Why exponents of variables?!

$\begin{cases} x^{x+y} = y^{12} \text{\_\_(1)}\\ y^{x+y} = x^{3} \text{\_\_(2)}\\ \end{cases}$

From equation (2) we get $\displaystyle x = y^{\displaystyle \frac{x+y}{3}}$

Substitute in (1) we get

$y^{\displaystyle \frac{(x+y)^{2}}{3}} = y^{12}$

---

There're 4 cases possible for the equation given

$a^{n} = a^{m}$

1. $a = 0$ if $m,n \neq 0$ .
2. $a = 1$
3. $a = -1$ if $m,n$ are both odd or even.
4. $m = n$

---

Case 1 : $y = 0$ . Substitute in (1) we get

$x^{x} = 0$

Which gives no solution.

Case 2 : $y = 1$ . Substitute in (1) we get

$x^{x+1} = 1$

Which gives $x = 1,-1$ .

Check the solution in (2) we get that $(-1,1)$ doesn't satisfy the equation.

$\therefore (x,y) = (1,1)$ .

Case 3 : $y = -1$ if $\displaystyle \frac{(x+y)^{2}}{3}$ is even (same as 12).

Substitute in (1) we get

$x^{x-1} = 1$

Which gives $x = 1,-1$ .

Check the solution in (2) we get $(-1,-1)$ doesn't satisfy the equation. And we checked that $(1,-1)$ makes $\frac{(x+y)^{2}}{3} = 0$ becomes even.

$\therefore, (x,y) = (1,-1)$

Case 4 : $\frac{(x+y)^{2}}{3} = 12$ we get $x+y = \pm 6$ .

• If $x+y = 6$ , from (2) we get $y^{6} = x^{3}$

Which means $x = y^{2}$ .

$x+y = y^{2}+y = 6$ .

Gives $y = 2,-3$ .

$x = 4,9$ .

Check the answers that $(4,2),(9,-3)$ both satisfy the solution.

$\therefore (x,y) = (4,2),(9,-3)$

• If $x+y = -6$ , from (2) we get $y^{-6} = x^{3}$

Which means $x = y^{-2}$

$x+y = y^{-2}+y = -6$

Which gives no integer solutions.

Therefore, $\boxed{(x,y) = (1,1), (1,-1), (4,2), (9,-3)}$ ~~~