Solve $x^3=y^2-7$?

Question

How to solve $x^3=y^2-7$?

I can see that $x$ cannot be even. Since $x$ even, implies $y$ odd, hence LHS is congruent $0 \mod 8$ and RHS $2 \mod 8$. Contradiction.

So $x$ is odd and $y$ is even. Maybe helpful: $x^3+8 = (x+2)(x^2-2x+4)$ and the RHS is thus congruent $3 \mod 4$. But I don't know how to proceed. Can someone help me? Thanks in advance.

If you have a prime $q \equiv 3 \pmod 4,$ and $u^2 + v^2 \equiv 0 \pmod q,$ what does that say about $u,v?$ — Will Jagy, Jun 01 '17 at 18:20
Uhm, I don't know immediately. I know that if $q=u^2+v^2$, then there are no solutions since $q \equiv 3 \mod 4$. — bob, Jun 01 '17 at 18:29
Oh I know, then $-1$ is a quadratic residu modulo $q$. So you get a contradiction if $q \equiv 3 \mod 4$ — bob, Jun 01 '17 at 18:31
@WillJagy How can this help me to continue? Sorry, but I don't see how to proceed. — bob, Jun 01 '17 at 18:51
@TobErnack Okay thank you, but why is $y+\sqrt{7} = (a+b\sqrt{7})^3$? I thought $y+\sqrt{7} = u (a + b \sqrt{7})^3$ for some unit $u$ in $\mathbb{Z}[\sqrt{7}]$, and there are infinitely many such units in $\mathbb{Z}[\sqrt{7}]$, namely powers of the fundamental one $8+3\sqrt{7}$. So I don't know why you can take $u=1$? Or am I wrong? — bob, Jun 01 '17 at 19:33
@bob The solutions in Conrad's notes (linked in the duplicate) are very simple. Have you seen it? — Dietrich Burde, Jun 01 '17 at 19:51
Oh really sorry! I didn't looked at that one! My mistake. Many thanks! — bob, Jun 01 '17 at 19:54
I put a complete proof of the thing about primes and the discriminant, in your case $u^2 + v^2$ and primes $q \equiv 3 \pmod 4,$ anyway https://math.stackexchange.com/questions/2291070/prime-divisors-of-k2k12/2291224#2291224 — Will Jagy, Jun 01 '17 at 23:57
There are an infinite number of rational solutions, perhaps you are looking for whole number solutions? — , Jun 02 '17 at 01:01

Will Jagy · Accepted Answer · 2017-06-02T00:12:17.243

Look at the prime factorizations below, these are $u^2 + 1$ for integers $u.$ What do you see, or, more to the point, what do you not see? Can you prove that?

see general phenomenon, with proof, my answer at Prime divisors of $k^2+(k+1)^2$

The thing you do not see in the output below is any prime factors $q$ with $q \equiv 3 \pmod 4.$ Those would be $3,7,11,19,23,31,43,...$ Not there. The odd prime factors are $1 \pmod 4,$ as $5,13,17,29,37,41,...$

============================

  u  0  u^2 + 1  1 =  1 
  u  1  u^2 + 1  2 = 2
  u  2  u^2 + 1  5 = 5
  u  3  u^2 + 1  10 = 2 * 5
  u  4  u^2 + 1  17 = 17
  u  5  u^2 + 1  26 = 2 * 13
  u  6  u^2 + 1  37 = 37
  u  7  u^2 + 1  50 = 2 * 5^2
  u  8  u^2 + 1  65 = 5 * 13
  u  9  u^2 + 1  82 = 2 * 41
  u  10  u^2 + 1  101 = 101
  u  11  u^2 + 1  122 = 2 * 61
  u  12  u^2 + 1  145 = 5 * 29
  u  13  u^2 + 1  170 = 2 * 5 * 17
  u  14  u^2 + 1  197 = 197
  u  15  u^2 + 1  226 = 2 * 113
  u  16  u^2 + 1  257 = 257
  u  17  u^2 + 1  290 = 2 * 5 * 29
  u  18  u^2 + 1  325 = 5^2 * 13
  u  19  u^2 + 1  362 = 2 * 181
  u  20  u^2 + 1  401 = 401
  u  21  u^2 + 1  442 = 2 * 13 * 17
  u  22  u^2 + 1  485 = 5 * 97
  u  23  u^2 + 1  530 = 2 * 5 * 53
  u  24  u^2 + 1  577 = 577
  u  25  u^2 + 1  626 = 2 * 313
  u  26  u^2 + 1  677 = 677
  u  27  u^2 + 1  730 = 2 * 5 * 73
  u  28  u^2 + 1  785 = 5 * 157
  u  29  u^2 + 1  842 = 2 * 421
  u  30  u^2 + 1  901 = 17 * 53

===========================================

Solve $x^3=y^2-7$?

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