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We are learning about the the Pythagorean Theorem in class. It says that $a^2+b^2 = c^2$. My homework problem says the following:

Find integers $a,b$ and $c$ such that $a^3+b^3 = c^3$.

How do I solve this equation?

I've been starting with $(3,4,5), (4,4,5)$ etc. Basically I am starting Pythogrean triples.

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  • $\begingroup$ Are you sure it says that? There are not such numbers (apart from treivial ones). $\endgroup$ Commented Mar 14, 2013 at 15:42
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    $\begingroup$ By trial and error. It has only very trivial solutions. $\endgroup$ Commented Mar 14, 2013 at 15:42
  • $\begingroup$ @BrianM.Scott right, just realized that and edited my comment. $\endgroup$ Commented Mar 14, 2013 at 15:43
  • $\begingroup$ How trivial are we talking about? $\endgroup$
    – Valtteri
    Commented Mar 14, 2013 at 15:43
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    $\begingroup$ @Valtteri: To answer that would pretty much give away the solution. I’m trying to think of a good hint. $\endgroup$ Commented Mar 14, 2013 at 15:44

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HINT: It is known that the equation $a^3+b^3=c^3$ has no solutions in which $a,b$, and $c$ are all positive integers. It does have infinitely many solutions in integers, but all of them have one of two or three basic forms (depending on how you count) and are rather trivial.

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    $\begingroup$ It is $(a,-a,0)$? $\endgroup$ Commented Mar 14, 2013 at 15:56
  • $\begingroup$ @loxststudent: Oops! You found a second form; very good. Yes, that works, but there’s another set of solutions with the $0$ elsewhere. $\endgroup$ Commented Mar 14, 2013 at 16:10
  • $\begingroup$ i think it would be $(a,0,a)$? $\endgroup$ Commented Mar 14, 2013 at 16:11
  • $\begingroup$ @loxststudent: That’s right; also $\langle 0,a,a\rangle$, if you want to count that case separately from $\langle a,0,a\rangle$. $\endgroup$ Commented Mar 14, 2013 at 16:14
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If $a=0,b^2=c^2\implies b=\pm c$ and $b^3=c^3\implies b=c$

$\implies b=c$

So, $(0,b,b)$ is a solution

Similarly, $(b,0,b)$ is a solution

If $c\ne0, \left(\frac ac\right)^2+\left(\frac bc\right)^2=1$

$\implies \frac ac<1\implies \left(\frac ac\right)^2>\left(\frac ac\right)^3$

Similarly, $\left(\frac bc\right)^2>\left(\frac bc\right)^3$

$\implies \left(\frac ac\right)^3+\left(\frac bc\right)^3<1$

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