## Task 15/90

It is to be investigated whether the equation $3^{n}=10^{4} m+1$ with $m ; n \in N, m ; n>0$ has solutions!
In the decimal system, there are $10^{4}$ different possibilities for the last four digits of a number. Therefore, among $10^{4}+1$ powers of the number 3, there are at least two different ones that end with the same four digits. Let these be $3^{k}$ and $3^{l}$ with $k ; l \in N$, where W.L.O.G. $00$. Since $3^{k}$ and $10^{4}$ (except 1) have no common divisor, $3^{k}$ is a divisor of $r: r=3^{k} m$ with $m \in N, m>0$. Thus, it follows that

$$
3 k\left(3^{k-l}-1=10^{4} r=10^{4} 3^{k} \cdot m\right.
$$

and because $3^{k} \neq 0$ with $l-k=n>0: 3^{n}-1=10^{4} m$ with $m ; n \in N, m ; n>0$.