<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] A small variation on the 'official' solution: instead of proving that the equation $12r+1=31k$ has infinitely many positive integer solutions we find a particular one, let's say $k=7,r=18$; it follows that $x=2^{4\cdot 18},y=2^{3\cdot 18},z=2^7$ is a particular solution for our equation and from here it is easy to notice that 
$$x=2^{4\cdot 18}\cdot k^{4\cdot 31},y=2^{3\cdot 18}\cdot k^{3\cdot 31},z=2^{7}\cdot k^{3\cdot 4}$$
is also a solution for the given equation, for any positive integer $k$.
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