A number theory problem by tannistha mondal

$a^2+84a+2008=b^2$ can be rewritten as $(a+42)^2=b^2-244$

or, $b^2-(a+42)^2=244$

or, $(b+a+42)(b-a-42)=244$

As $a,b$ are positive integers. $a,b$ both must be even or both must be odd.

Hence we take $b+a+42=122$ & $b-a-42=2$

From these we can say $a+b=\boxed{80}$