Prove that the equation $x^2-y^{10}+z^5=6$ has no integer solutions

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.

• It's a youth mathematics magazine from Romania:) – user85046 Sep 20 '13 at 19:46
• Why was the problem deleted before? – Will Jagy Sep 20 '13 at 22:47
• Hello! I thought nobody will return to this question! Well...at the end of this month. – user85046 Sep 21 '13 at 20:22
• CFG, in that case post that as an answer here once it appears. Meanwhile, you would have a better experience at artofproblemsolving.com and/or awesomemath.org – Will Jagy Sep 21 '13 at 20:30
• Avoid subjective titles, make them informative. – Pedro Tamaroff Sep 21 '13 at 20:32

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 $g=1$.