<s>[INST] Solve $a^5 - b^5 - c^5 = 30abc$ and $a^2 = 2(b + c)$ over positive integers. I have found this system of equations in a Romanian Mathematical magazine:
Solve over positive integers:
$a^5 - b^5 - c^5 = 30abc$ and $a^2 = 2(b+c)$. I tried multiplying the first equation with an $a$ and then substituting $a^6$ with $8(b+c)^3$, but then it becomes very messy. I tried some factoring, but couldn't get rid of the others $a$. Do you have any other ideas?
 [/INST] By Fermat's little theorem $a^5 \equiv a \pmod {5}$. And since $a^5 - a = a(a^2 + 1)(a + 1)(a - 1)$, $a^5 - a \equiv 0 \pmod 3$, $\pmod 2$. Since $2,3,5$ are all coprime, this yields $a^5 \equiv a \pmod {30}$.
If $a-b-c = 0$, then $a^2 = 2(b+c) = 2a \implies a = 2$. $(a,b,c) = (2,1,1)$ is the only case to check given the condition $a^2=2(b+c)$, which is false.
Now let $a>b≥c$, or otherwise the left-hand side becomes negative or zero while the RHS is always positive. From the above, $a-b-c ≥ 30 \implies a-30≥b+c$. So $a-30≥a^2/2$, which is never true as $\Delta < 0$.
Hence there are no positive integer solutions.
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