TITLE: Prove that the equation $x^2-y^{10}+z^5=6$ has no integer solutions
QUESTION [8 upvotes]: I have a nice diophantine equation which I tried to solve since march but no solution.
Tried modulo 11, tried to write it in some way to figure out a solution...
I posted this a few months ago, but it was removed.
The problem: Prove that the equation $x^2-y^{10}+z^5=6$ has no integer solutions( positive, negative). Maybe another idea if YOU have or a solution! I would appreciate it very much!
EDIT, Will Jagy: If $$ x \equiv 4,7;  \; \; \; y \equiv 0; \; \; \; z \equiv 1,3,4,5,9 \pmod {11}  $$ THEN
$$x^2-y^{10}+z^5 \equiv 6 \pmod {11}$$ which you can just check. At the level of school mathematics, there are two tricks I know for this type of problem. Neither works, but other people have found dozens to hundreds of tricks that I don't know. I also know of published problems with no answer because the book screwed up. Apparently the answer will be published in late September.

REPLY [2 votes]: Partial answer:
According to Fermat's little theorem, for any prime $p$ and any integer a, $a^p\equiv a \pmod p$.
Working mod 2:
\begin{align}
x^2-y^{10}+z^5 &\equiv 6 \pmod2\\
x-y+z &\equiv 0\pmod2\\
y &\equiv x+z\pmod 2
\end{align}
Working mod 3:
\begin{align}
x^2-y^{10}+z^5 &\equiv 6 \pmod3\\
x^2-(y^3)^3y+z^3z^2 &\equiv 0 \pmod 3\\
z &\equiv y^2-x^2 \pmod3
\end{align}
Now, any square is 0 or 1 mod 3. So if $3\nmid x$ and $3\nmid y$ then $3 \mid z$, or $3 \mid xyz$.
Suppose that $g$ is the greatest common divisor of $x$, $y$ and $z$. Then $g^2 \mid 6$, implies $g=1$.