If $a^{2}+84a+2008=b^{2}$ what is $a+b$ let $a, b$ are two positive integer satisfy the condition $a^{2}+84a+2008=b^{2}$. Find out $a+b$
My Solution
$a^{2}+84a+2008=b^{2} \implies (a+42)^{2}+244=b^{2} \implies (b+a+42)(b-a-42)=2^{2}61$. Considering (244,1) , (122,2) ,(61,4) we observe only (244,1) give integer $(a,b)=(18,62) \implies a+b=80$
My Question
Is there any other way to solve this problem?

With the quadratic formula you get
$$a=\frac{-84\pm \sqrt{84^2-4(2008-b^2)}}{2}=-42\pm \sqrt{b^2-244}$$
Now with this, it is possible to obtain the root since $b^2-244=x^2$, and as $b$ and $a$ must be positive integers, that root must be too, so neccesarily $x$ too. This lets
$$(b+x)(b-x)=244$$
And that implies that:
$$b+x=244\ \ \ \ \ \ \ \ \ b-x=1 \ \ \Rightarrow \ \ 2b=245 \mbox{ not integer}$$
$$b+x=122\ \ \ \ \ \ \ \ \ b-x=2 \ \ \Rightarrow \ \ 2b=124 \Rightarrow b=62$$
$$b+x=61\ \ \ \ \ \ \ \ \ b-x=4 \ \ \Rightarrow \ \ 2b=65 \mbox{ not integer}$$
$$b+x=1\ \ \ \ \ \ \ \ \ b-x=244 \ \ \Rightarrow \ \ 2b=245 \mbox{ not integer}$$
Since $b>0$, this lets $b=62$, so $a=-42\pm 2(62)$ but only is possible $a=-42+124=18$, so $a=18$ and $a+b=80$.
