A problem involving the addition of two cubes

[Charles Link]

TL;DR

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?

 

[fresh_42]

Euler, 1753, but I don't know how he did it.

 

[Charles Link]

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]

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]

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]

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

 

[fresh_42]

There are even infinitely many solutions for $a^3+b^3=c^3$.

 

[Svein]

There are even infinitely many solutions for $a^3+b^3=c^3$.

?

 

[fresh_42]

?

Read the OP carefully. All triples $(a,0,a)$, $(a,-a,0)$ are integer solutions!

 

[Svein]

Read the OP carefully. All triples $(a,0,a)$, $(a,-a,0)$ are integer solutions!

Of course! I just added the Fermat condition mentally - the integers should be >0!

 

[ali PMPAINT]

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

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 $a^{3}+b^{3}+c^{3}=d^{3}$ ??

 

[Vanadium 50]

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

Ha.
Ha.
Ha.