<s>[INST] Proof that $x^3+y^4=z^{31}$ has infinitely many solutions This is a question from RMO 2015.
Show that there are infinitely many triples (x,y,z) of integers such that $x^3+y^4=z^{31}.$
This is how I did my proof:
Suppose $z=0,$ which is possible because $0$ is an integer. Then $x^3+y^4=0 \Rightarrow y^4=-x^3.$ Now, suppose $y$ is a perfect cube such that it is of the form $a^3$ where $a$ is any integer. Then $(a^3)^4=-x^3 \Rightarrow a^{12}=-x^3,$ whereby $x=-(-a)^4.$ Hence there exists infinitely many triples {x,y,z}={$-(-a)^4, a^3, 0$}, which satisfy $x^3+y^4=z^{31}$ for every integer $a$.
However the solution that they have given is quite different from mine. What I want to know is that, is my solution valid, and is this a convincing method to do proofs of such kind?
Thanks in advance!
 [/INST] This is the solution they gave...
Choose $x = 2^{4r}$ and $y = 2^{3r}$. Then the left side is $2^{12r+1}$. If we take$ z = 2^k$,
then we get$ 2^{12r+1} = 2^{31k}$. Thus it is sufficient to prove that the equation 12r + 1 = 31k
has infinitely many solutions in integers. Observe that$ (12.18) + 1 = 31.7$. If we choose
$r = 31l + 18$ and $k = 12l + 7$, we get
$12(31l + 18) + 1 = 31(12l + 7)$;
for all$ l$. Choosing $l\in N$, we get infinitely many$ r = 31l + 18$ and$ k = 12l + 7$ such that
$12r + 1 = 31k$. Going back we have infinitely many (x; y; z) of integers satisfying the given
equation.
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