# Variation of Fermat's Last Theorem for n = 5

It is known that there is no solution in N to

x^5 + y^5 = z^5 (FLT),

but there is a solution to

x^5 + y^5 + u^5 + v^5 = w^5

This was the first counterexample to Euler's conjecture.

What about sum of three fifth powers

x^5 + y^5 + z^5 = t^5

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Many solutions exist, but all the known ones are trivial, e.g., $(x,y,z,t)=(17,42,-17,42)$. –  Gerry Myerson Aug 9 '11 at 12:51
It's also been known for some time that there are number fields $K$, such as ${\bf Q}(i)$, over which there are infinitely many nontrivial solutions. The nice way to see this is to intersect the quintic projective surface $w^5+x^5+y^5+z^5=0$ with the plane $w+x+y+z=0$: the resulting quintic curve contains the three lines $w+x=y+z=0$, $w+y=x+z=0$, $w+z=x+y=0$, and thus a residual conic which contains nontrivial rational points. Alas this conic has no real points, let alone rational ones, so a necessary condition on $K$ for this construction to work over $K$ is that $K$ be totally imaginary. –  Noam D. Elkies Aug 9 '11 at 17:18
Thank you for your answer Noam. Could you imagine there is some astronomical solution like the one you gave for the fourth powers or now it is more reasonable to look for a non-existence proof? –  Martin Aug 9 '11 at 17:58
@Martin: you're welcome. I'm afraid the short answer to your next question is that it's exceedingly unlikely that there's a nontrivial rational solution (unlike the fourth-power case, where it was actually likely a priori that there would be infinitely many primitive solutions), but we have no techniques to prove such a result. I might post an answer later detailing why, but have some things to tend to first before dinner... –  Noam D. Elkies Aug 9 '11 at 18:15