<s>[INST] 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?
 [/INST] 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 $p \neq 3.$
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