Suppose $a,b$ are positive real numbers such that $a+a^2 = 1$ and $b^2+b^4=1$.  Compute $a^2+b^2$.

[i]Proposed by Thomas Lam[/i]
1. First, solve for \(a\) from the equation \(a + a^2 = 1\):
   \[
   a^2 + a - 1 = 0
   \]
   This is a quadratic equation in the standard form \(ax^2 + bx + c = 0\). We can solve it using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = -1\):
   \[
   a = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}
   \]
   Since \(a\) is a positive real number, we take the positive root:
   \[
   a = \frac{-1 + \sqrt{5}}{2}
   \]

2. Next, solve for \(b\) from the equation \(b^2 + b^4 = 1\):
   \[
   b^4 + b^2 - 1 = 0
   \]
   Let \(x = b^2\). Then the equation becomes:
   \[
   x^2 + x - 1 = 0
   \]
   This is another quadratic equation. Solving it using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = -1\):
   \[
   x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2}
   \]
   Since \(b^2\) is a positive real number, we take the positive root:
   \[
   b^2 = \frac{-1 + \sqrt{5}}{2}
   \]

3. Now, compute \(a^2 + b^2\):
   \[
   a^2 = \left( \frac{-1 + \sqrt{5}}{2} \right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}
   \]
   \[
   b^2 = \frac{-1 + \sqrt{5}}{2}
   \]
   Adding these together:
   \[
   a^2 + b^2 = \frac{3 - \sqrt{5}}{2} + \frac{-1 + \sqrt{5}}{2} = \frac{3 - \sqrt{5} - 1 + \sqrt{5}}{2} = \frac{2}{2} = 1
   \]

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