An interesting system of simultaneous equations with two unknowns. $ \text{We are going to solve the system (E):}$
$\left\{\begin{aligned}    \displaystyle \quad x^{n}+y^{n}&=1 \cdots(1) \\x^{n+1}+y^{n+1}&=1 \cdots(2)\end{aligned}\right., \quad $ where $m,n\in N.$
$\displaystyle \begin{aligned}x&=\frac{x^{n+1}}{x^{n}}\\&=\frac{1-y^{n+1}}{1-y^{n}}\\&=\frac{1-y^{n}+y^{n}-y^{n+1}}{1-y^{n}} \\ \displaystyle &=1+\frac{y^{n}(1-y)}{(1-y)\left(1+y+y^{2}+\cdots+y^{n-1}\right)} \\ \displaystyle &=1+\frac{y^{n}}{1+y+y^{2}+\cdots+y^{n-1}} \cdots \text { (3) } \end{aligned} \tag*{}  $
$\textrm{Note that either $n$ or $n+1$ is odd,}\\$
$\displaystyle \quad x^m+y^m=1 \Rightarrow \textrm{ one of x and y must be non-negative  for any odd integer m.}$
WLOG, let $y\geq 0. $ $\textrm{ However, if y is positive, then by (3)},$
$$\displaystyle x=1+\frac{y^{2 n}}{1+y+y^{2}+\cdots+y^{2 n-2}}>1\Rightarrow \quad x^n+y^n>1, $$
$\text{which contradicts to one of the equation (1) or (2).}$
$\textrm{Hence }y=0 \textrm{ and } x=1.$
$\textrm{By symmetry, the real solutions are }$$ (1,0)\text{ and }(0,1).$
My question
What can we say for the system $$(F):\left\{\begin{aligned}    \displaystyle  x^{m}+y^{m}&=1  \cdots(3) \\x^{n}+y^{n}&=1  \cdots(4)\end{aligned}\right.,$$
where $m,n\in N.$
 A: I believe, there is more simple solution than mine.
My solution is based on evident consequence of system: $(1-x^m)^n=(1-x^n)^m$.
Lemma. If $0\leq x \leq 1$ and $m > n$ then $(1-x^m)^n=(1-x^n)^m$ has only two solutions $x=0$ and $x=1$.
Proof. The fact that $x=0$ and $x=1$ are solutions is easy to check. Let's consider case $0 < x < 1$, then $0 < x^m < x^n <1$, $1 > 1-x^m > 1-x^n > 0$, $1 > (1-x^m)^n > (1-x^m)^m > (1-x^n)^m > 0$. So $(1-x^m)^n \neq (1-x^n)^m$ for $0<x<1$.
Solution. Let's consider system $x^m+y^m=1$, $x^n+y^n=1$, $m,n\in\mathbb{N}$. Without loss of generality $m > n$. Case $m=n$ has infinitely many solutions so we don't consider it.
There are three possible cases: $m$, $n$ are both even, one of them is odd, both $m$ and $n$ are odd.
Case 1. $m$, $n$ are even. Then $|x|^m+|y|^m=1$, $|x|^n+|y|^n=1$. $|y|^m \geq 0 \Rightarrow |x|^m \leq 1 \Rightarrow |x| \leq 1$. $|x|$ should satisfy equation $(1-|x|^m)^n=(1-|x|^n)^m$. Using lemma, we can say that $|x|=0$ or $|x|=1$. Putting this into equations, we can get four solutions $(0;1)$, $(1;0)$, $(0;-1)$, $(-1;0)$.
Case 2. One of $m$ and $n$ is odd. Let's mark this odd number as $k$ and even number as $l$l. $|x|^l+|y|^l=1$, $|x|^l \geq 0 \Rightarrow |y|^l \leq 1 \Rightarrow |y| \leq 1$. $x^k+y^k=1 \Rightarrow y^k=1-x^k \Rightarrow |y|^k =|1-x^k|$. $|y|\leq 1 \Rightarrow |y|^k \leq 1 \Rightarrow |1-x^k| \leq 1 \Rightarrow x^k \geq 0 \Rightarrow x \geq 0$. $|x|^l+|y|^l=1$, $|y|^l\geq 0 \Rightarrow |x|^l\leq 1 \Rightarrow |x|\leq 1 \Rightarrow 0\leq x \leq 1$. $x$ must satisfy equation $(1-x^m)^n=(1-x^n)^m$. Using lemma, we can say that $x=0$ or $x=1$. Putting this into equations, we can get two solutions $(0;1)$ and $(1;0)$.
Case 3. Both $m$ and $n$ are odd. Consider three subcases $x > 1$, $x < 0$ and $0 \geq x \geq 1$.
When $x>1$: $-\left(\frac{y}{x}\right)^{mn}=(1-\left(\frac{1}{x})^m\right)^n=(1-\left(\frac{1}{x}\right)^n)^m$. Then $u=\frac{1}{x}$ must satisfy equation $(1-u^m)^n=(1-u^n)^m$ and $0 < u < 1$. Using lemma, we can say that this subcase has no solutions.
When $x < 0$: $y^m=1-x^m > 1 \Rightarrow y > 1$. Then we can use the same consideration as for subcase $x > 1$.
When $0 \leq  x \leq 1$ we can use lemma and get $x=0$ or $x=1$. Putting into equations, we can get two solutions $(0;1)$ and $1;0$.
Final answer: when $m$ and $n$ are both even, there are four solutions $(0;1)$, $(1;0)$, $(0;-1)$, $(-1;0)$, otherwise there are two solutions $(0;1)$ and $1;0$.
