# Still another diophantine equation

Can any of you guys provide a hint for thew following exercise?

Exercise. There is no $3$-tuple $(x,y,z) \in \mathbb{Z}^{3}$ such that $x^{10}+y^{10} = z^{10}+23$.

Thanks a lot for your insightful replies.

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## 1 Answer

The first thing to do is to check for local obstructions, that is some prime modulus $p$ where the equation is impossible modulo $p$ (sometimes $p^2$, $p^3$, etc.).

In this particular case, a tempting modulus to try is $11$, since $x^{10}$ is always $0$ or $1$ mod $11$. Unfortunately, this equation is feasible mod $11$ as $$1^{10} + 1^{10} \equiv 1^{10} + 23 \pmod{11}.$$ Another modulus worth trying is $5^2$, since $x^{10}$ is always $0,\pm1$ mod $25$. Unforunately, there is still a solution mod $25$.

Can you think of another modulus that places strong constraints on $x^{10}$? Hint: you will want the order of the multiplicative group $(\mathbb Z/p\mathbb Z)^\times$ to have a large common factor with $10$.

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Found an "appropriate" modulus (31). Nevertheless, I must confess that I didn't really get the "to have a large common factor with 10" part. Thanks a bunch... – absalon Jul 21 '12 at 1:31
@absalon The idea is that the non-zero values taken by $x^{d}$ are the $(d,p-1)$th powers. The larger $(d,p-1)$ is, the rarer those values will be, which is helpful to your cause. In the case that $d \mid p-1$ this is as good as it gets. $x^{10}$ is always $0,1,5,25$ mod $31$. – Erick Wong Jul 21 '12 at 4:41