We know that $2^4 = 4^2$ and $(2)^{4} = (4)^{2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?
6 Answers
This is a classic (and well known problem).
The general solution of $x^y = y^x$ is given by
$$\begin{align*}x &= (1+1/u)^u \\ y &= (1+1/u)^{u+1}\end{align*}$$
It can be shown that if $x$ and $y$ are rational, then $u$ must be an integer.
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9I know about this curve, but I haven't seen that parametrization before. +1 for that! – J. M. ain't a mathematician Nov 09 '10 at 02:19

4The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See [this](http://www.jstor.org/pss/4145040) for instance.) – J. M. ain't a mathematician Dec 04 '11 at 13:20

@J.M.: Thanks..! – Aryabhata Dec 20 '11 at 20:58

Dude, long time no see! Sure hope you're fine. – J. M. ain't a mathematician Dec 20 '11 at 23:52

@J.M: I am fine (was away on vacation!) :) Thanks for asking! – Aryabhata Dec 21 '11 at 15:52

8interesting... as u goes to infinity, x and y go to e – user541686 Apr 08 '13 at 19:11

@Mehrdad: If you plot $x^{1/x}$, you will see why :) – Aryabhata Apr 08 '13 at 19:16

I just tried it, but I don't see why unfortunately... It seems to approach 1 as x approaches infinity but how is that related? – user541686 Apr 08 '13 at 20:40

@Mehrdad: To clarify what I meant: the roots of $x^y = y^x$ satisfy $x^{1/x} = y^{1/y}$. $x^{1/x}$ is increasing till $e$ and then decreases (which you can see by plotting or other means). $x,y$ must be on either side of $e$, and as $u \to \infty$ they move closer... – Aryabhata Apr 08 '13 at 20:56

@Aryabhata: Ah okay cool, thanks – user541686 Apr 08 '13 at 21:47
For every integer $n$, $x = y = n$ is a solution. So assume $x \neq y$.
Suppose $n^m = m^n$. Then $n^{1/n} = m^{1/m}$. Now the function $x \mapsto x^{1/x}$ reaches its maximum at $e$, and is otherwise monotone. Thus (assuming $n < m$) we must have $n < e$, i.e. $n = 1$ or $n = 2$.
If $n = 1$ then $n^m = 1$ and so $m = 1$, so it's a trivial solution.
If $n = 2$ then $n^m$ is a power of $2$, and so (since $m > 0$) $m$ must also be a power of $2$, say $m = 2^k$. Then $n^m = 2^{2^k}$ and $m^n = 2^{2k}$, so that $2^k = 2k$ or $2^{k1} = k$. Now $2^{31} > 3$, and so an easy induction shows that $k \leq 2$. If $k = 1$ then $n = m$, and $k = 2$ corresponds to $2^4 = 4^2$.
EDIT: Up till now we considered $n,m>0$. We now go over all other cases. The solution $n = m = 0$ is trivial (whatever value we give to $0^0$).
If $n=0$ and $m \neq 0$ then $n^m = 0$ whereas $m^n = 1$, so this is not a solution.
If $n > 0$ and $m < 0$ then $0 < n^m \leq 1$ whereas $m^n \geq 1$. Hence necessarily $n^m = 1$ so that $n = 1$. It follows that $m^1 = 1^m = 1$. In particular, there's no solution with opposite signs.
If $n,m < 0$ then $(1)^m (n)^m = n^m = m^n = (1)^n (m)^n$, so that $n,m$ must have the same parity. Taking inverses, we get $(n)^{m} = (m)^{n}$, so that $n,m$ is a solution for positive integers. The only nontrivial positive solution $2,4$ yields the only nontrivial negative solution $2,4$.
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2@Paulo: Yuval answered your question adequately. He showed that the pair $(2,4)$ are the only distinct integers which satisfy, and that any pair $(x,x), x \in \mathbb{N}$ is a solution. – Brandon Carter Nov 09 '10 at 01:17

