# If $a, b, c,$ and $d$ are integers and $\frac{1}{a} + \frac{1}{b} = \frac{1}{c}+\frac{1}{d}$ where both sides of the equation be the same

If we have integers $$a$$, $$b$$, $$c$$, $$d$$ and $$\frac{1}{a} + \frac{1}{b} = \frac{1}{c}+\frac{1}{d}$$,when will both sides of the equation be the equal.

So what that means is that for integers $$a, b, c,$$ and $$d$$ will it be possible to come up with $$c$$ and $$d$$ so that they are not the same as $$a$$ and $$b$$. Can somebody please tell me if this is possible and also give me an example if it is.

For instance, $$\frac13+\frac13=\frac12+\frac16$$
\begin{align}{1\over16}+{1\over48}&={1\over21}+{1\over28}={1\over18}+{1\over36}=\\{1\over20}+{1\over30}&={1\over15}+{1\over60}={1\over14}+{1\over84}\end{align}
In Theorem 2 of this paper it is shown that the number of solutions in positive integers $$x,y$$ to $${1\over x}+{1\over y}={1\over n}$$ is $$\tau(n^2)$$, where $$\tau(n)$$ is the number of positive divisors of $$n$$ so that there are indeed arbitrarily large sets of two-term Egyptian fractions with the same sum.
• If $(a,b,c,d)$ is one such $4$-tuple, then certainly $(ma,mb,mc,md)$ is another. – Saucy O'Path May 19 at 12:16