1. Find all pairs $a, b$ of non-negative real numbers for which

$$
\sqrt{a^{2}+b}+\sqrt{b^{2}+a}=\sqrt{a^{2}+b^{2}}+\sqrt{a+b} .
$$
1. By squaring the equation with non-negative sides and other equivalent transformations, we gradually obtain

$$
\begin{aligned}
a^{2}+b+2 \sqrt{\left(a^{2}+b\right)\left(b^{2}+a\right)}+b^{2}+a & =a^{2}+b^{2}+2 \sqrt{\left(a^{2}+b^{2}\right)(a+b)}+a+b, \\
\sqrt{\left(a^{2}+b\right)\left(b^{2}+a\right)} & =\sqrt{\left(a^{2}+b^{2}\right)(a+b)}, \\
\left(a^{2}+b\right)\left(b^{2}+a\right) & =\left(a^{2}+b^{2}\right)(a+b), \\
a^{2} b^{2}+a^{3}+b^{3}+a b & =a^{3}+a b^{2}+b a^{2}+b^{3}, \\
a b(a b+1-a-b) & =0, \\
a b(a-1)(b-1) & =0 .
\end{aligned}
$$

Thus, the sought pairs of non-negative numbers $a, b$ are those that satisfy at least one of the conditions $a=0, b=0, a=1$ or $b=1$.

Full solution is worth 6 points.