@Brandon Carter: (2)^(4)= (4)(2). I just edit the question. Thank you for the comment. – Paulo Argolo Nov 09 '10 at 01:22
Say $x^y = y^x$, and $x > y > 0$. Taking logs, $y \log x = x \log y$; rearranging, $(\log x)/x = (\log y)/y$. Let $f(x) = (\log x)/x$; then this is $f(x) = f(y)$.
Now, $f^\prime(x) = (1\log x)/x^2$, so $f$ is increasing for $x<e$ and decreasing for $x>e$. So if $x^y = y^x$ has a solution, then $x > e > y$. So $y$ must be $1$ or $2$. But $y = 1$ doesn't work. $y=2$ gives $x=4$.
(I've always thought of this as the ``standard'' solution to this problem and I'm a bit surprised nobody has posted it yet.)
If $x > 0 > y$, then $0 < x^y < 1$ and $y^x$ is an integer, so there are no such solutions.
If $0 > x > y$, then $x^y = y^x$ implies $x$ and $y$ must have the same parity. Also, taking reciprocals, $x^{y} = y^{x}$. Then $(x)^{y} = (y)^{x}$ since $x$ and $y$ have the same parity. (The number of factors of $1$ we introduced to each side differs by $xy$, which is even.) So solving the problem where $x$ and $y$ are negative reduces to solving it when $x$ and $y$ are positive.
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Although this thing has already been answered, here a shorter proof
Because $x^y = y^x $ is symmetric we first demand that $x>y$ Then we proceed simply this way:
$ x^y = y^x $
$ x = y^{\frac x y } $
$ \frac x y = y^{\frac x y 1} $
$ \frac x y 1 = y^{\frac x y 1}  1 $
Now we expand the rhs into its wellknown exponentialseries
$ \frac x y 1 = \ln(y)*(\frac x y 1) + \frac {((\ln(y)*(\frac x y 1))^2}{2!} + ... $
Here by the definition x>y the lhs is positive, so if $ \ln(y) $ >=1 we had lhs $\lt$ rhs Thus $ \ln(y) $ must be smaller than 1, and the only integer y>1 whose log is smaller than 1 is y=2, so there is the only possibility $y = 2$ and we are done.
[update] Well, after having determined $y=2$ the same procedure can be used to show, that after manyally checking $x=3$ (impossible) $x=4$ (possible) no $x>4$ can be chosen.
We ask for $x=4^{1+\delta} ,\delta > 0 $ inserting the value 2 for y:
$ 4^{(1+\delta)*2}=2^{4^{(1+\delta)}} $
Take log to base 2:
$ (1+\delta)*4=4^{(1+\delta)} $
$ \delta =4^{\delta}  1 $
$ \delta = \ln(4)*\delta + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $
$ 0 = (\ln(4)1)*\delta + \frac { (\ln(4)*\delta)^2 }{2!} + \ldots $
Because $ \ln(4)1 >0 $ this can only be satisfied if $ \delta =0 $
So indeed the only solutions, assuming x>y, is $ (x,y) = (4,2)$ .
[end update]
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I've collected some references, feel free to add more of them. (Some of them are taken from other answers. And, of course, some of them can contain further interesting references.)
Online:
Nick's Mathematical Puzzles  Solution to puzzle 48: Exponential equation (Wayback Machine)
On Torsten Sillke's page: http://www.mathematik.unibielefeld.de/~sillke/PUZZLES/x%5Eyx%5Ey (Wayback Machine)
Wikipedia: Equation $x^y=y^x$
Papers:
Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to $x^y = y^{mx}$ : A NumberTheoretic Excursion, The American Mathematical Monthly , Vol. 111, No. 1 (Jan., 2004), pp. 1321; available at jstor, arxiv or at author's homepage.
Marta Sved: On the Rational Solutions of $x^y = y^x$, Mathematics Magazine, Vol. 63, No. 1 (Feb., 1990), pp. 3033, available at jstor. It is mentioned here, that this problem appeared in 1960 Putnam Competition (for integers)
