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$$y^2=x^3+23$$ Are there any easy ways to solve this problem with number theory, abstract algebra, etc.? (trial and error for mods by the way)

Bart Michels
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Frknx
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    You can solve special cases of the Mordell curve using UFD's. – Edward Jiang Mar 12 '15 at 20:56
  • I would like to study this kind of equations.I know undergraduate level abstract algebra and number theory.Can you suggest some advanced books about equations like this ? – Frknx Mar 12 '15 at 21:10

2 Answers2

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$$ y^2 + 4 = x^3 + 27 = (x+3)(x^2 - 3 x + 9)$$ If $x$ were even, we would have $y^2 \equiv 3 \pmod 4,$ impossible.

So, $x$ is odd, $y$ is even. Note that $y^2 + 4 $ is divisible by $4.$ Since $x^2 - 3 x + 9$ is odd, this means $x+3$ is divisible by $4.$ That is $x \equiv 1 \pmod 4.$

As a result, since $-3 \equiv 1 \pmod 4,$ we have $-3 x \equiv 1 \pmod 4,$ so $$ x^2 - 3 x + 9 \equiv 3 \pmod 4. $$

Finally, this means that there is some prime $q \equiv 3 \pmod 4$ such that $q | (x^2 - 3 x + 9).$

The contradiction comes because that means $q |(y^2 + 4),$ this being impossible. Indeed, the lemma is that, given prime $q \equiv 3 \pmod 4,$ if $q |(u^2 + v^2),$ then both $q|u$ and $q |v.$ In this case, we have $v^2 = 4$ and $v = 2,$ and we cannot have $q | 2.$

$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$

Will Jagy
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  • Very nice. An added comment on the lemma: $-1$ is not a quadratic residue mod $q$ if $q \equiv 3 (\operatorname{mod} 4)$, so $k$ is a quadratic residue if and only if $-k$ is not. Therefore $q\mid (u^2+v^2)$ implies $q \mid u$ and $q \mid v$. – aes Mar 13 '15 at 09:36
  • @aes, from time to time I put the full lemma, no-one ever seems to care; given $\Delta = b^2 - 4 a c$ and odd prime $q$ that does not divide $\Delta,$ also $(\Delta |q) = -1,$ whenever $q$ divides $ax^2 + b xy + c y^2$ then $q$ divides both $x$ and $y.$ – Will Jagy Mar 13 '15 at 18:50
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If you consider the equation mod 4, we have $y^2-x^3 \equiv 3 \bmod 4$. Since $y^2 \equiv (0,1)$ and $x^3 \equiv (0,1,3)$, we must have $y$ even and $x\equiv 1 \bmod 4$.

Then setting $y = 2m$ and $x=4k+1$, we can rewrite: $$ \begin{align} (2m)^2 &= (4k+1)^3+23\\ 4m^2 &= 64k^3+48k^2+12k+1+23\\ m^2 &= 16k^3+12k^2+3k+6\\ \end{align}$$

... and that doesn't obviously get us much further forward.

So I tried a little hunting around for other modular bases - $96$ is relatively sparse for solutions, and that gave me a chance to brute-force my way to a non-solution... $x=4489, y=300763$. (And managed to fool myself, even though I knew that $y$ needed to be even...)

Having that near-miss solution might provide some way forward for either finding real solutions or closing the door...

Joffan
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    your values are simply $67^2$ and $67^3,$ so the cube of the first is the square of the second is $67^6.$ What is missing is the added 23 – Will Jagy Mar 12 '15 at 22:05
  • Yes... my mistake, I realized about 15 minutes after I first posted. – Joffan Mar 12 '15 at 22:55