let $y=\frac{m}{n}\cdot x$, where $(m,n)=1,\{m,n\}\subset\mathbb N.$ Hence$x^{\frac{m}{n}\cdot x}=(\frac{m}{n}\cdot x)^{x-\frac{m}{n}\cdot x}.$ Hence $x=(\frac{m}{n})^{\frac{n-m}{2m-n}}$ and $y=(\frac{m}{n})^{\frac{m}{2m-n}}.$ 1) $\frac{n-m}{2m-n}>0.$ Hence, $n=1$ and $\frac{1-m}{2m-1}>0.$ Hence, $\frac{1}{2}<m<1.$ This is contradiction. 2) $\frac{n-m}{2m-n}=0.$ Hence $m=n$ and $x=y=1$ 3) $\frac{n-m}{2m-n}<0.$ Hence,$x=(\frac{n}{m})^{\frac{n-m}{n-2m}}, y=(\frac{n}{m})^{\frac{m}{n-2m}},n>2m.$ Hence $m=1$ and $x=n^{\frac{n-1}{n-2}},y=n^{\frac{1}{n-2}},n\geq3.$ Let $f(t)=t^{\frac{1}{t-2}}, t>2.$ Hence, $f'(t)=f(t)\cdot\frac{\frac{t-2}{t}-lnt}{(t-2)^2}<0.$ Hence, $f(t)\leq f(3)=3$ and $y\leq3.$ Let $n=3.$ Hence $y=3$ and $x=9.$ Let $n=4.$ Hence $y=2$ and $x=8.$ Let $n>4.$ Hence $1<y<2.$ This is contradiction. Well, $\{(1,1),(9,3),(8,2)\}.$

Can someone post the original sollution please

edit: oops ! , a typo!

In reference with post # 4; Kyoshiro must have made a mistake in typing the answer at the last. He should have written, as the solutions are $ x = y = 1 ; y = 3,x = 9 ;y = 2,x = 8$ , is: $ (x,y) = (1,1),(8,2),(9,3)$

Step 1: First, If we take the $ y$ root of both sides, we have the equation: $ x = y^\frac{x - y}{y}$. The left side of obviously an integer. In order for the right side to be an integer, the power of $ y$ must be a positive integer, or $ \frac{x - y}{y}$ must be a positive integer. So, $ x-y$ must be a multiple of $ y$. Therefore $ x$ has to be a multiple of $ y$. If $ x$ is a multiple of $ y$, then we have the equation "$ x$ is equal to $ y$ times a number", or $ x=ay$. Step 2: Now we substitute $ x$ for $ ay$ in the original equation: $ x^y = y^{x - y}$. $ (ay)^y = y^{ay - y}$. $ a^yy^y = \frac{y^{ay}}{y^y}$. $ a^yy^{2y} = y^{ay}$ $ ay^2=y^a$ $ a=\frac{y^a}{y^2}$ and it finally reduces to... $ a=y^{a-2}$ Step 3: We now have an equation that allows us to easily find the corresponding $ x$ value for a given $ y$ value. We use logic to finish the problem. Evidently, if we plug in $ 1$ for $ y$, $ a$ equals $ 1$ also. So our $ x$ $ y$ ordered pair is $ (1,1)$. If we plug in $ 2$ for $ y$, $ a$ equals $ 4$ also. So our $ x$ $ y$ ordered pair is $ (2,8)$. If we plug in $ 3$ for $ y$, $ a$ equals $ 3$ also. So our $ x$ $ y$ ordered pair is $ (9,3)$. I'm not going to explain the rest, but if you use logical reasoning, you can tell that no other value for $ y$ greater than $ 3$ will have a positive integer value for $ a$. So we have three ordered pairs, and we're done. If you want me to explain anything, just post or pm me.

liltobe9 wrote: Evidently, if we plug in $ 1$ for $ y$, $ a$ equals $ 1$ also. So our $ x$ $ y$ ordered pair is $ (1,1)$. If we plug in $ 2$ for $ y$, $ a$ equals $ 4$ also. So our $ x$ $ y$ ordered pair is $ (2,8)$. If we plug in $ 3$ for $ y$, $ a$ equals $ 3$ also. So our $ x$ $ y$ ordered pair is $ (9,3)$. liltobe9, your second ordered pair needs to be changed to $ (8,2)$.

liltobe9 wrote: First, If we take the $ y$ root of both sides, we have the equation: $ x = y^\frac {x - y}{y}$. The left side of obviously an integer. In order for the right side to be an integer, the power of $ y$ must be a positive integer, or $ \frac {x - y}{y}$ must be a positive integer. Why does the power of $ y$ have to be a positive integer? $ 4^{\frac{1}{2}} = 2$.

