A Knot of Congruences

Problem:
--------
Prove (or disprove?) that the only solutions of

                   ab = c (mod a+b)
                   ac = b (mod a+c)
                   bc = a (mod b+c)

in positive coprime integers a,b,c are {1,1,1} and {5,7,11}.

Discussion:
----------- 
Note that if a,b,c are not required to be positive then there are many
coprime solutions, such as {299,-49,-401}.  Also, if a,b,c are positive
but not necessarily coprime, then there are many solutions, such as
{69,99,111}.

Since the congruences are symmetrical we can assign the parameter labels 
so that a < b < c.  In explicit form the system of congruences can be
written as 
                   ab = c + x(a+b)
                   ac = b + y(a+c)
                   bc = a + z(b+c)

where x,y,z are integers.  These equations imply that

  / a-1 \  / b-1 \  / c-1 \     / a-1 \     / b-1 \     / c-1 \
 ( ----- )( ----- )( ----- ) - ( ----- ) - ( ----- ) - ( ----- ) = 2
  \  z  /  \  y  /  \  x  /     \  z  /     \  y  /     \  x  /

An equation of this form, i.e., ABC-A-B-C = 2, has the positive
integer solutions {2,2,2}, {1,2,5}, and {1,3,3}, and the integer 
solution {-1,-1,-1}.  However, not all triples {a,b,c} that satisfy 
the original set of congruences lead to integer quantities in the 
brackets.

Another interesting implication of the three coupled equations is

                         / c-b \     / b-a \
                        ( ----- ) + ( ----- )
       / c-a \           \ az  /     \  cx /
      ( ----- )  =   --------------------------
       \  by /                / c-b \  / b-a \
                       1  +  ( ----- )( ----- )
                              \  az /  \  cx /

which shows that the "normalized" distances [a to b] and [b to c] add
up to the total distance [a to c] in accord with the addition rule for
velocities in special relativity.  (This can be seen most clearly by
setting x=1/c, y=1/b, and z=1/a.)

For any integer solution A,B,C of the original congruences, define

     g = GCD(A,B,C)      a = A/g       b = B/g        c = C/g

and put M = LCM(a+b,a+c,b+c).  It's clear that a,b,c are pairwise
coprime and that infinitely many other solution triples are given by

         A' = a(Mk+g)      B' = b(Mk+g)      C' = c(Mk+g)

where k is any integer.  A solution {ag,bg,cg} with g < M is called
a minimal solution.

Examination of the minimal solutions with g=1 shows that certain values
of (a+b+c) occur frequently.  For example, the following triples all
have (a+b+c) = -361

                    a     b     c
                  ----- ----- -----
                    47   -53  -355
                   109  -169  -301
                   119  -131  -349
                   143  -145  -359
                   149  -121  -389
                  1619  -179 -1801

The basic equations imply that if a,b,c are coprime then a divides 
xy-1, b divides xz-1, and c divides yz-1, but this doesn't seem to be 
sufficient to prove that {1,1,1} and {5,7,11} are the only positive
coprime solutions.  It appears that for most (all?) g > 1 there are
infinitely many minimal solutions, but I can't prove that either.