y$ is odd. Since $x^2-y^3=23$ is odd, $x$ and $y$ are not of same parity. Therefore $x$ is even and $y$ is odd.

The equation (1) is equivalent to

Since $x$ is even and $y$ is odd, we write $x=2u,y=2v+1$. Then (2) becomes

so

This equation shows that $v$ is even, thus $v=2w$ for some integer $w$, and

The factor $16w^2-4w-1$ is of the form $4k-1$, where $k=4w^2-1\in \mathbb{Z}$. If all prime factors of $16w^2-4w-1$ were of the form $4K-1$, their product would be of the same form, which is false. Therefore there is some prime factor $q\equiv -1(\mathit{mod}4)$ of $16w^2-4w-1$, and $q\mid u^2+1$.

But then $\left(\frac{-1}{q}\right)=1$, thus ${(-1)}^{(q-1)∕2}=1$, $(q-1)∕2$ is even, thus $q\equiv 1(\mathit{mod}4)$. This is a contradiction, which proves that $x^2=y^3+23$ has no solution in integers. □