Find all the natural numbers $m$ and $n$, such that the square of $m$ minus the product of $n$ with $k$, is 2, where the number $k$ is obtained from $n$ by writing 1 on the left of the decimal notation of $n$.
Let $t$ be the number of digits of $n$. Then $k=10^{t}+n$. So

$$
\mathrm{m}^{2}=n\left(10^{t}+\mathrm{n}\right)+2, \text{ i.e. } \mathrm{m}^{2}-\mathrm{n}^{2}=10^{t} n+2
$$

This implies that $\mathrm{m}, \mathrm{n}$ are even and both $\mathrm{m}, \mathrm{n}$ are odd.

If $t=1$, then, 4 is a divisor of $10 n+2$, so, $n$ is odd. We check that the only solution in this case is $\mathrm{m}=11$ and $\mathrm{n}=7$.

If $t>1$, then 4 is a divisor of $\mathrm{m}^{2}-\mathrm{n}^{2}$, but 4 is not a divisor of $10^{t}+2$.

Hence the only solution is $\mathrm{m}=11$ and $\mathrm{n}=7$.