Exponential equation in x,y,z

Question

Find all positive integers x, y, z satisfying $$x^{y^{z}} \cdot y^{z^{x}} \cdot z^{x^{y}}=5xyz$$ I took log on both sides, which led me to $$(\log x)(y^{z}-1)+(\log y)(z^{x}-1)+(\log z)(x^{y}-1)=\log 5$$ However, further simplification is not clear. Please guide me.

Answer 1

If $x, y, z>1$, then $y^z, z^x, x^y \geq 2$ so $x^2y^2z^2 \mid x^{y^z}y^{z^x}z^{x^y}=5xyz$ so $xyz \mid 5$, impossible since $x, y, z \geq 2$.

Thus at least one of $x, y, z$ is $1$. The equation is cyclic, so we may assume $z=1$, so the equation becomes $x^yy=5xy$ so $x^{y-1}=5$. It is now clear that we must have $x=5, y=2$.

Therefore all solutions are given by $(x, y, z)=(5, 2, 1), (2, 1, 5), (1, 5, 2)$

Answer 2

given that the product of the (integral) powers of $x$ $y$ and $z$ is a multiplicand of $5$ means we can assume $x=5$

so $5^{y^{z}}\cdot y^{z^{5}}\cdot z^{5^{y}}=25yz$

then take $y=2$

$5^{2^{z}}\cdot 2^{z^{5}}\cdot z^{25}=50z$

and then we can take $z=1$

$5^{2}\cdot 2^{1}\cdot 1^{25}=25\cdot 2=50$

then try the same again with $x=10$

Answer 3

It's easy to conclude that one of the variables is multiple of 5. WLOG let's $x=5k$, then we have:

$$(5k)^{y^z} \cdot y^{z^5k} \cdot z^{(5k)^y} = 5^{y^z} \cdot k^{y^z} \cdot y^{z^{5k}} \cdot z^{(5k)^y} = 25kyz$$

Now we have three cases:

Case 1: $y \neq km, z \neq kn, k \neq 1$

We have $k^{y^z}$ at the LHS, but at the RHS we have $k$, and from the restriction we've made, we can only have:

$$k^{y^z} = k \iff y^z = 1 \iff y = 1$$

Then the equation will look like:

$$5k \cdot z^{5k} = 25kz$$ $$z^{5k} = 5z$$ $$z^{5k-1} = 5$$

From this we conclude $z=5$ and $k = \frac 25$

This is impossible, becuase $k$ has to be a solution and $x$ multiple of $5$.

Case 2: $y = kn, z \neq k, k \neq 1$

WLOG let's $y=kn$. For the sake of simplicity we'll take $y=k$

Simularly as Case 1, we'll have:

$$k^{k^z} \cdot k^{z^{5k}} = k^2$$ $$k^{k^z + z^{5k}} = k^2 \iff k^z + z^{5k} = 2$$

Because all numbers are natural numbers, we have $k^z = 1$ and $z^{5k} = 1$. This implies $k=z=1$, but this violates the condition.

Case 3: $k = 1$

Then the equation looks like:

$$5^{y^z} \cdot y^{z^5} \cdot z^{5^y} = 25yz$$

Note that $y$ and $z$ can't be multiple of $5$ because if one of them is, let's say $y=5$, then because $x,y,z \in \mathbb{N}$ the first term will be at least multiple of $5^5$, and the RHS will be $5^4z$. For other multiple of $5$ the difference is even bigger, because the exponential increase far more than then polynomial.

So from this we get that RHS has $5$ as multiple $2$ times. Because we concluded that only $x$ can be a multiple of $5$, then we must have $5^{y^z} = 5^2 \iff y^z = 2$. This is only posible for $y=2, z=1$

The only solution is $(5,2,1)$, but because the equation is symetric, other solutions are $(2,1,5)$ and $(1,5,2)$. You can obtain them if you assume $z=5k$ or $y=5k$, instead of $x=5k$