Somehow Fermat's Last Theorem but not really

Question

Remember the equation

$(a)^n$ + $ (b)^n$ $\ne$ $ (c)^n$.

Well then, what would happen if we allow the powers $\textit{n}$ to be three distinct odd primes such that

$ (a)^{p1}$ + $ (b)^{p2}$ = $ (c)^{p3}$.

When we do that, can we find three integers $\textit{a}$, $\textit{b}$ and $\textit{c}$ satisfying the above equation?

Answer 1

I'm just stealing Wojowu's answer

Let $r$ be an inverse of $p_3$ modulo $p_1p_2$ let $m=rp_3-1$ then a solution is $$(2^{m/p_1})^{p_1}+(2^{m/p_2})^{p_2} = (2^{(m+1)/p_3})^{p_3}$$

I don't know about the problem when we restrict to $a,b,c$ coprime, I'm quite sure there exist modifications making the problem as hard as Fermat last theorem.