8. The set of positive integer solutions $(a, b)$ that satisfy $a^{2}+a b+b^{2}=2018$ is
If $a, b$ are both odd or one odd and one even, then $a^{2}+a b+b^{2}$ is odd, and cannot equal 2018; if $a, b$ are both even, then $a^{2}+a b+b^{2}$ is divisible by 4, while 2018 is not divisible by 4, therefore the solution set of the equation $a^{2}+a b+b^{2}=2018$ is $\varnothing$.