Find all natural numbers $x,y,z$ such that $x^7+y^7=7^z$

Question

Find all natural numbers $x,y,z \in \mathbb{N}$ such that $x^7+y^7=7^z$.
Firstly, I note that given that the right hand side only has factors of $7$, it must be odd and so $x^7+y^7$ must be odd i.e $x^7$ and $y^7$ have opposite parity, which then means $x$ and $y$ have opposite parity. Also, I have got the following: $$ x^7+y^7=7^z \implies x^7+y^7\equiv0\pmod{7} $$ By Fermat's Little Theorem, we have $x^7\equiv x\pmod{7}$ and similarly for $y$, $y^7\equiv y\pmod{7}$, so then $x^7+y^7\equiv0\pmod{7}\implies x+y\equiv0\pmod{7}$. I have also factorised $x^7+y^7$: $$ x^7+y^7=(x+y)(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)=7^z $$ So not only does $7$ divide $x+y$, we also have that $7\mid(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6)$ but at this point I feel like I am just not making any real progress. Any help is appreciated.

Answer 1

If a solution exists where $7\mid x$ and $7\mid y$ then another solution can be generated by dividing $x$ and $y$ by $7$.

Suppose a solution exists where $x$ and $y$ are positive integers that are not divisible by $7$. We have that $$x^7+y^7=(x+y)\left(x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6\right)$$As mentioned in the OP, by Fermat's Little Theorem, $$x^7+y^7\equiv x+y\equiv0\pmod7,$$ so let $x+y=7k$ where $k\in\mathbb{N}$.

Since $$x^6-x^5y+x^4y^2-x^3y^3+x^2y^4-xy^5+y^6=(x+y)^6-7xy(x+y)^4+14(xy)^2(x+y)^2-7(xy)^3,$$$x^7+y^7$ can be rewritten as $$\begin{align}&(x+y)\left[(x+y)^6-7xy(x+y)^4+14(xy)^2(x+y)^2-7(xy)^3\right]\\&=(x+y)\left[7^6k^6-7^5k^4xy+2\cdot7^3k^2(xy)^2-7(xy)^3\right]\\&=7(x+y)\left[7^5k^6-7^4k^4xy+2\cdot7^2k^2(xy)^2-(xy)^3\right]\end{align}$$ Since $7\mid7^5k^6$, $7\mid7^4k^4xy$ and $7\mid2\cdot7^2k^2(xy)^2$ but $7\nmid(xy)^3$, $7\nmid7^5k^6-7^4k^4xy+2\cdot7^2k^2(xy)^2-(xy)^3$. However, for $x^7+y^7$ to be a perfect power of $7$, all of its factors must also be perfect powers of $7$ and so $7^5k^6-7^4k^4xy+2\cdot7^2k^2(xy)^2-(xy)^3$ must equal $1$. It follows that $$x^7+y^7=7(x+y)$$If $x>1$ then $x^7>7x$, so at least one of $x$ and $y$ must be $1$. If $x=1$, $y^7-7y-6=0$, which has no positive integer solutions.

Therefore, there are no solutions where $x$ and $y$ are both positive integers and the only possible natural number solutions are if one of $x$ and $y$ is $0$. If $x=0$ then $y^7=7^z$, so $y=7^k\,\forall \,k\in\mathbb{Z}$.

This means that the only solutions are $(x,y,z)=(0,7^k,7k),(7^k,0,7k)$.