I have a different way to show that $ x=ay$ for integer $ a$. Clearly $ x,y$ must have the same primes dividing them. Let $ p$ be a prime dividiing $ x,y$, and let $ m,n$ be the exponents of $ p$ in $ x,y$ respectively. In order for the equation to hold, it must be true that the exponents of $ p$ on both sides of the equation are the same. Then we have: $ my = n(x-y)$ $ \frac{m}{n} = \frac{x-y}{y}$ Case 1: $ x>2y$ so $ \frac{m}{n} > 1$. This tells us that for all $ p$, the exponent of $ p$ in $ x$ is greater than that in $ y$. So in this case, $ y=ax$. Substituting this and it is fairly clear there are no solutions. Case 2: $ x=2y$ so $ \frac{m}{n} = 1$. This means $ x=y$ and it is easy to see that the only solution is $ (x,y)=(1,1)$ Case 3: $ x < 2y$ so $ \frac{m}{n} < 1$. This means that for all $ p$, the exponent of $ p$ in $ x$ is less than that in $ y$. So in this case, $ x=ay$. We then continue in the same way as litobe9 to get that: $ (x,y) = (1,1), (8,2), (9,3)$

Zhero wrote: liltobe9 wrote: First, If we take the $ y$ root of both sides, we have the equation: $ x = y^\frac {x - y}{y}$. The left side of obviously an integer. In order for the right side to be an integer, the power of $ y$ must be a positive integer, or $ \frac {x - y}{y}$ must be a positive integer. Why does the power of $ y$ have to be a positive integer? $ 4^{\frac {1}{2}} = 2$. Zhero: you didn't look at the equation properly. You're right if $ y$ is a perfect square and we have the square root, or if $ y$ is a perfect cube and we have the cube root. However, we have the $ y$ root of $ y$. There is no value of $ y$ for which the $ y$ root of $ y$ is an integer, except for $ 1$. In this instance, the power of $ y$ must be a positive integer or else $ y^\frac {x - y}{y}$ cannot be a positive integer. Note that if $ y$ is $ 1$, the power is still an integer. We cannot use $ 4^{\frac {1}{2}}$, it would have to be $ 4^{\frac {1}{4}}$.

$ 4^{\frac{6 - 4}{4}} \in \mathbb{Z}$

As for the logical reasoning described in liltobe9's answer, one could say the following: If $a\neq2, a=y^{a-2} \Rightarrow y=\sqrt[a-2]{a}$ which also means that $a-2>0 \Rightarrow a>2$ and since $a\in Z \Rightarrow a\ge3$ However we know from the AM-GM inequality that $ \frac{a+1+1+...+1}{a-2}\ge\sqrt[a-2]{a}$. (There are $a-3$ 1's in the LHS) So now we have that $\frac{2a-3}{a-2}\ge\sqrt[a-2]{a}=y \Rightarrow 2+\frac{1}{a-2}\ge{y} (*)$ Since $3\le{a}<\infty \Rightarrow 1\le{a}-2<\infty \Rightarrow 1\ge\frac{1}{a-2}>0 \Rightarrow$ $3\ge2+\frac{1}{a-2}>0$ and from $(*) \Rightarrow 3\ge2+\frac{1}{a-2}\ge{y} \Rightarrow y\le{3}$

andreass wrote: As for the logical reasoning described in liltobe9's answer, one could say the following: If $a\neq2, a=y^{a-2} \Rightarrow y=\sqrt[a-2]{a}$ which also means that $a-2>0 \Rightarrow a>2$ Since $(x, y) = (1, 1)$ is a solution with $a = 1 < 2$, probably you should reconsider this argument.

