find all the integral solution for $y^2 + 31 = x^3$

Question

Find all the integral solution for $y^2 + 31 = x^3$. I am reading Ireland's book 'a classical introduction to modern number theory', this is one of the exercises. The hint behind the book is $y^2 + 4 = x^3 - 27$, I don't know how to use it. I use python tried below $1000$ there is no solution. And also there is no prime $p$ below $10000$ such that $y^2 + 31 = x^3 (\text{mod} p)$ not solvable. Thank you

This is Mordell's equation $y^2=x^3+k$. There are several posts and solutions here, e.g. this one, or this one. Concerning the hint, see the proof of Theorem $2.2$ and $2.3$ in Conrad's notes. Also here. — Dietrich Burde, Apr 01 '21 at 08:32
@DietrichBurde sorry is there a link where can I find Conrad’s notes? — no lemon no melon, Apr 01 '21 at 11:41
@nolemonnomelon Sure, see my answer here, or the first comment above. By the way, there is no integer solution of $y^2=x^3-31$. — Dietrich Burde, Apr 01 '21 at 11:53
@DietrichBurde How did you show that this is unsolvable? The method that Conrad adopts in his notes when $k$ is odd for $y^2+k=x^3$ is based on the fact that $k\equiv1\mod4$ and hence $x$ can't be even. But that logic doesn't work here. — Martund, May 08 '21 at 06:52
@Martund There are so many proofs given at this site. For a reference, see for example the table and the references here, where $31$ is listed as having no solution. Even if Conrad's notes may not cover the case, it has been done here. The question $y^2=x^3+k$ reappears here, for each singe $k$ a new question. I find this too much. — Dietrich Burde, May 08 '21 at 07:50
@DietrichBurde, The Apostol reference that the OEIS site gives again proves only that special case which I told you above. See this screenshot that I just took from there. It doesn't apply in our case. — Martund, May 08 '21 at 11:21
@Martund I found the solution of this problem in "introductory algebraic number theory" by Saban Alaca page 399 Theorem 14.2.5 . The key point is h(-31) = 3 and prime ideal over 2 is not principle — Holybear, Jul 06 '21 at 08:15
@Servaes Thank you. I will write the solution when I get home:) — Holybear, Jul 29 '21 at 02:36

score 2 · Answer 1 · edited Jan 29 '22 at 20:18

I found the solution of this problem in Introductory Algebraic Number Theory by Saban Alaca, page 399, Theorem 14.2.5. Here is the solution.

First we prove $x$ is odd impossible. If $x$ is odd, then $y$ is even, by $y^2+4=x^3-27=(x-3)(x^2+3x+9)$, $y^2+4=4(\frac{y}{2}+1)^2$ then any odd prime $p$ divides $y^2+4$ must have $p \equiv 1 \bmod 4$. If $x \equiv 1 \bmod4$, then $y^2 \equiv 2 \bmod4$, contradiction. If $x \equiv 3 \bmod4$, then $x^2 + 3x + 9 \equiv -1 \bmod4$ and greater than 1. Hence there is a prime $q \equiv -1 \bmod4$ divides $x^2+3x+9$, this is a contradiction to we have known that any odd prime $p$ divides $y^2+4$ must have $p \equiv 1 \bmod4$.

Second we prove $x$ is even impossible. Let $K = Q(\sqrt{-31})$. Consider prime 2 in $O_K$. $$2O_K = (2, \frac{3+\sqrt{-31}}{2})(2, \frac{3-\sqrt{-31}}{2}).$$ $$(\frac{y+\sqrt{-31}}{2})(\frac{y-\sqrt{-31}}{2}) = (2, \frac{3+\sqrt{-31}}{2})(2, \frac{3-\sqrt{-31}}{2})(\frac{x}{2})^3.$$ We can prove ideal $(2, \frac{3+\sqrt{-31}}{2})$ is not principal by considering its ideal norm. Also the ideals $(\frac{y+\sqrt{-31}}{2})$ and $(\frac{y-\sqrt{-31}}{2})$ are coprime since the common prime ideal divides them is $(\sqrt{-31})$, which is impossible. Then by $(\frac{y+\sqrt{-31}}{2})(\frac{y-\sqrt{-31}}{2}) = (2, \frac{3+\sqrt{-31}}{2})(2, \frac{3-\sqrt{-31}}{2})(\frac{x}{2})^3$ we know there is an ideal $A$ which has $(\frac{y+\sqrt{-31}}{2})=(2, \frac{3+\sqrt{-31}}{2})A^3$ or $(\frac{y-\sqrt{-31}}{2})=(2, \frac{3+\sqrt{-31}}{2})A^3$. By class number of $K$ equals 3, $A^3$ must be principal. Then $(2, \frac{3+\sqrt{-31}}{2})$ is principal. But we already use ideal norm shows it's not.

find all the integral solution for $y^2 + 31 = x^3$

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