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.

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