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Show that $a^2 + b^2 = c^3 $ has infinitely many solutions in $ \{ (a,b,c) \in \Bbb Z ^3 | (a,b)=1, (a,c)=1, (b,c)=1 \}$ .
Describe all these solutions.

I don't know how to approach this question. I'm thinking about something like $a^2+b^2=(a-bi)(a+bi)$, so we get $xy=c^3$ where $x,y \in \Bbb Z [i]$. I'm not sure it's the right way to do this.

user26857
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Kuba
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  • Do you know that $\mathbf{Z} [i]$ is a factorial ring etc ?... – Olórin Feb 15 '15 at 11:07
  • the solution can be written in the following way. http://www.artofproblemsolving.com/blog/109353 http://www.artofproblemsolving.com/blog/103853 http://www.artofproblemsolving.com/blog/103009 http://www.artofproblemsolving.com/blog/102042 – individ Feb 15 '15 at 11:13
  • Take any prime $p$ of the form $4k+1$, write it as $p = x^2 + y^2$ and let $a + ib = (x+iy)^3$. you will find $\gcd(a,b) = \gcd(a,p) = \gcd(b,p) = 1$ and $a^2 + b^2 = p^3$. – achille hui Feb 15 '15 at 12:03
  • Primes are a red herring, I think. Just let $(x+iy)^3=a+ib$, and see what happens. – Gerry Myerson Feb 15 '15 at 12:04
  • http://math.stackexchange.com/questions/885181/how-to-prove-c-b-2-3cb-x3-has-no-nonzero-integer-solutions – Bumblebee Feb 15 '15 at 13:00

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