Clearly $ x\ge\ 2y $ for $ x,y\ge\ 1 $. Let $ gcd(x,y)=d $ and we have $ x=dx_1, y=dy_1, gcd(x_1,y_1)=1 $. So the given equation is equivalent to $ d^yx_1^y=d^{(x-y)}y_1^{(x-y)} $ or $ x_1^y=d^{(x-2y)}y_1^{(x-y)} $. Now if $ y_1 $ has a prime factor $ p $, $ p $ must divide $ y_1^{(x-y)} $ a contradiction. So $ y_1 $ must be $ 1 $, or $ y=d $. The give equation now is equivalent with this equation $ x_1=d^{(x_1-2)} \ge\ 2^{(x_1-2)} $. By trivial induction, for $ x_1\ge\ 4, 2^{(x_1-2 })\ge\ x_1 $ and now is easy to see that the solutions are $ (x,y) = (1,1),(8,2),(9,3) $

andrejilievski wrote: Clearly $ x\ge\ 2y $. Since (1, 1) is a solution, clearly not.

brothers I am posting my solution. it is clear that x>=y case 1: x=y, x^x = 1 or, x = 1, so (x,y) = (1,1) case 2: x>y lemma : x = a*y where a>1 and a is positive integer. proof : x^y = y^(x - y) or, x = y^[(x - y)/y] as x>y and x,y are integers, (x - y)/y is integer so, y|x x = a*y where a>1 and a is positive integer.(proved) back to our main problem (a*y)^y = y^[y*(a - 1)] or, (a^y)*y^(2*y) = y^(a*y) or, a*y^2 = y^a or, y^(a - 2) = a; it is clear that a>2 if y>4 by induction we find that y^(a - 2)>a so, y<4 then casework

So arqady's solution can be strengthened with the following theorem: For positive integer x,y,a,b we have that $x^a = y^b \Rightarrow $ x|y or y|x. From this we split the problem into two cases, namely x|y and y|x. The former case yields that x=y, which yields (1,1). For the latter case, we let x = ky which yields $ y = k^{\frac{1}{k-2}} $, which gives us the rest of the solutions since we only have to check k = 3,4 (which both work) and can then use x > $2^x-2$ to prove that these are the only solutions. (k = 3 and 4 give the other two solutions and the inequality with 2's in it holds only for x>4, which essentially eliminates every other possibility for k.)

Find all pairs of positive reals $(x,y)$ satisfying the equation. Find all pairs of positive rationals $(x,y)$ satisfying the equation.

Another way to show $x$ is a multiple of $y$ when $x>y$: Let $x=y+a, a>0$. Then $(y+a)^y=y^a \Rightarrow y^y+\cdots +a^y=y^a$. So $a>y$ and $y | a^y$. So $y|a$, hence $y |x$.

Main equation is \begin{align} x^y = y^{x-y} \end{align} If $x=y$, then $(x,y)=(1,1)$ is a solution. If $x<y$, then $y^{x-y} \not\in \mathbb Z^+$ and there is no solution. Now, let's examine the case: $x>y>1$. In this case, $x^y = y^{x-y} < x^{x-y}$ and $y<x-y \implies x>2y$. If $\gcd(x,y)=d$, then $x=ad$, $y=bd$, $\gcd(a,b)=1$ such that $a,b \in \mathbb Z^+$. Rewrite these in $(1)$, $a^yd^y = b^{x-y}d^{x-y} \implies a^y = b^{x-y}d^{x-2y}$. Therefore $b \mid a^y$. On the other hand, $1=(a,b)=(a^y, b)= b$ and we yields $b=1$, $y=d$. Hence we find that $$ a=d^{a-2}, \qquad x=d^{a-1}, \qquad y=d $$ Our main idea: Because of $d^{a-2}$ will be very fast increasing, equation $a=d^{a-2}$ has very limited solutions. We have to search the solutions of $a=d^{a-2}$ for $d\geq 1$ and $a>2$. $\bullet$ For $d=1$, $a>2$, there is no solution. $\bullet$ For $d=2$, $a\in \{ 4, 8, 16, 32, \dots \}$. Only $a=4$ satisfy. We find the solution $(x,y)=(8,2)$. $\bullet$ For $d=3$, $a\in \{ 3, 9, 27, 81, \dots \}$. Only $a=3$ satisfy. We find the solution $(x,y)=(9,3)$. Now, let's examine the case: $d\geq 4$. Then, $a=d^{a-2}\geq 4^{a-2}$ and $16a \geq 4^a$. For $a=3$, $16\cdot 3 \geq 4^3$ is false. Also, for $a\geq 4$, inequality $16a \geq 4^a$ can't satisfy. Because $4^a$ faster increase than $16a$. Therefore, there is no solution for the case $d\geq 4$. Hence, all $(x,y)$ solutions of the main equation are $(1,1)$, $(8,2)$, $(9,3)$.