$a^2=b^3+23$ , solve the equation in positive integer

Question

Find all positive integer solutions $(a,b)$ such that $a^2=b^3+23$

I think there is no solution, yet I don't know how to prove it

Using Magma with folllowing command: IntegralPoints(EllipticCurve([0,23]));. It says the elliptic curve $y^2 = x^3 + 23$ doesn't have any integral solutions. — achille hui, Feb 03 '14 at 10:27
As it is over the positive integers, just plug in all integers up to 23. Not an elegant approach, but an easy proof none the less... — user1729, Feb 03 '14 at 10:31
@user1729, I don't think so. $a^2=b^3+k$ can have solutions in which $a$ and $b$ exceed $k$ (why not?). — Gerry Myerson, Feb 03 '14 at 11:08
@Gerry oh, sorry, got the position of the plus and equals mixed up! — user1729, Feb 03 '14 at 11:19
Perhaps a modulo $5$ approach to $(a-5)(a+5)=b^3-2$ might prove useful ? See also elliptic curve. — Lucian, Feb 03 '14 at 13:05
@user1729 Dozens of examples of $a$ and $b$ exceeding $k$, actually. — Balarka Sen, Feb 03 '14 at 14:16
What irritates me is that this does not seem to have even a rational point. Any proof? — Balarka Sen, Feb 03 '14 at 14:17

Balarka Sen · Answer 1 · 2014-02-03T14:24:51.877

This, again, is a special case of Mordell's equation. Some questions never die, do they?

$$y^2 = x^3 + 23$$

If $y$ is odd, then $x$ is $2 \pmod 4$, which is impossible as there is no such cube element in $\Bbb Z/4 \Bbb Z$, hence $y$ is even, and $x$ is odd. Precisely, $x$ is $1 \pmod 4$.

The general trick (after a lot of searching and fining no such tuple) is to make the right hand side factor on $\Bbb Z[x]$ :

$$y^2 = x^3 + 23$$

$$\Rightarrow y^2 + 4 = x^3 + 27 = (x + 3)(x^2 - 3x + 9)$$

The factor $x^2 - 3x + 9$ is obviously $3 \pmod 4$, and thus it has a prime factor of $3 \pmod 4$. But then $y^2 = -4 \pmod p$, which is impossible$(*)$

Sorry, but a necessary addition to convince myself I am not homework helping someone :

Exercise : Prove $(*)$

$a^2=b^3+23$ , solve the equation in positive integer

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