2. Determine the triples $(x, y, z)$ of natural numbers that satisfy the relation

$$
\left(2^{x}+1\right) \cdot\left(3^{y}+4\right) \cdot\left(4^{z}-3\right)=2015
$$

Ciocîrlan Ioana
## Solution.

The number 2015 can be factored into prime factors as follows: $2015=5 \cdot 13 \cdot 31$. (1p)

Let $x, y, z \in \mathbb{N}$ such that $\left(2^{x}+1\right) \cdot\left(3^{y}+4\right) \cdot\left(4^{z}-3\right)=2015$.

Considering the set of divisors of 2015, we deduce that $2^{x}+1=5$ or $2^{x}+1=65$. (2p)

1) If $2^{x}+1=5$, then $x=2$ and $\left(3^{y}+4\right) \cdot\left(4^{z}-3\right)=13 \cdot 31=403$. Then, the only possibility is that $4^{z}-3=13$. This results in $z=2$ and $y=3$. Therefore, the triplet $(2,3,2)$ satisfies the equation in the statement. (2p)
2) If $2^{x}+1=65$, then $x=6$ and $\left(3^{y}+4\right) \cdot\left(4^{z}-3\right)=31 \cdot 31$. We get $z=1$ and $y=3$, so the triplet $(6,3,1)$ satisfies the equation in the statement. (2p)