7

Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$

I guess, that x cannot be even. because we can apply mod 4 test

say $x=2k$ then $y^2=8k^3+23\equiv3\mod(4)$

but, this is not possible, since the square of an integer is congruent to $0$ or $1$ $\mod (4)$

so $x$ is of the form $4k+3$ or $4k+1$

$x=4k+3$ is also not possible for the same reason but for the last case the test fails.

any hints ?

derivative
  • 2,450

1 Answers1

12

There are no integer solutions.

You noted already that $x$ cannot be even, because the equation would then fail mod $4$; so we need only consider $x$ odd. Then $x^3+23$ is even, so must be a multiple of $4$, whence $x \equiv 1 \bmod 4$.

We cannot obtain a contradiction from further congruence considerations. But since $23 = 3^3 - 2^2$ we try adding $4$ on both sides, getting $$ y^2 + 4 = x^3 + 27 = (x+3) (x^2-3x+9), $$ and observe that $x \equiv 1 \bmod 4$ implies $x^2 - 3x + 9 \equiv 1 - 3 + 9 \equiv 3 \bmod 4$. Since also $x^2-3x+9 > 0$ we deduce that $x^2-3x+9$ has a prime factor $p \equiv 3 \bmod 4$. But this is impossible by quadratic reciprocity: $y$ would be a square root of $-4 \bmod p$. QED

(In fact mwrank reports that there aren't even any rational solutions, but that's not an elementary proof.)

[Added later: See Keith Conrad's Examples of Mordell's Equation for further examples of elementary but nontrivial proofs that certain equations of this form $y^2 = x^3 + k$ have no integer solutions, and also some examples where a nonempty list of solutions of $y^2 = x^3 + k$ can be proved complete.]

KCd
  • 46,062
Noam D. Elkies
  • 25,917
  • 1
  • 64
  • 82
  • Can`t we conclude from the fact that if $x^3+23$ is a multiple of 4, so is $x^3+27$. If we factorize, like you did, $x+3$ must be $0mod4$ (if $x≡1mod4$) therefore $x^2−3x+9$ is multiple of 4, but this is always $3mod4$, which is a contradiction – derivative Nov 25 '13 at 14:16
  • How do you conclude from $x+3 \equiv 0 \bmod 4$ that $x^2-3x+9$ is a multiple of 4? – Noam D. Elkies Nov 25 '13 at 17:24
  • Sorry it was nonsense. – derivative Nov 25 '13 at 18:16
  • Why should there be a prime factor of the form $p = 3\bmod4$, also why is $y$ a square root of $-4\bmod p$? – Aurora Borealis Nov 15 '17 at 10:56
  • @AuroraBorealis Odd primes are in the residue class ${1, 3} \mod 4$, which is closed under multiplication. Notice $1 \cdot 1 \equiv 1 \mod 4$. Thus, at least one of the primes in the prime expansion of $3 \mod 4$ must be congruent to $3 \mod 4$. – Mateen Ulhaq Aug 02 '18 at 07:46
  • @AuroraBorealis Since $np=(x+3)(x^2−3x+9)$, we have $y^2+4 \equiv np \equiv 0 \mod p$. – Mateen Ulhaq Aug 02 '18 at 12:49
  • @AuroraBorealis Finally, $\left(\frac{-4}{p}\right) = \left(\frac{-1}{p}\right) \left(\frac{2}{p}\right)^2 = -1$ using the formula here for $\left(\frac{-1}{p}\right) $. Thus, $-4$ is a non-residue mod $p$. – Mateen Ulhaq Aug 02 '18 at 13:06