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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 $4$?

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    $\begingroup$ Let $k$ be any integer. And let $z = k^k$. Then $x^x = z$ will have an integer solution $k$. $\endgroup$
    – fleablood
    Sep 17, 2018 at 22:26
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    $\begingroup$ "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. $\endgroup$
    – fleablood
    Sep 17, 2018 at 22:32

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$$ 27 = 3^3 $$ $$ 3125 = 5^5 $$

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  • $\begingroup$ Not to mention $1^1 = 1$. $\endgroup$
    – fleablood
    Sep 17, 2018 at 22:28
  • $\begingroup$ I kind of confused integers with primes. When i saw what i wrote it was too late. there's even a 4 in the title, it's embarrassing to look at this question. I'm sorry. $\endgroup$
    – M.Silva
    Sep 17, 2018 at 22:33
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    $\begingroup$ @M.Silva I guess the lesson is to try to check statements you come across, perhaps by trying examples. For now, this will often be possible, problems will generally be intended for solution by hand. You might enjoy one I could not do for months math.stackexchange.com/questions/2895089/… I even had trouble thinking of a way to investigate as far as true/false. Finally I found a way to do a good example. $\endgroup$
    – Will Jagy
    Sep 18, 2018 at 0:13
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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 $a$.

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