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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

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$b$ or $y{}{}{}{}$? –  Nancy Rutkowskie Feb 3 at 10:25
    
$b$. It's a typo. –  b00n heT Feb 3 at 10:26
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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 3 at 10:27
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interested name Magma,i was thinking that it is lava :D –  dato datuashvili Feb 3 at 10:28
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A table here: oeis.org/A054504 –  Frank Science Feb 3 at 10:29

1 Answer 1

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 $(*)$

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