# Positive rationals satisfying $a+b+c+d=abcd$?

Victor has posted a couple of problems involving finding real and rational solutions of $a+b+c=abc$. Two techniques have been given: using triangles, and using scaling. Neither seems to work for the following problem.

How can one easily (without brute force) characterize and produce the quadruples of positive rational numbers such that $a+b+c+d=abcd$?

The triangle technique doesn't seem to work for $n=4$. The scaling technique works, but doesn't necessarily give rational numbers.

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Why you can't just do $a + b + c = (abc - 1)d$ and so $d = (a+b+c)/(abc-1)$ for any $a$, $b$, and $c$? – Rahul Feb 19 '12 at 20:18
@RahulNarain: Yep! I showed this to my 10-year-old, and that's what she did. D'oh. Much simpler than the techniques we came up with in the other versions. – Ben Crowell Feb 19 '12 at 20:23
@RahulNarain: Since you posted first, maybe you could convert your comment into an answer, so I can accept it. – Ben Crowell Feb 19 '12 at 20:27
Since the answer is so trivial for the real and rational cases, I wonder if there is any nontrivial variation on this. The $n=4$ case over the integers might be hard. – Ben Crowell Feb 19 '12 at 20:31
It might not be so hard. The right side grows pretty fast, so that prevents the variables from getting very big. Certainly you can't have all of them be larger in absolute value than $\pm1$, for example. Beyond that, there's a lot of cases, but I imagine one could work through them in finite time. For what it's worth, there's an infinite family of integer solutions with $a=0$, $b+c+d = 0$. – Rahul Feb 19 '12 at 20:57

Write the equation as $a(bcd-1)=b+c+d$. For any positive rationals $b,c,d$ with $bcd>1$, $a=(b+c+d)/(bcd-1)$ is a positive rational and this gives you a solution. On the other hand, there is no positive rational solution with $bcd\le 1$.