solutions to $x^{x+y} = y^4, y^{x+y} = x$ in integers Three of the elements in the solution set of the simultaneous system $$ x^{x+y} = y^4, \qquad y^{x+y} = x $$ are ordered pairs of integers $(x, y)$. Find these ordered pairs.
Substitution leads to nothing (well infinity), and I have no idea on how to start this.  Any help is appreciated!
 A: If $y \geq 0 \to x \geq 0$, if $y < 0, x+y \text{ is odd} \to x < 0 \to y^4 < 0 \text{ contradiction}\to x+y \text{ is even}\to x > 0\to x+y \geq 0 \text{ or else $y^4$ is not an interger}$. Thus there are two scenarios:
$1)$: $y \geq 0$: $x+y = \dfrac{4\ln y}{\ln x} = \dfrac{\ln x}{\ln y} \to \ln x = \pm 2\ln y\to x+y = \pm 2, x = \pm y^2$. Can you continue?
$2)$: $y < 0 \to y = - y', y' < 0$. Can you take it from here as well?
A: Whether $(x,y)=(0,0)$ is a solution depends on your definition of $0^0$, which is sometimes made to be $1$. I'll assume here you don't define it. Let $xy\neq 0$.
$$(y^{x+y})^{x+y}=y^4\iff y^{(x+y)^2-4}=1,$$   
so either $y\in\{-1,1\}$, giving us $(x,y)=(1,1),(1,-1)$
or $$(x+y)^2-4=0\iff x+y=\pm 2$$ with $y\neq 0$.   


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*If $(x,y)=(a,2-a)$ with $a\neq 2$, then $(2-a)^2=a$, so $(x,y)=(4,-2),(1,1)$, and after checking they work.

*If $(x,y)=(a,-2-a)$ with $a\neq -2$, then $$(-2-a)^{-2} = a\iff a(-2-a)^2=1$$
mod $a$ gives $a\in\{-1,1\}$, so $(x,y)=(1,-1),(1,1)$.
Answer: $(x,y)=(1,-1),(1,1),(4,-2)$.
