Given $x$, $y$, $z$ which are positive integers so that $(x+y)^4 + 5z = 63x$, find the value of $x+y+z$

Question

Given $x$, $y$, $z$ which are positive integers so that $(x+y)^4 + 5z = 63x$, find the value of $x+y+z$.

I might be wrong about this but I'm thinking that this is some sort of very complicated non-linear Diophantine equation because of the conditions of the variables. I've put the equation into WolframAlpha and saw that there are exact solutions to these, so I assume it can be solved. Personally, I have very little experience with these and I'd like to see how it can be solved.

Answer 1

$(x+y)^4+5z$ grows much faster than $63x$ – the minimum value the former expression can assume for given $x$ is $(x+1)^4+5$, and if this is greater than $63x$ there can be no solution for that $x$. Hence $x=1$ or $x=2$, and we can quickly enumerate all solutions from there:

• If $x=1$ then $(1+y)^4+5z=63$, from which $y$ can only be $1$ (for the same reasons as $x$ is constrained). But then $5z=63-16=47$, which is impossible.
• If $x=2$ then $(2+y)^4+5z=126$, and once again $y$ must be $1$. Now $5z=126-81=45$ and thus $z=9$.

Thus the only solution is $(x,y,z)=(2,1,9)$ with sum $12$.