Does the equation 1/a + 1/b + 1/c + 1/(abc) = m/(a + b + c) have infinitely many solutions in positive integers a, b, c for any positive integer m?
Answer
Yes, eg m = 12.
Solution
Consider m = 12. Note that 1/a + 1/b + 1/c + 1/abc - 12/(a+b+c) = p(a,b,c)/(abc(a+b+c)), where p(a,b,c) = a2(b+c) + b2(c+a) + c2(a+b) + a + b + c - 9abc. We can regard this as a quadratic in a. The product of the roots is (b2c + c2b + b + c)/(b+c) = (bc + 1), so if one root is a, then the other is (bc + 1)/a. Moreover if a <= b <= c, then (bc+1)/a > c(b/a) >= c. Thus provided (bc+1)/a is integral, the solution (a,b,c) leads to another solution (b,c,(bc+1)/a).
We can start with the obvious solution (1, 1, 1). So define the sequence a1, a2, a3, ... defined by a1 = a2 = a3 = 1, an+3 = (an+2an+1 + 1)/an. Provided we can show that (an+2an+1 + 1)/an is always integral, we are home, because any three consecutive terms of the sequence give a solution.
It is convenient to enlarge the induction hypothesis to: an divides an+2an+1 + 1 (1); an+1 divides an + an+2 (2); and an+2 divides anan+1 + 1 (3). It is obviously true for n = 1. Suppose it is true for n.
(1) implies an and an+1 are coprime. We have an+2an+3 + 1 = an+2(an+1an+2 + 1)/an + 1 = k/an, where k = (an+2 + an) + an+1an+22. By (2) an+1 divides k, and by (1) an does, so they both do, which establishes (1) for n+1.
Similarly, an+1 + an+3 = an+1 + (an+1an+2 + 1)/an = (an+1an + 1) + an+1an+2 and an+2, an are coprime by (1), so an+2 divides an+1 + an+3, which establishes (2) for n+1. Finally, (3) for n+1 is obvious.
So we have established that a1, a2, a3, ... is an integer sequence. A trivial induction shows it is strictly increasing from a4 onwards, so it generates infinitely many solutions.
(C) John Scholes
jscholes@kalva.demon.co.uk
8 Aug 2003
Last corrected/updated 8 Aug 2003