Solve the following system of equations $$\sin^2{x} + \cos^2{y} = y^4. $$ $$\sin^2{y} + \cos^2{x} = x^2. $$

[i]A. Khrabov[/i]
1. We start with the given system of equations:
   \[
   \sin^2{x} + \cos^2{y} = y^4
   \]
   \[
   \sin^2{y} + \cos^2{x} = x^2
   \]

2. We claim that \((x, y) = (\pm 1, \pm 1)\) is the only solution. Note that if \((x, y)\) is a solution, then \((\pm x, \pm y)\) are also solutions due to the symmetry of the trigonometric functions and the even powers.

3. Assume \(x, y \geq 0\). Summing the two equations, we get:
   \[
   \sin^2{x} + \cos^2{y} + \sin^2{y} + \cos^2{x} = y^4 + x^2
   \]
   Using the Pythagorean identities \(\sin^2{x} + \cos^2{x} = 1\) and \(\sin^2{y} + \cos^2{y} = 1\), we have:
   \[
   1 + 1 = y^4 + x^2
   \]
   \[
   2 = y^4 + x^2
   \]
   Therefore:
   \[
   x^2 + y^4 = 2
   \]

4. Since \(x^2 + y^4 = 2\), we have \(y^4 \leq 2\). This implies \(0 \leq y < \sqrt[4]{2}\). Given that \(y\) is in the range \([0, \pi/2)\), we can use the fact that \(\sin(\cdot)\) and \(\cos(\cdot)\) are increasing and decreasing functions respectively in this interval.

5. Express \(x\) in terms of \(y\):
   \[
   x = \sqrt{2 - y^4}
   \]

6. Substitute \(x = \sqrt{2 - y^4}\) into the second equation:
   \[
   \sin^2{y} + \cos^2{\sqrt{2 - y^4}} = 2 - y^4
   \]

7. Consider the function \(f(y) = \sin^2{y} + \cos^2{\sqrt{2 - y^4}}\). We need to check if this function equals \(2 - y^4\) for \(0 \leq y < \sqrt[4]{2}\).

8. Notice that if \(y^4 > 1\), then \(y > 1\). This would imply \(\sqrt{2 - y^4} < y\), which is a contradiction since \(y^4 + y^2 - 2 < 0\) would not hold. Hence, \(y < 1\).

9. If \(y^4 < 1\), then \(y < 1\). This implies \(y = 1\) is the only possible solution for \(y > 0\).

10. Substituting \(y = 1\) into the equation \(x^2 + y^4 = 2\), we get:
    \[
    x^2 + 1 = 2
    \]
    \[
    x^2 = 1
    \]
    \[
    x = 1
    \]

11. Therefore, the only solution in the positive quadrant is \((x, y) = (1, 1)\). By symmetry, the solutions are \((\pm 1, \pm 1)\).

The final answer is \( \boxed{ (\pm 1, \pm 1) } \).