# $x^y = y^x$ for integers $x$ and $y$

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

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,easy-to-prove 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:

$y|x \to {\alpha_i}\ge{\beta_i}$ or vice versa

Using the Theorem #1 we can get that $y|x$ 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={(y-1+1)}^t={(y-1)}^t+...+\binom{t}{2} {(y-1)}^2 + \binom{t}{1}(y-1) +1 \gt \binom{t}{2} {(y-1)}^2 + t(y-1) +1$But we have$y-1\gt1$so:$y^t \gt \binom{t}{2} {(y-1)}^2 + t(y-1) +1= \frac {t(t-1)}{2} -t +1 +yt= \frac {(t-2)(t-1)}{2} + yt \gt yt$So it is proved that for$t\ge3$is not valid anymore.$\bullet$P.S: The equation is solved for positive integers yet the solution for all the integers is quite the same!(took me an hour to write this all,hope you like my solution) - I've collected some references, feel free to add more of them. (Some of them are taken from other answers. And, of course, some of them can contain further interesting references.) Online: Papers: • Michael A. Bennett and Bruce Reznick: Positive Rational Solutions to$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$, The American Mathematical Monthly, Vol. 74, No. 3 (Mar., 1967), pp. 298-300. jstor • Joel Anderson: Iterated Exponentials, The American Mathematical Monthly , Vol. 111, No. 8 (Oct., 2004), pp. 668-679. jstor • R. Arthur Knoebel, Exponentials Reiterated. The American Mathematical Monthly , Vol. 88, No. 4 (Apr., 1981), pp. 235-252 jstor, link Books: • Andrew M. Gleason, R. E. Greenwood, Leroy Milton Kelly: William Lowell Putnam mathematical competition problems and solutions 1938-1964, ISBN 0883854287, p.59 and p.538 • Titu Andreescu, Dorin Andrica, Ion Cucurezeanu: An Introduction to Diophantine Equations: A Problem-Based Approach, Springer, New York, 2010. Page 209 Searches: The reason I've added this is that it can be somewhat tricky to search for a formula or an equation. So any interesting idea which could help finding interesting references may be of interest. - Say$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. - 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)$. [end update] - For every integer$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$-2,-4$. - I edited the question. – Paulo Argolo Nov 9 '10 at 1:12 @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