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The question is to find all integer solutions to the equation $$ x^2+y^2=z^3+1. $$ This equation obviously has infinitely many integer solutions (take, for example, $(x,y,z)=(1,u^3,u^2)$ for any integer $u$) but the question is to describe all integer solutions. A parametric representation (like this answer to Solve for integers $x, y$ and $z$: $x^2 + y^2 = z^3.$ for equation $x^2+y^2=z^3$) would of course be ideal, but, in general, by "describe" I mean any algorithm that explicitly produces all solutions without performing any search by trial and error. See my previous question Solve in integers: $y(x^2+1)=z^2+1$ for an example of such algorithm. After that equation has been solved, the equation in question is one of the smallest/simplest ones for which I am not aware about any reasonable method/algorithm to describe all solutions, hence the question.

Update 04.02.2022: After the equations $y^2+z^2 = 2x^2 \pm 1 $ has been solved by individ in the comment to this question, we can now explicitly describe the solution sets to all equations of size $h \leq 17$, where $h$ is defined here What is the smallest unsolved diophantine equation? , except for the following equations of size $17$:

$$ y^2+z^2 = x^3+1 $$ $$ y^2+z^2 = x^3-1 $$ $$ y(x^2-y)=z^2-1 $$

All these equations have infinitely many integer solutions, but the question is to find all solutions. If you know any explicit way to describe all solutions to any of these equations, please answer.

Bogdan Grechuk
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    You linked to an MSE question, but appeared to intend to link to the answer; I edited accordingly. I would also encourage you to word your questions as questions or requests, rather than as imperatives (e.g., "How does one solve this equation?" rather than "Solve this equation"). – LSpice Feb 01 '22 at 21:30
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    Ok, title reformulated as a question, as suggested. – Bogdan Grechuk Feb 01 '22 at 21:52
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    If a prime $p$ divides $z^3+1$, then $z=kp-1$ for some integer $k$. Conversely, if $z=kp-1$ with $p\ne3$ prime, then $z^3+1=kp(k^2p^2-3kp+3)$ is a multiple of $p$ but not of $p^2$. So the question of whether there's a solution is very closely related to the question of whether $z+1$ is a multiple of some prime $p\equiv3\bmod4$. – Gerry Myerson Feb 01 '22 at 22:58
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    Hm, a prime $7$ divides $3^3+1=28$, but $3$ is not of the form $7k-1$ for some integer $k$. – Bogdan Grechuk Feb 01 '22 at 23:07
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    What Bogdan said — the fact that $z^3+1$ only has the real root $z=-1$ over $\mathbb{R}$ doesn't mean that it can't have three roots in $\mathbb{Z}/p\mathbb{Z}$ for some $p$; it depends on whether $-3$ is a quadratic residue mod $p$ or not,. – Steven Stadnicki Feb 01 '22 at 23:15
  • For such a small equation, higher degrees can also be mentioned. https://artofproblemsolving.com/community/c3046h1186088_one_simple_formula https://artofproblemsolving.com/community/c3046h1047806__ – individ Feb 02 '22 at 05:17
  • By using say the quadratic sieve to factor $z^3+1$ for $z=1,\dots,N$, one can enumerate all solutions with $|z| \leq N$ in time $O(N^{1+o(1)})$, which is best possible up to $o(1)$ errors given that the number of solutions with $|z| \leq N$ is indeed expected to be $O(N^{1+o(1)})$. – Terry Tao Feb 04 '22 at 16:12
  • (One also needs a fast way to write any prime $p=1 \hbox{ mod } 4$ as the sum of two squares, but such algorithms exist, see e.g., https://en.wikipedia.org/wiki/Fermat%27s_theorem_on_sums_of_two_squares#Algorithm ) – Terry Tao Feb 04 '22 at 16:16
  • For an equation of the form. $aX^2+bY^2=cZ^2+q$ https://mathoverflow.net/questions/263153/indefinite-quadratic-form-universal-over-negative-integers – individ Feb 04 '22 at 16:27
  • @Terry Tao: Thank you for the comments. First, I know how to fast represent number as a sum of squares, but not sure how to efficiently find ALL such representations. Second, if we just try all $z=1,2,\dots$, does this really counts as a "solution", even if the algorithm is efficient? Do you know any consistent and non-trivial definition saying that "let us call the equation solved if there is an algorithm for listing all its solutions up to a given bound N in near optimal time" but in more formal way? – Bogdan Grechuk Feb 04 '22 at 17:08
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    An equation of this kind is almost linear. $$x^2+y^2=2z^2+q$$ It is enough to factor the number $2k^2-q$ and the solutions can be written down. $$x=\frac{2k^2-q}{2t}+\frac{t}{2}-2k$$ $$y=\frac{2k^2-q}{2t}-\frac{t}{2}$$ $$z=\frac{2k^2-q}{2t}+\frac{t}{2}-k$$ – individ Feb 04 '22 at 17:11
  • Thank you! But what is the proof that this formula indeed represents ALL solutions? – Bogdan Grechuk Feb 04 '22 at 17:53
  • Ok, I now found the proof and will mark this equation as solved, thank you again. But my proof is by direct algebraic manipulations and gives no hint how to derive these formulas for this and similar equations, so more details would still be appreciated. – Bogdan Grechuk Feb 04 '22 at 18:38
  • @BogdanGrechuk Once one can factor a number $n$ into primes, and can represent every prime $p = 1 \hbox{ mod } 4$ as the sum of two squares, one has factored $n$ into Gaussian primes, and from this one can easily obtain all representations $n=x^2+y^2 = (x+iy)(x-iy)$ as sums of two squares since the Gaussian integers have unique factorisation. I think the definition of "solved" you proposed is already quite formal; in fact more so than any alternative notion of "solved" that I can think of (excepting perhaps that of a birational parameterisation, which is not expected in general). – Terry Tao Feb 04 '22 at 18:56
  • Ok, if we accept near optimal algorithm as an answer, then it is left to prove its optimality, that is, prove that the number of solutions with $|z|\leq N$ is indeed $O(N^{1+o(1)})$. – Bogdan Grechuk Feb 05 '22 at 08:03

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