Are there any non-zero integer solutions to the equation $x^5+y^5=7z^5$? I am unsure how to approach this.
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4it is homogeneous, you may demand $\gcd(x,y,z) = 1.$ Then try $\pmod {11},$ that may be enough. – Will Jagy Nov 22 '15 at 01:45
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@WillJagy, how did you know that $11$ would work? – lhf Nov 22 '15 at 01:53
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@lhf, the thing is restrictive because 5 divides 11-1; the fifth power map is five to one in geeral, values $0,1,-1.$ The bad news is that this does not suffice to finish the question. I would like to know the source of the question. ...We could have $$ x^5 \equiv 1, ; ; y^5 \equiv -1, ; ; z^5 \equiv 0 \pmod {11} $$ – Will Jagy Nov 22 '15 at 01:59
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Yes I had already observed that small congruences don't rule out much. (There are also many candidates mod 31 and 41 it appears.) On the other hand, a computer search rules out solutions with $\max(|x|,|y|,|z|)<10^4$ – pre-kidney Nov 22 '15 at 07:35
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This question has been solved on Mathoverflow.
The integer points are given by $(x,-x,0)$ for $x\in \mathbb Z$.
The trick is to reduce the equation to a form that has been studied more extensively. For any solution $(x,y,z)$ to the equation with $z\not=0$, let $X=-xy/z^2$ and $Y=\frac{x^5}{z^5}-\frac{7}{2}$. Then $Y^2-\frac{49}{4}=X^5$, which is a hyperelliptic curve, and there are techniques for finding all rational solutions to such an equation.
pre-kidney
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