F. Gerrish: 76.25 $a^{b}=b^{a}$: The Positive Integer Solution, The Mathematical Gazette, Vol. 76, No. 477 (Nov., 1992), p. 403. Jstor link
Solomon Hurwitz: On the Rational Solutions of $m^n=n^m$ with $m\ne n$, The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp. 298300. jstor
Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668679. jstor
R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235252 jstor, link
Books:
Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 19381964, ISBN 0883854287, p.59 and p.538
Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A ProblemBased Approach, Springer, New York, 2010. Page 209
Searches: The reason I've added this is that it can be somewhat tricky to search for a formula or an equation. So any interesting idea which could help finding interesting references may be of interest.
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Well I finally found an answer relating to some number theory I suppose !
Assume that : $x={p_1}^{\alpha _ 1}.{p_2}^{\alpha _ 2}...{p_k}^{\alpha _ k}$ it is clear that number y prime factors are the same as number x but with different powers i.e: $y={p_1}^{\beta _ 1}.{p_2}^{\beta _ 2}...{p_k}^{\beta _ k}$ replacing the first equation we get:
${({p_1}^{\alpha _ 1}.{p_2}^{\alpha _ 2}...{p_k}^{\alpha _ k})}^y={({p_1}^{\beta _ 1}.{p_2}^{\beta _ 2}...{p_k}^{\beta _ k})}^x$ i.e: ${p_1}^{{\alpha _ 1}y}.{p_2}^{{\alpha _ 2}y}...{p_k}^{{\alpha _ k}y}={p_1}^{{\beta _ 1}x}.{p_2}^{{\beta _ 2}x}...{p_k}^{{\beta _ k}x}$
Since the the powers ought to be equal we know for each $1\le i \le k$ we have:${\alpha_i}y={\beta_i}x$ i.e: ${\alpha_i}/{\beta_i}=x/y$
Considering that the equation is symmetric we can assume that $x \le y$ but we have ${\alpha_i}/{\beta_i} = x/y \ge 1$ hence ${\alpha_i} \ge {\beta_i}$
Assume this obvious,easytoprove theorem:
Theorem #1
Consider $x,y \in \mathbb{N}$ such that $x={p_1}^{\alpha _ 1}.{p_2}^{\alpha _ 2}...{p_k}^{\alpha _ k}$ $y={p_1}^{\beta _ 1}.{p_2}^{\beta _ 2}...{p_k}^{\beta _ k}$ for each $1\le i \le k$ we have:
$yx \to {\alpha_i}\ge{\beta_i}$ or vice versa
Using the Theorem #1 we can get that $yx$ i.e $x=yt$ replacing in the main equation we get:
$x^y=y^x \to ({yt})^y=y^{({yt})} \to yt=y^t$
Now we must find the answers to the equation $yt=y^t$ for $t=1$ it is obvious that for every $y \in \mathbb{N}$ the equation is valid.so one answer is $x=y$
Yet again for $t=2$ we must have $2y=y^2$ i.e $y=2$ and we can conclude that $x=4$ (using the equation $x=yt$)so another answer is $x=4$ $\land$ y=2$ (or vice versa)
We show that for $t\ge3$ the equation is not valid anymore.
If $t\ge3$ then $y\gt2$ we prove that with these terms the inequality $y^t \gt yt$ stands.
$y^t={(y1+1)}^t={(y1)}^t+...+\binom{t}{2} {(y1)}^2 + \binom{t}{1}(y1) +1 \gt \binom{t}{2} {(y1)}^2 + t(y1) +1$
But we have $y1\gt1$ so:
$y^t \gt \binom{t}{2} {(y1)}^2 + t(y1) +1= \frac {t(t1)}{2} t +1 +yt= \frac {(t2)(t1)}{2} + yt \gt yt$
So it is proved that for $t\ge3$ is not valid anymore.$\bullet$
P.S: The equation is solved for positive integers yet the solution for all the integers is quite the same!(took me an hour to write this all,hope you like my solution)
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