Saudi Arabia Mathematical Competitions

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

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

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