= Solution
Choose any pair $(a,b)$ from part (i), and put $A=2^a3^b$. The divisibilities of the exponents make
$$
\boxed{x=2^{a/3}3^{b/3},\qquad
y=2^{(a+1)/5}3^{b/5},\qquad
z=2^{a/7}3^{(b+1)/7}}
$$
positive <integers>. Their powers are
$$
x^3=A,\qquad y^5=2A,\qquad z^7=3A.
$$
Thus $x^3+y^5=z^7$, 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 \b[infinitely many distinct solutions]. The construction uses <modular congruences> to turn the elementary identity $A+2A=3A$ into three prescribed perfect powers.
Back to article page
