<s>[INST] System of Exponential Equations in $x$ and $y$ Three of the elements in the solution set of the simultaneous system $x^{x+y} = y^4, y^{x+y} = x$ are ordered pairs of integers $(x, y)$. Find these ordered pairs. 
I found the trivial solution at $(1,1)$, but was unable to find the other two solutions.  Could I get help with the others?  Thanks!
 [/INST] $$1) \quad x^{x+y} = y^4$$
$$2) \quad y^{x+y}=x$$

$$x^{x+y}=(y^{x+y})^{x+y}=y^{(x+y)^2}=y^4\implies (x+y)^2=4\implies x+y =\pm 2$$
Else $(x+y)^2$ need not equal $4$ if $y=-1,0,1$ as Meelo has commented(but $(x+y)^2$ need be even for $y=-1$ case.)

Clearly they have designed this so$x=y=0$ and $(-1,-1)$ solutions are inconsistent. 
$$x^2=y^4\implies(x,y)=(\pm n^2,\pm n),n\in\Bbb Z$$ of these consider $(\pm 4, \pm 2)$: $x+y=2$ has only the soltuion here  $(4,-2)$ only solution. We can see from $2)$, $y^{x+y}=x$, here ${-2}^2 = 4$ holds. Thus this is a solution.

$$x+y=-2,x^{-2}=y^4\implies (\pm1,1)$$ are these consistent with $y^{x+y}=x$, only $(1,1)$ is.

Consider the options of $y=-1,0,1$, we have already found $y=1$ and ruled out $y=0$, consider $y=-1$:
$$y=-1\implies x^{x-1}=1,-1^{x+y}=x, x=1,$$
$$(x,y)=(1,-1)$$
Solutions $(1,\pm1),(4,-2)$
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