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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