Saudi Arabian Mathematical Competitions

It is clear that $2a + 1 - 2b$ is an odd integer, hence
$$|2a + 1 - 2b| \geq 1$$
If $(a, b)$ is a pair of nonnegative integers satisfying
$$a + 2b - b^{2} = \sqrt{2a + a^{2} + |2a + 1 - 2b|}$$
then we get
$$a + 2b - b^{2} \geq \sqrt{2a + a^{2} + 1} = \sqrt{(a + 1)^{2}} = a + 1$$
Therefore $0 \geq (b - 1)^{2}$, that is $b = 1$.

For $b = 1$, the equation becomes
$$a + 1 = \sqrt{2a + a^{2} + |2a - 1|} \tag{1}$$
Equation (1) is equivalent to $|2a - 1| = 1$. We get $a = 0$ or $a = 1$.

The desired pairs are $(a, b) = (0, 1), (1, 1)$.