 Solution

Choose any pair $(a,b)$ from part (i), and put $A=2^a{3}^b$. The divisibilities of the exponents makepositive integers. Their powers areThus $x^3+y^5=z^7$, with all three powers nonzero and distinct. For example, take $a=84+105t$, $b=90+105t$ for every nonnegative integer $t$. Then $A$ strictly increases, giving infinitely many distinct solutions. The construction uses modular congruences to turn the elementary identity $A+2A=3A$ into three prescribed perfect powers.