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How to find all real solution to satisfy this equation without casework or bruteforce?

a+b+c=abc

Thanks in advance!

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It would more fun if you would tell us something about where you came across a problem and why you're interested in it, and it would also be nice if you would tell us what thoughts you've had about how to solve it, e.g., "I tried (...), but it didn't work." –  Ben Crowell Feb 19 '12 at 17:09
    
Rahul Narain and my daughter independently came up with a much simpler method, which works for all $n$ and for both real and rational cases: math.stackexchange.com/questions/111040/… –  Ben Crowell Feb 19 '12 at 20:25

2 Answers 2

up vote 6 down vote accepted

Since $\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$ for $\alpha$, $\beta$, $\gamma$ being three angles in a triangle, pick any triangle and take the tangents of its three angles and you'll have a solution to your equation over the reals: $(\tan\alpha,\tan\beta,\tan\gamma)$.

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Since $\alpha$, $\beta$, and $\gamma$ are the three angles in a triangle, $\gamma=\pi-\alpha-\beta$, so $\tan\gamma=$ $\tan(\pi-\alpha-\beta)=$ $-\tan(\alpha+\beta)=$ $-\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$. Substitute that for $\tan\gamma$ on both sides of the identity and do some manipulation and the result falls out. –  Isaac Feb 19 '12 at 4:14
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Note that there are solutions such as $(0,0,0)$, $(-\sqrt{3},-\sqrt{3},-\sqrt{3})$, and $(\sqrt3,-1/\sqrt{3},-1/\sqrt{3})$ that won't be realized by taking tangents of the interior angles of a triangle. –  Ben Crowell Feb 19 '12 at 17:07

Start with any set of three numbers $a$, $b$, and $c$, let $s=\sqrt{(a+b+c)/(abc)}$, and let $a'=sa$, $b'=sb$, and $c='sc$. Then $(a',b',c')$ is a solution. The only solution that can't be found by this method is the trivial one, $(0,0,0)$.

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How are you able to derive this method? –  Victor Feb 19 '12 at 20:02
    
You can easily prove it by substitution. The idea is simply that the RHS and LHS scale differently, so by picking the scaling factor appropriately, you can always get them to be equal. –  Ben Crowell Feb 19 '12 at 20:24

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