## Task 4.

Determine all pairs of natural numbers $(m, n)$ such that

$$
3 \cdot 5^{m} - 2 \cdot 6^{n} = 3
$$
## First Solution. (Ivan Kokan)

If $n=1$, then $3 \cdot 5^{m}-12=3$, from which we get one solution $(m, n)=(1,1)$.

For $n=n_{0}+1 \geqslant 2, n_{0} \in \mathbb{N}$, the equation becomes $5^{m}-2 \cdot 2^{n_{0}+1} \cdot 3^{n_{0}}=1$, or equivalently

$$
5^{m}-1=4 \cdot 6^{n_{0}}
$$

Since the right-hand side is divisible by 6, it follows that $m$ is an even number, i.e., $m=2 m_{0}, m_{0} \in \mathbb{N}$. Then we have

$$
\left(5^{m_{0}}-1\right)\left(5^{m_{0}}+1\right)=4 \cdot 6^{n_{0}}
$$

Since $M\left(5^{m_{0}}+1,5^{m_{0}}-1\right)=M\left(5^{m_{0}}-1,2\right)=2$, we can write $5^{m_{0}}+1=2 \cdot 3^{n_{0}}$, $5^{m_{0}}-1=2^{n_{0}+1}$.

By subtracting and dividing by 2, we get $1=3^{n_{0}}-2^{n_{0}}$, which is only possible for $n_{0}=1$. For $n_{0}>1$, the difference on the right-hand side is greater than 1. For $n_{0}=1$, we have $m_{0}=1$, i.e., the solution $(m, n)=(2,2)$.