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.