2. The set of positive integer solutions $(a, b)$ that satisfy $a^{2}+a b+b^{2}=2010$ is $\qquad$
2. $\varnothing$.

If $a$ and $b$ are both odd or one is odd and the other is even, then $a^{2}+a b+b^{2}$ is odd, which contradicts the fact that 2010 is even;

If $a$ and $b$ are both even, then $a^{2}+a b+b^{2}$ is a multiple of 4, which contradicts the fact that 2010 is congruent to 2 modulo 4.
Therefore, the original indeterminate equation has no solution.