x^5+y^5+z^5=2004

Bojan Basic

Prove (or disprove) that $x^5+y^5+z^5=2004$ has no integer solutions.

**Posted:** Mon Mar 14, 2005 3:57 pm

djimenez

**Posted:** Tue Mar 15, 2005 12:02 pm

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Mildorf

Under mod 25, the only possible 5th power residues are $0, 1, 7, 18$ , and $24$ . The unique way to add three and get 4 is where $x$ , $y$ , and $z$ are all of the form $5k + 3$ . It seems that this would make it possible to argue that the spacing of 5th powers is too sparse if we must increment by 5, but the rationals are dense - we will need more than a few small calculations.

**Posted:** Wed Mar 23, 2005 7:45 pm

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It must be possible because it's almost impossible to prove it's impossible.

al.M.V.

I do think that something like finding all solutions to x ⁵ + y ⁵ + z ⁵ = 2005 could be on some MO, but then this one is much easier... Mr. Green

**Posted:** Sat Mar 26, 2005 11:48 am

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