$2+$ [ Factorization ]

Find at least one integer solution to the equation $a^{2} b^{2}+a^{2}+b^{2}+1=2005$.

#
$\left(a^{2}+1\right)\left(b^{2}+1\right)=a^{2} b^{2}+a^{2}+b^{2}+1=2005=5 \cdot 401=1 \cdot 2005$. Since the number 2004 is not a perfect square, we get 8 solutions; all of them are derived from $(2,20)$ by permutation and changing signs.

## Answer

For example, $a=2, b=20$.