# The solutions for $x^x = z, z \in \mathbb{Z^+ - \{4\}}$ are irrational

Suppose that the function $x^x = z, z \in \mathbb{Z^+ - \{4\}}$ has rational solutions. Then $x^x = (\frac{a}{b})^{{\frac{a}{b}}}$ where $a$ and $b$ are coprimes. Now let's look at the equation $(\frac{a}{b})^{{\frac{a}{b}}} = z$, clearly $(\frac{a}{b})^a = z^b$, so $\sqrt[a]{\frac{a^a}{b^a}} = \sqrt[a]{z^b}$. We conclude that $\frac{a}{b} = \sqrt[a]{z^b}$, this imples that both sides are rational, but the right side can only be rational if $z^{\frac{b}{a}}$ is rational, since $b$ and $a$ are coprimes, this cannot be true, so the assumption that the solutions are rational must be false.

This is my attempt to prove, it's clearly missing something. Does exist a more "technical" proof of the statement that every solution for this equation is irrational whenever $z$ is different of $4$?

• Let $k$ be any integer. And let $z = k^k$. Then $x^x = z$ will have an integer solution $k$. Sep 17, 2018 at 22:26
• "since b and a are coprime" If $b = 1$ and the solution is an integer this is not an issue. And indeed $x^x = k^k$ for any integer obviously has solutions. But if the rest of the proof is correct you may have prove for any integer not in the form $k^k$ there is no rational solution. I'm pretty sure that's true and your proof may be valid. Sep 17, 2018 at 22:32

$$27 = 3^3$$ $$3125 = 5^5$$
• Not to mention $1^1 = 1$. Sep 17, 2018 at 22:28
Your relation can be written $$a^a=b^az^b$$ If $a=1$, then we conclude $a=b=z=1$.
Let's assume $a>1$. Then $a$ is divisible by a prime and we conclude $b=1$, because no prime dividing $b$ can divide $a$. Thus the relation becomes $a^a=z$, which is certainly possible: just take any positive integer $a\ne2$ and define $z=a^a$.
What is true is that the equation $x^x=z$ has no rational solution if $z$ is not of the form $a^a$ for an integer $a$.