18

Here's the problem:

Find all pairs of rational numbers $(a,b)$ such that $0<a<b$ with $a^a=b^b$.

Using some calculus, I was able to find that $0<a<\frac{1}{e}<b<1$, and I also found one solution $(a,b) = (\frac{1}{4}, \frac{1}{2})$. I thought it would be the only solution, but I couldn't prove it. How can I go further and solve this problem?

user62498
  • 3,556
number0316
  • 369
  • 2
    It's not the only solution: $(a,b)=(\frac{8}{27},\frac49)$ works. – A. Goodier Apr 22 '23 at 09:05
  • 9
    Writing $b=ax$ for some $x>1$ and taking logs, we can solve for $a$ and $b$ in terms of $x$ and parameterise the solution by $a=x^{\frac{x}{1-x}}, b=x^{\frac{1}{1-x}}$. Since $x=b/a$, we know $x$ is rational, so an equivalent problem is to find the rational $x>1$ for which $x^{\frac{1}{1-x}}$ is rational. Letting $x=(n+1)/n$ for $n$ a positive integer, we have infinitely many solutions. – A. Goodier Apr 22 '23 at 09:14
  • Interestingly enough, there seems to be a close connection to a cousin problem $x^y = y^x$, however, I am still in progress to figure it out mathematically... Nonetheless, have play around on Desmos: (https://www.desmos.com/calculator/cqejfjupdm) – Dstarred Apr 22 '23 at 10:23
  • Isn't this a duplicate of https://math.stackexchange.com/questions/4029626/what-are-all-real-solutions-of-equation-xx-yy-what-about-integer-solution ? – Gerry Myerson May 18 '23 at 23:09

2 Answers2

22

Write $b=ax$ for some $x>1$. Then $x=b/a$ so $x$ is rational.

Then taking logs, we get $$a\log a=b\log b=ax\log(ax)\implies \log a=\frac{x\log x}{1-x} \implies a=x^{\frac{x}{1-x}}, b=x^{\frac{1}{1-x}}$$ since $x\neq1$ and $a\neq0$.

Since $x$ is rational, it suffices to find the values of $x>1$ for which $x^{\frac{1}{1-x}}$ is rational.

We claim that the only such values of $x$ are $x=\frac{n+1}{n}$ where $n$ is an integer.

We may write $x=1+\frac1n$ where $n\in\mathbb{Q}$, $n>0$. It is easy to see if $n$ is a positive integer that $x^{\frac{1}{1-x}}$ is rational. If $n$ is not an integer, write $n=p/q$ where $p,q\in\mathbb{Z}^+$ are coprime with $q>1$. We have $x^{\frac{1}{1-x}}=\left(\frac{n}{n+1}\right)^n=\left(\frac{p}{p+q}\right)^{p/q}$. Hence $p$ and $p+q$ must be $q$th powers.

No two $q$th powers can differ by $q$ since for positive integers $u,v,q$, we have by the Binomial Theorem, $$(u+v)^q-u^q=qu^{q-1}v+\ldots+v^q>q\,.$$ Therefore, if $n$ is not an integer, $x^{\frac{1}{1-x}}$ is not rational, so the only rational solutions are given by $$a=\left(\frac{n}{n+1}\right)^{n+1}, b=\left(\frac{n}{n+1}\right)^n, n\in\mathbb{Z}^+.$$

A. Goodier
  • 10,964
  • 3
    Interestingly, $a,b$ can be rewritten as $a_n= (1-\frac{1}{n+1})^{n+1}$ and $b_n=(1-\frac{1}{n+1})^n$ and $\lim_{n\to \infty} a_n =\lim_{n\to \infty} b_n = \frac{1}{e}$ which connects with the observation of OP that $a<1/e<b$. – Surb Apr 22 '23 at 09:43
  • 1
    It is clear from your proof that if $x$ is of the form $\frac{n+1}{n}$, then $n$ needs to be an integer; but is there a reason why $x$ needs to be of that form? Without that condition the two exponents $\frac{1}{1-x}$ and $\frac{x}{1-x}$ would not be integers; does that suffice to rule out any such solutions completely? – A.J. Apr 22 '23 at 23:14
  • @A.J. $x$ is a rational number greater than $1$, so there is a positive rational number $n$ such that $x=1+\frac1n$ . – A. Goodier Apr 23 '23 at 08:06
  • Ah, of course, thanks; that's the link I was missing. – A.J. Apr 23 '23 at 19:17
8

A Sequence of Solutions

In this answer, all the points where $x^y=y^x\iff x^{1/x}=y^{1/y}\iff (1/x)^{1/x}=(1/y)^{1/y}$ are given when $x=t^{\frac1{t-1}}$ and $y=t^{\frac t{t-1}}$. This gives $a$ and $b$, where $a^a=b^b$ when $a=t^{\frac1{1-t}}$ and $b=t^{\frac t{1-t}}$. If we set $t=\frac{n-1}n$, we get $$ \begin{align} a_n&=\left(\frac{n-1}n\right)^n\tag{1a}\\ b_n&=\left(\frac{n-1}n\right)^{n-1}\tag{1b} \end{align} $$ which are rational if $n\in\mathbb{N}$, and $$ {a_n}^{a_n}={b_n}^{b_n}=\left(\frac{n-1}n\right)^{n\left(\frac{n-1}n\right)^n}\tag2 $$

Monotonic Convergence

Furthermore, in this answer it is shown that $\left(\frac{n+1}n\right)^n$ is increasing and $\left(\frac{n+1}n\right)^{n+1}$ is decreasing. Taking reciprocals and substituting $n\mapsto n-1$, we get $a_n=\left(\frac{n-1}n\right)^n$ is increasing and $b_n=\left(\frac{n-1}n\right)^{n-1}$ is decreasing. That is, $$ a_n\lt\frac1e\lt b_n\tag3 $$

Completeness

We wish to show that the solutions given in $(1)$, with $n\in\mathbb{N}$, are all the rational solutions.

If both $a_n$ and $b_n$ are rational, then $\frac{n-1}n=\frac{a_n}{b_n}\in\mathbb{Q}$, and therefore, $n=\frac{b_n}{b_n-a_n}\in\mathbb{Q}$.

Let $n=\frac pq$ where $p,q\in\mathbb{N}$ and $(p,q)=1$. $\text{(1a)}$ becomes $$ a_n=\left(\frac{p-q}p\right)^{p/q}\tag4 $$ Bezout says there are $c,d\in\mathbb{Z}$ so that $cp+dq=1$. Since $a_n$ and $\frac{p-q}p$ are rational, $$ a_n^c\left(\frac{p-q}p\right)^d=\left(\frac{p-q}p\right)^{1/q}=\frac uv\tag5 $$ where $u,v\in\mathbb{N}$ and $(u,v)=1$. Since $0\lt\frac uv\lt1$, we have $0\lt u\lt v$.

$(5)$ implies $p=v^q$ and $p-q=u^q$. Since $u\ge1$ and $v-u\ge1$, we have $v\ge2$ and $$ \begin{align} q &=v^q-u^q\tag{6a}\\ &=\underbrace{\ (v-u)\ \vphantom{\left(q^1\right)}}_{\ge1}\underbrace{\left(v^{q-1}+uv^{q-2}+\dots+u^{q-1}\right)}_{\ge 2^q-1}\tag{6b} \end{align} $$ $q\ge2^q-1\implies q=1\implies n\in\mathbb{N}$.

robjohn
  • 345,667