B A problem involving the addition of two cubes

Charles Link

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Hypothetical situation: An algebra teacher was looking for examples of how the cubes of two integers might add to be the cube of a 3rd integer. Can you help the algebra teacher?
This post is meant to be a fun one. It is at the beginner level, because I think most people at the intermediate level would know what the solution is, but they might also find it entertaining as well. ## \\ ## A hypothetical situation: An algebra teacher observed that ## 3^2+4^2=5^2 ## and ## 5^2+12^2=13^2 ## and quite a number of others. The teacher wanted to give the class a little practice with some 3rd power arithmetic, and was looking for some integer examples where ##a^3+b^3=c^3 ##. Can you help the algebra teacher find a couple?
 

Charles Link

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Euler, 1753, but I don't know how he did it.
The first of many. I was thinking more on the lines of Fermat, but yes, I think Euler did it for the cubes. And the question that still remains unanswered, though many have their guesses, is whether Fermat had a proof for it.
 

Vanadium 50

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and was looking for some integer examples where a3+b3=c3 .
Huh? Doesn't Wiles' Theorem (formerly Fermat's Last Theorem) say there are no such triples?
 

Charles Link

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Huh? Doesn't Wiles' Theorem (formerly Fermat's Last Theorem) say there are no such triples?
That's why I made it a thread for beginners. I wanted to see if they knew that was the case. It was intended to be semi-educational.
 

Svein

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But there exists integers a, b, c and d such that [itex]a^{3}+b^{3}+c^{3}=d^{3} [/itex]. Example: [itex]3^{3}+4^{3}+5^{3}=6^{3} [/itex].
 

fresh_42

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There are even infinitely many solutions for ##a^3+b^3=c^3##.
 
But there exists integers a, b, c and d such that [itex]a^{3}+b^{3}+c^{3}=d^{3} [/itex]. Example: [itex]3^{3}+4^{3}+5^{3}=6^{3} [/itex].
So, we know that for a^2+b^2=c^2, we can write it as:
a=m^2-n^2,b=2mn and therefore c=m^2+n^2, and then we can write it for all positive integers. Can we do the same for [itex]a^{3}+b^{3}+c^{3}=d^{3} [/itex] ??
 

Vanadium 50

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Congratulations. You got me (and others) with your trick question.

Ha.
Ha.
Ha.
 

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