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.
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