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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 $x^y = y^x$?

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5 Answers

up vote 33 down vote accepted

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.

For more details, see this and this.

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I know about this curve, but I haven't seen that parametrization before. +1 for that! –  J. M. Nov 9 '10 at 2:19
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The parametrization, it turns out, is due to Christian Goldbach, which he did in 1843. (See this for instance.) –  J. M. Dec 4 '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. 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
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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 $2^{k-1} = k$. Now $2^{3-1} > 3$, 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 non-trivial positive solution $2,4$ yields the only non-trivial negative solution $-2,-4$.

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I edited the question. –  Paulo Argolo Nov 9 '10 at 1:12
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@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 9 '10 at 1:17
    
Thank you very much! –  Paulo Argolo Nov 9 '10 at 1:18
    
@Brandon Carter: (-2)^(-4)= (-4)(-2). I just edit the question. Thank you for the comment. –  Paulo Argolo Nov 9 '10 at 1:22
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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 well-known exponential-series

$ \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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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 $x-y$, 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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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:

Papers:

  • Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to $x^y = y^{mx}$ : A Number-Theoretic Excursion, The American Mathematical Monthly , Vol. 111, No. 1 (Jan., 2004), pp. 13-21; 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. 30-33, 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. 298-300. jstor

  • Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668-679. jstor

  • R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235-252 jstor, link

Books:

  • Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 1938-1964, ISBN 0883854287, p.59 and p.538

  • Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A Problem-Based 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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