Given that $a$, $b$, $c$ and $d$ are distinct integers. Is $$a^3+b^3+c^3=d^3$$ possible?
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4$ (9t^3 + 1)^3 + (9t^4)^3 + (-9t^4 - 3t)^3 = 1$, for example. – lulu Apr 06 '17 at 15:48
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1See e.g. this page, section 2. – Robert Israel Apr 06 '17 at 15:54
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1http://math.stackexchange.com/questions/469151/find-all-integer-solutions-to-diophantine-equation-x3y3z3-w3/776918#776918 – individ Apr 06 '17 at 16:20
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There exist infinitely many solutions $(a,b,c,d), a,b,c,d \in Z$ such that $a^3 + b^3 + c^3 = d^3$. specifically, just consider the 4-tuples of the form $(a,0,0,a)$. Note that the they trivially satisfy the equation. Hence proved.
Lelouch
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1@Someone I strongly suggest you amend your question something like this: > EDIT: Although I did not say so at first, I intend the choice of integers to be such that $abcd \neq 0$. – Mr. Brooks Apr 06 '17 at 22:41
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However, as I think more about this, Lelouch, your answer nevertheless fails to answer Someone's question as originally written, since "distinct integers" means $a \neq d$. – Mr. Brooks Apr 06 '17 at 22:44
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Oh, wait a minute, though, now I see that @MyGlasses's edit actually changes the meaning of the original question. Jeez! – Mr. Brooks Apr 06 '17 at 22:46
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Yeah, the previous question was trivially true. Thats all i posted here. – Lelouch Apr 07 '17 at 02:26