Find number of integers satisfying $$x^{y^z} \times y^{z^x} \times z^{x^y}=5xyz$$
My try:
we can rewrite the equation as
$$x^{y^z-1} \times y^{z^x-1} \times z^{x^y-1}=5$$
Then since all are integers we get
$x=5$, $y=1$ ,$z=1$ and like wise we get other two triplets.
is there a way to do formally?
On
Finding all of the solutions may be tricky, but counting how many there are is very easy! Notice that if one variable is $0$ and the other variables are positive, then both sides are obviously equal to $0$. So this gives infinitely many different solutions, since the other two variables can be any positive integers. (More precisely, there are $\aleph_0$ solutions.)
The factor of $5$ must come from somewhere, so $x, y$ or $z$ must have a factor of $5$ (let's just assume it's $x$, as the expression is symmetric). The power of $x$, i.e. $y^z - 1$, must be strictly positive. If there were any other prime factor of $x$, because $x$ is raised to a positive integer power, it would have to divide $5$, hence $5$ is the only prime factor of $x$. Moreover, we can similarly see that $25$ does not divide $x$, so $x$ is $5$ (or $y$ or $z$, when you account for symmetry).
Since $x = 5$, we must have $y^z - 1 = 1$. We already knew it had to be strictly larger than $0$, but if it were at least $2$, then $25$ would divide the left side but not the right side. Thus we have $y^z = 2$.
Similarly, this means $y = 2$. We know via the same arguments that $2$ must divide $y$, that $2$ is the only prime factor of $y$, and if $y$ were a power of $2$ greater than $2$ itself, then $y^z$ would be divisible by $4$. Similarly, $z = 1$.
So, our only possible solutions are $(x, y, z) = (5, 2, 1), (2, 1, 5), (1, 5, 2)$, by symmetry. Are they solutions?
No. There are no solutions.