Solution to Problem 248

Correct solutions came from BU students Jacob Seymour, Kylee Lorte, Linde Jones.  Correct solutions were also sent in by Philippe Fondanaiche, France; Farid Lian, Colombia; K. Sengupta, India; Ahron Teitelman, Israel; Lou Cairoli, USA.

There are infinitely many solutions.  Of course, we can take a = b = c, but where's the fun in that.  

This problem is much more difficult than I originally thought.  One of our regular correspondents, long-time wit, and pack-rat remembered first seeing the problem as an undergraduate; his professor referred the class to a solution by Paul Erdös and Ivan Niven of Problem E 682 in the American Mathematical Monthly, April 1946.  Here is that solution, somewhat filled in.

We prove there are infinitely many solutions in integers a,b,c.  First note that dividing a,b,c by their greatest common divisor provides another solution; without loss we can assume that a,b,c share no common divisor.  

Putting all terms over a common denominator, the fractions become
.
If p is a prime dividing one of a,b,c it must also divide another, since otherwise two ― and only two ― of the terms in the numerator are divisible by p, making the numerator not divisible by p and the fraction not an integer. Let's suppose that pα, α ≥ 0, is the largest power of p dividing a and pβ, β > 0, the largest power dividing b; note that  p doesn't divide c.  The fraction above takes the form
,
where the k's are integers not divisible by p.  The denominator is divisible by pα + β, so the numerator is divisible by at least  pβ.  The last two terms in the numerator are divisible by pβ so the first must be, too; it follows that 2α ≥ β.  If, on the other hand, 2α > β, then we may divide numerator and denominator by pβ to get the integer
;
visibly, the denominator is divisible by p while the numerator is not.  It follows that 2α = β; in particular, p2α divides b.  Let e be the greatest common divisor of a and b, it follows that e2 divides b.   Let f  be the greatest common divisor of b and c; g the greatest common divisor of c and a.  By symmetry we have
a = eg2, b = f e2, c = gf 2.
The original fraction becomes
.
We show this is an integer for infinitely many relatively prime integers e, f, g.  In fact, this is already true for g = 1 for if you have a solution with ef  dividing e3 + f 3 + 1, you get another solution from the pair of integers f , (f 3 + 1)/e, and this is a new solution whenever e < f .  

For example, start with the solution  e = 2, f = 3, g = 1 which produces a = 2, b = 12, c = 9.  This generates the solution with e = 3, f  = 14, g = 1 which produces a = 3, b = 126, c = 196, etc.

The Monthly article notes that the solution a = 1539, b = 1369634, c = 129549924, appeared in the Gazeta Matematica in 1931.  
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