# find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$

find all ordered pairs of natural numbers (x,y,z) that satisfies $$x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$$

My approach was $$x^{3(x^x-1)} \lt x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$$ and if x is 5 -> $$5^{3(5^5-1)}=5^{9372} \gt 1990^{1990}$$

so $$x\in \{1,2,3,4\}$$,and I tried to check in cases when x=1, x=2,x=3,x=4 but I couldn't proceed.

## 1 Answer

First note the prime factorization of $$1990 = 2 \times 5 \times 199$$.

Hence at least one of $$x,y,z$$ must be divisible by $$199$$, from the equation $$x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$$.

Ignore the order for now and suppose $$199 \mid x$$. Then $$x^{x^y-1} \ge 199^{199^y-1}$$ and $$z^{z^x-1} \ge z^{z^{199}-1}$$.

For $$y \ge 2$$, $$199^{199^y-1} \ge 199^{199^2-1} \gg 1990^{1990}$$ (by observing logarithms.) Hence $$y=1$$.

Similarly we see that for $$z \ge 2$$, $$z^{z^{199}-1} \ge 2^{2^{199}-1} \gg 1990^{1990}$$. Hence $$z = 1$$ as well.

This reduces the equation to

$$x^{x-1} = 1990^{1990}$$

which has no solutions, since $$2,5,199 \mid x$$.