If $a$ and $b$ are prime numbers greater than 7, then the expression

$$
\left(a^{2}-1\right)\left(b^{2}-1\right)\left(a^{6}-b^{6}\right)
$$

is divisible by 290304.
1. $a^{2}-1$ and $b^{2}-1$ are divisible by $2^{3} \cdot 3$. Since $a$ is an absolute prime number, in the expression $a^{2}-1=(a+1)(a-1)$, both $a+1$ and $a-1$ are divisible by 2, and one of the factors is divisible by 4, because among two consecutive even numbers, one is always a multiple of 4. Additionally, one of the factors is divisible by 3, because $a-1, a, a+1$ are three consecutive terms of the sequence of natural numbers. Therefore, $a^{2}-1$ and similarly $b^{2}-1$ are divisible by $2^{3} \cdot 3$.
2. $a^{6}-b^{6}$ is divisible by $2^{3} \cdot 3^{2}$. The difference of the squares of two absolute prime numbers greater than 3 is always divisible by $24=2^{3} \cdot 3$ (Math. Practice Book I. 134. problem), and if the numbers $a$ and $b$ are not divisible by 3, then $a^{6}-b^{6}$ is divisible by $9=3^{2}$ (K. M. L. XI. 21. 1.). Thus, $a^{6}-b^{6}$ is divisible by $2^{3} \cdot 3^{2}$.
3. $a^{6}-b^{6}$ is divisible by 7. According to the problem, $a$ and $b$ are of the form $7P \pm 1, 7P \pm 2, 7P \pm 3$. If we raise these numbers to the 6th power, we get expressions of the following form:

$$
\begin{gathered}
7q+1^{6}=7q+1 \\
7r+2^{6}=7r+64=7r^{\prime}+1 \\
7s+3^{6}=7s+729=7s^{\prime}+1
\end{gathered}
$$

From which it is clear that $a^{6}-b^{6}$ is always a multiple of 7.

4. Based on the above, the given expression is divisible by

$$
2^{9} \cdot 3^{4} \cdot 7=290304 \text{. }
$$

(Nándor Ehrenfeld, Nyitra.)