# Problems by Ramanujan

Submitted to the Journal of the Indian Mathematical Society (JIMS).
mathematics
Published

August 15, 2019

## 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.