Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power Find all positive integers $x, y, z$ so that $(x+2y)(y+2z)(z+2x)$ is equal to prime power.
My try
Since $x, y, z$ are positive integers it follows that $x+2y, y+2z, z+2x \geq 3$. We have following system of equations:
$$\begin{align}
x+2y &= p^r \tag{1}\label{eq1} \\
y+2z &= p^s \tag{2}\label{eq2} \\
z+2x &= p^t \tag{3}\label{eq3}
\end{align}$$
Where $p$ is prime number and $r, s, t$ are positive integers.
Solving this system of equations for $x, y, z$ gives us $x=\frac{4p^t-2p^s+p^r}{9}$, $y=\frac{4p^r-2p^t+p^s}{9}$, $z=\frac{4p^s-2p^r+p^t}{9}$. Since $x, y, z$ are integers following equations must hold:
$$\begin{align}
4p^t-2p^s+p^r &\equiv 0 \pmod 9 \\
4p^r-2p^t+p^s &\equiv 0 \pmod 9 \\
4p^s-2p^r+p^t &\equiv 0 \pmod 9
\end{align}$$
I don't know how to continue from there. Can somebody help?
 A: Suppose we have a solution $(x, y, z)$. Note that if $d = \gcd(x, y, z) > 1$, then $d$ must also be a power of $p$, and we get a smaller solution $(x', y', z') = (x/d, y/d, z/d)$ with $\gcd(x', y', z') = 1$. 
Thus without loss of generality we can assume our solution has $\gcd(x, y, z) = 1$. Now, if $p \neq 3$, then by the formulas $x = (4p^t - 2p^s + p^r)/9$, etc., we must have that $x, y, z$ are divisible by $p$, contradicting $\gcd(x, y, z) = 1$. It follows that there can only be solutions in the case $p = 3$. 
Now suppose without loss of generality that $z$ is the largest of $x, y, z$. Again write 
$$x + 2y = 3^r \qquad y + 2z = 3^s \qquad z + 2x = 3^t.$$
Since $y \leq z$, we have $3^s \leq 3z$, and similarly since $x \leq z$, we have $3^t \leq 3z$. But then $3^{s-1} \leq z < z + 2x = 3^t$, so $s \leq t$, and conversely, $2(3^{t-1}) \leq 2z < y + 2z = 3^s$, so $t \leq s$, and thus we must have $s = t$. This gives $y + 2z = z + 2x$, hence $z = 2x - y \leq 2x$, so $3^t \leq 4x$, and $x \geq 3^t/4$. It follows that $3^r = x + 2y > 3^t/4 > 3^{t-2}$, so $r \geq t-1$. In the other direction, since $z-x=x-y$ and $z \geq x$, we have $x \geq y$, hence $3^r = x + 2y \leq z + 2x = 3^t$, so $r \leq t$.
Thus our only solutions for $(r, s, t)$ are of the forms $(t, t, t)$ and $(t-1, t, t)$. In the case $(t, t, t)$, it immediately follows that $x = y = z$, hence $(x, y, z) = (1, 1, 1)$. Solving the second case for $x, y, z$ gives
$$(x, y, z) = (3^{t-3}(12-6+1), 3^{t-3}(4-6+3), 3^{t-3}(12-2+3)) = (3^{t-3} \cdot 7, 3^{t-3}, 3^{t-3} \cdot 13)$$
and since we assumed $\gcd(x, y, z) = 1$, our solution is $(7, 1, 13)$. 
Now relaxing the assumption that $z$ was the largest, so we allow for cyclic shifts, the solutions with $\gcd(x, y, z) = 1$ are exactly $(1, 1, 1), (7, 1, 13), (1, 13, 7), (13, 7, 1)$. Relaxing the other assumption that $\gcd(x, y, z) = 1$, the general solutions are exactly all scalings of these primitive solutions by a power of $3$.
A: Assume the prime is $p \neq 3.$ Note that we may keep (simultaneously) dividing out $p$ from $x,y,z$ until at least one of them is not divisible by $p\; , \;$ with the result that $$ \gcd(x,y,z) = 1 $$
We do have such primitive solutions when 
$$ (x,y,z) = (1,1,1) \; \; , \; \; (x,y,z) = (13,7,1) $$
when both primes are $3.$
Next, add up the three items that must be powers of $p,$ we find
$$ 3 (x+y+z) \equiv 0 \pmod p.  $$
With the assumption $p \neq 3,$ we have $x+y+z \equiv 0 \pmod p.$ Then subtract $x+2y \equiv 0 \pmod p,$ so that $y \equiv z \pmod p.$
Repeat with the other pairs, we get
$$ x \equiv y \equiv z \pmod p $$
However, we already had $x+y+z \equiv 0 \pmod p.$ We get $
3x \equiv 0 \pmod p,$ same for $y,z,$ contradicting $\gcd(x,y,z) = 1.$ In turn, this contradicts the assumption that $p \neq 3.$
A: This is not a full solution, but some thoughts.
There are two possibilities: $x=y=z$, or not.
Case $1$: $x=y=z$: This is equivalent to finding all $x$ such that $3x$ is a prime power, which happens only when $x=3^a$ for some $a\in\mathbb N\cup\{0\}$. 
Case $2$: Let $x+2y=p^a,\;y+2z=p^b,\;z+2x=p^c$, $a,b,c\in\mathbb N$, $a=\min(a,b,c),\;b=\max(a,b,c)$. Note that $3\cdot(x+y+z)=p^a(1+p^{b-a}+p^{c-a})$
Assume for now that $p\neq3$. Then, $p^{b-a}+p^{c-a}\equiv2\pmod3$. That means that $p^{b-a},p^{c-a}\equiv1\pmod3$. In particular, it tells us that $p\equiv1\pmod3$. 
So, $p=3$ or $p=6k+1$ for some $k\in\mathbb N$. Will edit as I get more ideas.
