I have a problem said that : solve in $ \mathbb{Z} $ the equation $$ y^{2}+ x^{3}= 9 $$ I tried this way. $ y^{2}+ x^{3}= 9 \iff y= \pm \sqrt{9-x^2} $ Then $9-x^{2} \geq 0 $ so we get $(3, 0) $ and $ x^{3}=(y-3)(y+3) $ and get $(0, 3)\& (0, -3) $. My question is how I can get the other pairs that satisfy this equation?. Thanks

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    $\begingroup$ Well, and $(x,y)=(2,\pm1)$. $\endgroup$ – Mike Bennett Sep 8 '15 at 15:03
  • $\begingroup$ @MikeBennett yes this two pairs satisfy this equation. How we get them? $\endgroup$ – user257567 Sep 8 '15 at 15:08
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    $\begingroup$ check my answer, i get another solution. $\endgroup$ – Salihcyilmaz Sep 8 '15 at 15:22
  • $\begingroup$ @Salihcyilmaz nice solution thanks. I will study it carefully $\endgroup$ – user257567 Sep 8 '15 at 15:29
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    $\begingroup$ This is a Bachet-Mordell equation; for references see here. $\endgroup$ – Dietrich Burde Sep 8 '15 at 19:22

I get another solution $(-3,6)$. I don't know how many solution there are, but if i found more i will edit this post. Here is my approach:

If $x\geq0$ then we have $0\geq x\geq2.08$ since $y^2 \geq 0$ must be satisfied. Then by trial and error, we substitute $x=0, x=1, x=2$ one by one to see if there is corresponding $y$. In which we get the solutions you already gave, namely $(0,3)$ and $(0,-3)$ and two others $(2,-1)$ and $(2,1)$.

Now let us change our focus on next condition, where we assume $x<0$. Here we have $y^{2}= 9 - x^{3}$. Here there may be solutions for some (x,y) since $y^2\geq 0 $ is satisfied.

Now for simplicity, i will change the equation by letting $x=-x$, then we have $y^{2}= 9 + x^{3}$, for $x > 0 $. Carry constant 9 to other side, then we have $y^{2}- 9 = x^{3}$ which implies $(y-3)(y+3) = x^{3}$. Now i assumed that i can find a solution such that $$(y-3)^2=(y+3)$$ which could gave us the number we seek. Namely if such $y$ exist we have $(y-3)^3= x^3$. Solving the equation $$(y-3)^2=(y+3)$$ we find $y=6$, and $y=1$. Substituting back in the equations we get solutions $(-3,6)$ and $(2,1)$. But $|y|\geq 3$ and $x < 0 $ must be satisfied , then we find only one solution which is$$(-3,6)$$

  • $\begingroup$ Thousands of typos corrected $\endgroup$ – Salihcyilmaz Sep 8 '15 at 15:35
  • $\begingroup$ explain that please $\endgroup$ – user257567 Sep 8 '15 at 16:21
  • $\begingroup$ exactly what point you didn't understand? $\endgroup$ – Salihcyilmaz Sep 8 '15 at 16:22
  • $\begingroup$ sir you say above that there is thousands of pairs correct that satisfy the equation. I hope that you explain that? $\endgroup$ – user257567 Sep 8 '15 at 16:57
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    $\begingroup$ There were some typos. For instance, there was written $(6,-3)$ for solution in which i mean $x=-3$ and $y=6$. But (6,-3) does not implies $x=-3$ and $y=6$. Rather it implies that $x=6$ and $y=-3$, so i corrected it and write $(−3,6)$. Things like that. My brain plays similar games a lot to me. What i thought and what i write may become different sometimes. Things like that. Now the solution is corrected, there should not be any problem now. $\endgroup$ – Salihcyilmaz Sep 8 '15 at 17:02

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