635. What exact square is equal to the product of four consecutive odd numbers?
Let's denote the smallest of the odd numbers by $n$. Then

$$
n(n+2)(n+4)(n+6)=a^{2}
$$

where $a$ is an integer, and also odd. We transform the left side of this equation by multiplying the first factor by the fourth, and the second by the third:

$$
\left(n^{2}+6 n\right)\left(n^{2}+6 n+8\right)=a^{2}, \quad\left(n^{2}+6 n\right)^{2}+8\left(n^{2}+6 n\right)=a^{2}
$$

We complete the left side of the last equation to a perfect square. For this, we add 16 to both sides of the equation:

$$
\begin{gathered}
\left(n^{2}+6 n\right)^{2}+8\left(n^{2}+6 n\right)+16=a^{2}+16, \quad(n+6 n+4)^{2}=a^{2}+16 \\
\left(n^{2}+6 n+4\right)^{2}-a^{2}=16
\end{gathered}
$$

When do two perfect squares differ by 16? To answer this question, we list the sequence of perfect squares, starting from 0, until the difference between even consecutive squares becomes greater than 16:

$$
0 ; 1 ; 4 ; 9 ; 16 ; 25 ; 36 ; 49 ; 64 ; 81
$$

From this, we see that the difference between two perfect squares equals 16 in only two cases:

$$
16=16-0=25-9
$$

But since $a$ is odd, only the second option fits:

$$
\left(n^{2}+6 n+4\right)^{2}=25, \quad a^{2}=9
$$

We get two equations with $n$:

$$
n^{2}+6 n+4=5, \quad n^{2}+6 n+4=-5
$$

The first of these has non-integer roots, while from the second we find that $n=-3$.

Answer: $9=(-3) \cdot(-1) \cdot 1 \cdot 3$.