## Problem 666

Solve in positive rational numbers

$$x^y = y^x$$

For example ,$x=4, y =2; x=3\frac{3}{8}, y = 2\frac{1}{4}$.

### Solution

Put $x=ky$. It follows that

$$y^{k-1} = k$$

It is easy to see that $k$ is rational if and only if $k = 1 + 1/n$ for some positive integer $n$.

Thus,

$$y = \left(1+ \frac{1}{n}\right)^n \text{ and } x = \left(1+ \frac{1}{n}\right)^{n+1}$$

When $n=1, x=4$ and $y=2$; when $n=2, x=\frac{27}{8}$ and $y= \frac{9}{4}$.

## Problem 785

Show that

\begin{align} (3 \lbrace (a^3+b^3)^{\frac{1}{3}}-a \rbrace \lbrace (a^3+b^3)^{\frac{1}{3}}-b \rbrace)^{\frac{1}{3}} \nonumber\\ = (a+b)^{\frac{2}{3}}-(a^2-ab+b^2)^{\frac{1}{3}} \label{id1} \\ (2 \lbrace (a^2+b^2)^{\frac{1}{2}}-a \rbrace \lbrace (a^2+b^2)^{\frac{1}{2}}-b \rbrace)^{\frac{1}{2}} \nonumber\\ = (a+b)-(a^2+b^2)^{\frac{1}{2}} \label{id2} \end{align}

### Solution

To prove $\ref{id1}$, in the identity

$(a+b-r)^3 = (a+b)^3 - r^3 - 3r(a+b)^2 + 3r^2(a+b)$

Put $r^3 = a^3 + b^3$.

Thus,

\begin{align*} (a+b-r)^3&=&3ab(a+b) - 3r(a+b)^2 + 3r^2(a+b) \\ &=&3(a+b)(r-a) (r-b) \end{align*}

Then divide both sides by $(a+b)$ and take the cube root of both sides.

To prove $\ref{id2}$, in the identity

$(a+b-r)^2 = (a+b)^2 - 2r(a+b) + r^2$

Put $r^2 = a^2 + b^2$.

Thus,

\begin{align*} (a+b-r)^2&=&2r^{2} + 2ab -2r(a+b) \\ &=&2(r-a)(r-b) \end{align*}

Then take the square root of both sides.