POLYA003:  Integers of the form a/(bc) + b/(ca) + c/(ab) (proposed by Jean-Pierre Ehrmann, 11/7/01).

For which positive integers  n  can we find three positive integers a,b,c such that n = a/(bc) + b/(ca) + c/(ab) ?

Discussion. [SCL, 11/11/01]: Assume that a\le b\le c and that a<c. Write a/bc+b/ac+c/ab=n as
a^2+b^2+c^2=abcn (1)
and consider (1) as a quadratic equation in c. There is a second solution, d=abn-c, such that a^2+b^2+d^2=abdn. Now, looking at the quadratic formula applied to (1), the roots are (abn\pm \sqrt{D})/2, where D=(abn)^2-4(a^2+b^2). Since (1) has at least one real root (c was given), we have a^2+b^2 \le abn/2, and
c^2=abcn-(a^2+b^2) \ge abcn-anb/2 \ge abcn-abcn/2=abcn/2. (2)
Hence, c\ge abn/2. Note that equality in (2) only holds if c=1, and this corresponds to a=b=c=1, n=3. Thus, we may assume that c>abn/2 and then d=abn-c<c.
Thus, we have a proof, by descent (on a+b+c), that all solutions lead to solutions with a=b=c. The only two solutions with a=b=c are a=b=c=1, n=3, and a=b=c=3 and n=1. This follows from 3a^2=a^3n, thus 3=an, and a=1 a=3.

[JPE, 11/13/01]:  Stephen gave us a nice proof of the fact that n = 1 or 3. Now, taking n = 3, your descent method leads to the values of a,b,c. If we suppose a<=b<=c and a^2+b^2+c^2 = 3 abc, then necessarily a =1,2 : (there must be a solution with b = c) .
For a = 1, b and c are two consecutive terms of the sequence u_0=u_1=1; u_(n+1)=3 u_n - u_(n-1).
For a = 2, b and c are two consecutive terms of the sequence u_0=u_1=1; u_(n+1)=6 u_n - u_(n-1)
If we look for the values of a,b,c if n = 1, we get 3*A,3*B,3*C where A,B,C is any solution corresponding to n = 3.

[FvL, 11/14/01]: I am not quite satisfied with the solution Stephen gave. I miss the solution with a=b=1, c=2, but I don't see immediately any flaw. So let me drop what I had in mind: We have the equation a^2 + b^2 + c^2 - nabc = 0. We can start with the assumption that a,b,c be coprime. If gcd(a,b)>1 then
also gcd(a,c)=gcd(a,b). But if pa, pb, pc give a solution for some n, then a,b,c give a solution for pn. So we assume a,b,c coprime. Suppose that a,b,c>1. Then
x^2 - (bcn)x + b^2 + c^2 = 0 has roots x=a and x=(b^2+c^2)/a. Inserting the second root gives   (a+1)(b^2 + c^2) = na^2bc. This shows that bc|a+1. Similarly ac|b+1 and ab|c+1. But then from

a+1   b+1   c+1
--- + --- + --- = m
bc    ac    ab

together with the equation in question, we conclude that  1/bc + 1/ac + 1/ab = m - n is an integer. That is clearly not possible for a,b,c coprime and unequal to
1. Allowing one of a,b,c equal to one, say c=1, we get in a similar fashion  x^2 - bnx + b^2 + 1 = 0 has roots x=a and x=(b^2+1)/1. Inserting the latter gives
(a+1)(b^2+1)=na^2b and b|a+1. Idem a|b+1. So (a,b)=(2,3) or (3,2). That does not yield a solution of the original equation. Hence we should have two 1's. Now it is easy to see that the only coprime solutions are:   a=b=c=1 ---> N=3;   a=b=1 c=2 ---> N=3. If we drop the coprime requirement, we get the real divisors of earlier found N=3, thus N=1.   a=b=c=3 ---> N=1;  a=b=3, c=6 ---> N=1.

[SCL, 11/14/01]: I don't see a problem with a=b=1, c=2. For these values, we get the quadratic: 2+c^2=3c, which has two solutions, c=1 and c=2. (Where c=2 was given.)  The equation d=abn-c=3-c then gives d=1, and the solution a=1,b=1,d=1, n=3.  I didn't put it in the proof, but it is obvious that in all cases, d>0,
since d is the sum/difference in the quadratic formula of two quantities which are obviously of different magnitudes. (One could also see that we never get d=0, by realizing that the equation a^2+b^2+d^2=abdn, with d=0, forces a=b=0. If a,b>0 to begin with, and we don't change them, then d<>0.)

[FvL, 11/15/01]: I see now that my approach contains flaws (inserting the second root wrongly) while Stephen's approach is correct. Sorry.

Bibliography. [1] POLYA010: Integers of the form bc/a + ca/b + ab/c, proposed by Floor van Lamoen.