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.

  • $\begingroup$ Have you considered Fermat's little theorem (or Fermat-Euler)? $\endgroup$ – Dan Robertson Jun 21 '15 at 13:47
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    $\begingroup$ Small hint: Write $a=84$, $b=90$ rather than negative numbers. $\endgroup$ – wythagoras Jun 21 '15 at 13:49
  • $\begingroup$ Right, then $x$ is an integer, and moreover there are infinitely many integers of the form $x=2^a 3^b$, so maybe each of these integers are solution to the equation. But how do I show this? Putting $y,z$ in the same form does not help, since $z^7$ grows much quicker than $x^3$ or $y^5$, and addition operation does not compensate for that gap. $\endgroup$ – user160738 Jun 21 '15 at 13:57

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?

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  • $\begingroup$ Wow, I find that magical. Then I guess I can proceed as what Jef Laga said - thanks! $\endgroup$ – user160738 Jun 21 '15 at 14:10
  • $\begingroup$ You can multiply throughout by $k^{105}$ for any integer $k$ to get more solutions. $\endgroup$ – Macavity Jun 21 '15 at 14:13

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.

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