Austria 2014

We have $$a_n=a_0+\sum _{k=0}^{n-1}(a_{k+1}-a_k)=a_0+\sum _{k=0}^{n-1}2\cdot 3^k=a_0+2\frac{3^n-1}{3-1}=a_0-1+3^n.$$ If $a_0=1$, then $a_n=3^n$ and $$\frac{a_k^j}{a_j^k}=\frac{3^{kj}}{3^{jk}}=1,$$ so $a_0=1$ is clearly a solution.

We now assume that $a_0\ne 1$ and write $a_0-1=x/y$ with coprime integers $x$ and $y$ with $y>0$. We obtain $$\frac{a_k^j}{a_j^k}=\frac{{\left(\frac{x}{y}+3^k\right)}^j}{{\left(\frac{x}{y}+3^j\right)}^k}=\frac{(x+y{3}^k{)}^j}{(x+y{3}^j{)}^k}{y}^{k-j}\in \mathbb{Z}.$$ We have $\text{gcd}(x+y{3}^j,y)=\text{gcd}(x,y)=1$, which implies that $$\frac{(x+y{3}^k{)}^j}{(x+y{3}^j{)}^k}$$ is also an integer. This implies that $$(x+3^{k}y{)}^{k-j}\frac{(x+3^{k}y{)}^j}{(x+3^{j}y{)}^k}=\frac{(x+3^{k}y{)}^k}{(x+3^{j}y{)}^k}={\left(\frac{x+3^{k}y}{x+3^{j}y}\right)}^k$$ is an integer, too. If the $k$th power of a rational number is an integer, then the number itself has to be an integer. Therefore, $$\frac{x+3^{k}y}{x+3^{j}y}$$ is an integer. We now set $k=j+1$ and write $$\frac{x+3^{j+1}y}{x+3^{j}y}=3+\frac{-2x}{x+3^{j}y}.$$ As this is an integer by the above considerations, the second summand $$\frac{-2x}{x+3^{j}y}$$ is also an integer. As $y>0$ by construction, the denominator is unbounded for $j\to \infty$, which results in $x=0$, i.e., $a_0=1$.

We conclude that $a_0=1$ is the only suitable initial value. $\square$