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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.

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For instance, $$\frac13+\frac13=\frac12+\frac16$$

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$$\begin{align}{1\over16}+{1\over48}&={1\over21}+{1\over28}={1\over18}+{1\over36}=\\{1\over20}+{1\over30}&={1\over15}+{1\over60}={1\over14}+{1\over84}\end{align}$$

I wonder if there are arbitrarily large sets like this. I would guess yes.

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.

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  • $\begingroup$ If $(a,b,c,d)$ is one such $4$-tuple, then certainly $(ma,mb,mc,md)$ is another. $\endgroup$ – Saucy O'Path May 19 at 12:16
  • $\begingroup$ @SaucyO'Path Yes, but they don't have the same sum. $\endgroup$ – saulspatz May 19 at 12:21
  • $\begingroup$ Ah, ok. Arbitrarily large meant arbitrarily large sets ot pairs with fixed harmonic sum. $\endgroup$ – Saucy O'Path May 19 at 12:24

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