## Task 4.

Determine all pairs of natural numbers $(a, b)$ for which there exist natural numbers $x, y$ and $z$ such that

$$
a^{x}=(a+b)^{y}=(5 a+11 b)^{z}
$$
## Second Solution.

Let's prove that there exist a natural number $c$ and natural numbers $u<v<w$ such that $a=c^{u}, a+b=c^{v}$, $5 a+11 b=c^{w}$.

Let $t=\operatorname{gcd}(xy, xz, yz)$. We claim that $a=c^{yz/t}, a+b=c^{xz/t}, 5a+11b=c^{xy/t}$ for some natural number $c$. To do this, it is sufficient to prove that $\nu_{p}(a)$ is divisible by $yz/t$ for every prime number $p$.

Let $p_{1}, \ldots, p_{k}$ be all the prime numbers that divide any of $a, a+b, 5a+11b$. Then we have

\[
\begin{aligned}
a & =p_{1}^{\alpha_{1}} \ldots p_{k}^{\alpha_{k}} \\
a+b & =p_{1}^{\beta_{1}} \ldots p_{k}^{\beta_{k}} \\
5a+11b & =p_{1}^{\gamma_{1}} \ldots p_{k}^{\gamma_{k}}
\end{aligned}
\]

From the conditions, it follows that for every $i \in\{1, \ldots, k\}$,

\[
x \alpha_{i}=y \beta_{i}=z \gamma_{i}
\]

From this, it follows that for every $i$ there exists a number $u_{i}$ such that

\[
\alpha_{i} u_{i}=yz, \beta_{i} u_{i}=xy, \gamma_{i} u_{i}=xz
\]

Then $u_{i}$ divides $xy, yz$, and $zx$, so it also divides $t$, and we conclude that $\alpha_{i}=yz/u_{i}$ is divisible by $t$, and the claim is proved.

Notice that the converse is also true: if $a, a+b$, and $5a+11b$ are powers of the same natural number, then it is possible to find numbers $x, y, z$ such that $a^{x}=(a+b)^{y}=(5a+11b)^{z}$, so the problem is equivalent to solving the following system of equations in the set of natural numbers:

\[
\begin{aligned}
a & =c^{u} \\
a+b & =c^{v} \\
5a+11b & =c^{w}
\end{aligned}
\]

From the first two equations, it follows that $b=c^{v}-c^{u}$, and from the third, we get $11c^{v}-6c^{u}=c^{w}$. Then,

\[
11c^{v-u}-c^{w-u}=6
\]

From this, it follows that $c^{v-u}$ divides 6, so $v-u=1$, and $c \in\{1,2,3,6\}$.

If $c=1$, then $a=a+b$, a contradiction.

If $c=2$, then the equation becomes $22-2^{w-u}=6$, or $w-u=4$, so we get solutions $(u, v, w)=(n, n+1, n+4)$ for any natural number $n$, so $a=2^{n}$, $a+b=2^{n+1}$, $5a+11b=2^{n+4}$. In this case, we get the family of solutions $(a, b)=\left(2^{n}, 2^{n}\right)$ for $n \in \mathbb{N}$.

If $c=3$, then the equation becomes $33-3^{w-u}=6$, or $w=u+3$, and we get solutions $(u, v, w)=(n, n+1, n+3)$ for any natural number $n$, so $a=3^{n}$, $a+b=3^{n+1}$, $5a+11b=3^{n+4}$. In this case, we get the family of solutions $(a, b)=\left(3^{n}, 2 \cdot 3^{n}\right)$ for $n \in \mathbb{N}$.

If $c=6$, then $66-6^{w-u}=6$, but then 60 is a power of 6, a contradiction. Therefore, there are no solutions in this case.