Show that $a+bi$ and $a-bi$ are coprime gaussian integers and there are integers $m$ and $n$ such that $a = m^3 - 3mn^2$ and $b = 3m^2n - n^3$.
For this question, i just proved $a \not\equiv b$ and c is odd.
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"And there are integers $m$ and $n$ such that $a=m^3-3 m n^2$ and $b=3 m^2 n-n^3$"
Actually every $m$ and $n$ will work, by noticing that $(m^3-3 m n^2)^2+(3 m^2 n-n^3)^2 = (m^2+n^2)^3$
For the first statement of the question, also by substituting the same variables and GCD algorithm for Gaussian integers, you can arrive easily that the two numbers are Coprimes.
Ahmad
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Very nice. $c=m^2+n^2$ – Zubin Mukerjee May 12 '17 at 10:46
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@Zubin This follows much more conceptually from unique factorization - se my answer. – Bill Dubuque May 12 '17 at 14:58
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This is an immediate consequence of the fact that Gaussian integers are a UFD. Then by unique factorization, with $\,w = a+bi,$ and $\,ww' = c^3$ and $w,w'$ coprime it follows that, up to unit factors, $w$ is a cube, so $\,w = (m+ni)^3 = m^3-3mn^2 + (3m^2n-n^3)i\,$ as claimed.
Remark $ $ This is the cubic analog of the well-known derivation of the parametrization of Pythagorean triples using unique factorization of Gaussian integers. Follow the link for further discussion, including Weil's discussion of its relation with Fermat's method of infinite descent.
Bill Dubuque
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