Solution March 3, 2007

Prove that there exist no $x,y \in \mathbb{Z}$ such that $y^2 = x^3 + 23 \,$

Proof suggested by landen based on Rose, A Course in Number Theory. Proof by int-e.

This problem is a special case of the following theorem.

$y^2 = x^3 + m^3 -n^2\,$ has no solutions in two cases:

$m\equiv 3 \pmod{4}\,$ , $n\,$ is even, and $p\not\equiv 3 \pmod{4}\,$ when $p|n\,$
$m\equiv 2 \pmod{4}\,$ , $n\,$ is odd, and $p\not\equiv 3 \pmod{4}\,$ when $p|n\,$

The problem of the day is an instance of case 1, because $23 = 27-4\,$ .

Proof.

Assume we have a solution.

In both cases, $m^3 - n^2\equiv 3\pmod 4$ . If $x\,$ is even, we have $y^2 \equiv 3\pmod 4$ , a contradiction. So $x\,$ is odd and $y\,$ is even. It follows that $x\equiv 1\pmod 4\,$ .

We can rewrite the equation as $y^2 + n^2 = (x+m)(x^2 - xm + m^2)\,$ . Both factors are non-negative; $x^2 -xm + m^2 \geq 0$ is true for all $x,\,m$ , while $x+m\geq 0$ follows from $x^3+m^3 = y^2 + n^2 \geq 0$ .

In case 1, $t:=x^2 - xm + m^2\equiv 3\pmod 4$ , so $t\,$ has a prime divisor $p\equiv 3\pmod 4$ . Modulo $p\,$ , we have $y^2 \equiv -n^2\pmod p$ . $p\,$ can not divide $n\,$ because that would violate the conditions for $n\,$ so we can divide by $n^2\,$ . We find that $(y/n)^2 \equiv -1\pmod p$ , which is a contradiction because $-1\,$ is not a quadratic residue (http://en.wikipedia.org/wiki/Quadratic_residue) modulo $p\,$ .

In case 2, $t:=x+m\equiv 3\pmod 4$ and we get a contradiction in the same way. $\square$

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