$a^2+b=b^{1999}$ how many integer pairs $(a,b)$ are there that satisfy this equation.

$$a^2=b(b^{1998}-1)$$

As $(b,b^{1998}-1)=1$ if $a\ne0$ both have to be perfect square

So, we need $$b^{1998}-1=d^2\iff(b^{999}+d)(b^{999}-d)=1$$ for some integer $d$

Since $\gcd(b,b^{1998}-1)=1$ we know that $b$ is a square. Thus we want $k^2=b^{1998}-1$. If $|b|>1$ then $(b^{999})^2>b^{1998}-1>(b^{999}-1)^2$. Thus from square bounding $|b|\leq1$. Now case bash the remaining cases.