Show that there are infinitely many integer solutions to the equation $x^3+y^5=z^7$

Question

Show that there are infinitely many integers such that $$x^3+y^5=z^7$$ and where $x^3,y^5$ and $z^7$ are all non-zero and distinct.

The hint suggests to look at solutions of simultaneous equation

\begin{eqnarray*} a \equiv 0 \mod 21 \qquad b \equiv 0 \mod 15 \\ a \equiv -1 \mod 5 \qquad b \equiv -1 \mod 7 \end{eqnarray*}

which we can directly read off the solutions $a=-21, b=-15$ which must be unique up to modulo $105$.

Then set $x=2^a3^b$, but I'm not really sure how this hint is useful in any way.

Answer 1

Hint From what you have got, see why you can write $$2^{84}3^{90}+2^{85}3^{90}=2^{84}3^{91}$$

Now can you see how to generate more solutions?

Answer 2

If $(a,b,c)$ is a solution to your equation, then so is $(k^{35}a,k^{21}b,k^{15}c)$ for every integer $k$.

Now you only have to find one solution with the required conditions, which, considering your earlier efforts, shouldn't be too hard.