Find all triples $(a,b,c)$ giving $3$ powers of $3$ simultaneously [closed]

For which triples $(a,b,c)$ of positive integers are $2a+b,2b+c,2c+a$ all powers of $3$ ?

I found out that $a, b, c$ have the same remainder after division by $3$.

Solution 1:

There are no solutions except those already mentioned by Prometheus ($a=b=c=3^n$) and Peter ($1,7,13$ times a power of 3).

1. Suppose there is a solution where not all three numbers are equal; say, $c$ is the (non-strictly) biggest of them. Now, since $2c+a=3^n$, it follows that $3^{n-1}\leqslant c < 3^n/2$.

2. Then $2b+c$, which also has to be a power of $3$, is too big for $3^{n-1}$ and too small for $3^{n+1}$, so it has no other choice except to be $3^n$ too. So $2c+a=2b+c$, or $c+a=2b$, hence $b={a+c\over2}>c/2\geqslant3^{n-1}/2$.

3. Now $2a+b$ is too big for $3^{n-2}$ and too small for $3^n$, hence $2a+b=3^{n-1}$, or $\frac52a+\frac12c=3^{n-1}$. Also, $2c+a=3^n$. This solves to yield $(a,b,c)=3^{n-3}\cdot(1,7,13)$.

So it goes.