1. Find at least one integer solution to the equation

$$
a^{2} b^{2}+a^{2}+b^{2}+1=2005
$$
1. To solve the problem, it is sufficient to verify that the numbers $a=2, b=20$ satisfy our equation.

Let's explain how to find all integer solutions to this equation (although this was not required for solving the problem). Factorize the left-hand side of the equation:

$$
a^{2} b^{2}+a^{2}+b^{2}+1=a^{2}\left(b^{2}+1\right)+\left(b^{2}+1\right)=\left(a^{2}+1\right)\left(b^{2}+1\right)
$$

The equation will take the form $\left(a^{2}+1\right)\left(b^{2}+1\right)=2005$. Therefore, both $\left(a^{2}+1\right)$ and $\left(b^{2}+1\right)$ are divisors of the number 2005 (see fact 5).

Find all positive divisors of the number 2005. To do this, factorize this number into prime factors: $2005=$ $=5 \cdot 401$ (see fact 10). Now it is clear that all positive divisors of the number 2005 are 1, 5, 401, and 2005.

Note that $a^{2}+1$ cannot be equal to 2005: otherwise, $a^{2}=2004$, but 2004 is not a perfect square. Moreover, if $a^{2}+1=1$, then $b^{2}+1=2005$, which is also impossible. Therefore, the only possible cases are

$$
\left\{\begin{array}{l}
a^{2}+1=5, \\
b^{2}+1=401
\end{array} \quad \text { or } \quad \left\{\begin{array}{l}
a^{2}+1=401 \\
b^{2}+1=5 .
\end{array}\right.\right.
$$

Solving these equations, we find all solutions to the original equation: $a=$ $= \pm 2, b= \pm 20$ or $a= \pm 20, b= \pm 2$.