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Does any right triangle exist such that $a^3+b^3=c^3$? Does any triangle exist such that $a^3+b^3=c^3$?

I'm stuck on this problem; I tried applying the Pythagorean theorem in three dimensions, but in vain. Any tips?

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Blue
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rb3652
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    It would still need to obey the regular Pythagorean Theorem, too. Also, by Fermat's Last Theorem, $a,b,c$ are not all integers. – The Count Jun 27 '19 at 16:32
  • @TheCount Right; I actually used that to my advantage. I know (a/c)^2<(a/c)^3, and (b/c)^2<(b/c)^3. I added them to get 1>((a^3+b^3)/c^3), which simplifies to c^3>a^3+b^3. But I don't know where to go on from there. Sorry for the no latex. – rb3652 Jun 27 '19 at 16:34
  • I have a sneaking suspicion that no right triangle will obey this relation, and I am suspicious in general as well. Also, $(a/c)<1$ so $(a/c)^3<(a/c)^2$, I believe. – The Count Jun 27 '19 at 16:36
  • https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem – Seyed Jun 27 '19 at 17:47

2 Answers2

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We need $a\ne0$ and $b\ne0$. If it holds, then we have

$(a^3+b^3)^2=c^6=(a^2+b^2)^3$

$a^6+2a^3b^3+b^6=a^6+3a^4b^2+3a^2b^4+b^6$

$2ab=3a^2+3b^2$

$(a-b)^2+2a^2+2b^2=0$

$a=b=0$

CY Aries
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If $a^2+b^2=c^2,$ with $a,b,c$ positive, you'd have: $c>a,b>0$ and thus $ca^2>a^3$ and $cb^2>b^3.$ So you get:

$$c^3=c\cdot c^2=c(a^2+b^2)=ca^2+cb^2>a^3+b^3.$$

So $c^3>a^3+b^3.$

You can prove more generally for any triple $(a,b,c)$ of positive reals, there is at most one positive $n$ such that $a^n+b^n=c^n.$ If $a^n+b^n=c^n$ then $c>a,b>0$ and for $m>n$ you have:

$$\begin{align}c^m&=c^{m-n}\cdot c^n \\&=c^{m-n}(a^n+b^n)\\&=c^{m-n}a^n+c^{m-n}b^n\\ &>a^{m-n}a^n+b^{m-n}b^n\\ &=a^m+b^m.\end{align}$$

Thomas Andrews
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