196. Interesting square. Find a nine-digit integer of the form $\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}$, which represents the product of the squares of four different prime numbers, and

$$
\overline{b_{1} b_{2} b_{3}}=2 \overline{\left(a_{1} a_{2} a_{3}\right)} \quad\left(a_{1} \neq 0\right)
$$
196. Note that

$$
\begin{gathered}
N^{2}=\overline{a_{1} a_{2} a_{3} b_{1} b_{2} b_{3} a_{1} a_{2} a_{3}}=\overline{a_{1} a_{2} a_{3}} \cdot 1002001= \\
=\overline{a_{1} a_{2} a_{3}} \cdot 7^{2} \cdot 11^{2} \cdot 13^{2}
\end{gathered}
$$

Therefore, the number $\overline{a_{1} a_{2} a_{3}}$ is the square of a prime number $P$, different from 7, 11, and 13. Furthermore, $a \neq 0$, and $\overline{b_{1} b_{2} b_{3}}<1000$; hence, $10<P<23$. Therefore, $P$ is either 17 or 19; $\overline{a_{1} a_{2} a_{3}}=289$ or $361$; $N^{2}=289578289$, or 361722361. Both these cases provide a solution to our problem.

[P. N. Nagara, M. M., 24, 108 (November 1950).]