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I am a high school teacher and last week we were given some example problems for the new End of Course Exam for Algebra 1.

One such is "$a^b=b$; $a \ne b$ Find a rational solution."

I know I can take $a=b^{\frac{1}{b}}$ and use any value of $b=\frac{1}{k}$ where $k$ is a positive integer greater than 1. to find $a$. So if I let $k=5$ then $(a,b)=(\frac{1}{5^5},\frac{1}{5})=(\frac{1}{3125},\frac{1}{5})$

Is there a more intuitive way to solve this that I am not seeing?

John Tompson
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    Your solution is probably the simplest to find a solution: $(\frac14,\frac12)$ is the simplest of these. It does not answer the question of whether all solutions are of this form. – Henry Jul 07 '14 at 13:20
  • Henry: I attempted to find a general-form solution, and ended up at the $b = \frac{1}{k}$ case, so I think that's actually the only solution. – cloudfeet Jul 07 '14 at 13:54

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OK, I started by expressing $b = \frac{x}{y}$, where $x$ and $y$ integers with no common factors, $x > 0$ and $y > 0$. (We can assume $b$ is not negative unless we're going to deal with imaginary numbers etc.).

This means that $a = (\frac{x}{y})^{\frac{y}{x}}$, or written differently: $a = (\sqrt[x]{\frac{x}{y}})^y$

Now, since $x$ and $y$ share no common factors, for $a$ to be rational, we need $\sqrt[x]{\frac{x}{y}} = \frac{\sqrt[x]{x}}{\sqrt[x]{y}}$ to be rational. Again, since they share no common factors, this means we need both $\sqrt[x]{x}$ and $\sqrt[x]{y}$ to be rational.

So: when is $\sqrt[x]{x}$ rational (for integer $x$)? It doesn't work for $x > 1$ (the smallest square number is > 2, the smallest cube number is > 3, etc, so the $x^{th}$ root of $x$ is never going to be rational. This means that your only possible value for $x$ is 1.

Given $x = 1$, any $y$ you choose is rational. So, you wind up at exactly your solution (where $y = k$):

$$b = \frac{1}{y}, a = (\frac{1}{y})^y$$

I don't know whether this is more or less intuitive, but I believe this shows that your family of solutions is the only one.

As mentioned in the comments, it might not be obvious that $\frac{\sqrt[x]{x}}{\sqrt[x]{y}}$ being rational implies $\sqrt[x]x$ and $\sqrt[x]y$ are rational. It only works because we're looking at roots of co-prime integers:

Suppose $\frac{\sqrt[z]{x}}{\sqrt[z]{y}}$ is rational, but $\sqrt[z]y$ is not (for integer $z$, and co-prime integers $x$ and $y$).

Dissolve $y$ into prime factors: $y = {p_1^{m_1}}{p_2^{m_2}}...$ - for $\sqrt[z]y$ to be irrational, we must have at least one $m_k$ that is not a multiple of $z$.

However, since $\frac{\sqrt[z]{x}}{\sqrt[z]{y}}$ is rational, when we factorise $x$ into primes, then we must end up with $x = q_1^{n_1}q_2^{n_1}...{p_k^{m_k}}$, so that the $p_k^{m_k}$ is "cancelled out". However, this implies a common factor between $x$ and $y$, which is a contradiction.

cloudfeet
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  • Thanks for your time. I was really discouraged because this is supposed to be an Algebra 1 standardized test question. I thought maybe I was missing something elementary. Thanks again. – John Tompson Jul 07 '14 at 14:28
  • @Vee The argument has a gap: $\ \frac{\sqrt[x]{x}}{\sqrt[x]{y}}\in \Bbb Q,\Rightarrow, \sqrt[x]{x},, {\sqrt[x]{y}}\in\Bbb Q\ $ requires proof. – Bill Dubuque Jul 07 '14 at 14:32
  • @Bill: I left it out for brevity, but can expand in the question if it's not obvious. You dissolve $x$ and $y$ into prime factors, so $\sqrt[x]{y} = \sqrt[x]{p_1}\sqrt[x]{p_2}...$, then say "If $\sqrt[x]{y}$ is not rational, $y$ must have a prime factor not repeated $x$ times". However, as $\frac{\sqrt[x]{x}}{\sqrt[x]{y}}$ is rational, that means $\sqrt[x]{x}$ must cancel out those prime factors - but that implies a common factor, which is not allowed. – cloudfeet Jul 07 '14 at 14:53
  • You're right, it looks misleadingly like I'm saying "If $\frac{a}{b}$ is rational, then $a$ and $b$ are rational" which would be patently false - it only works because we're looking at roots of co-prime integers. I'll add something to the answer... – cloudfeet Jul 07 '14 at 14:58