(In reply to
Equation by Salil)
I was able to use your approach to come up with what I believe is an exhaustive proof that there are no solutions.
if we let
x=(a,b)
y=(a,c)
z=(b,c)
then we have
x^2+xy/(x,y)+xz/(x,z)+yz/(y,z)=199
now each of the terms in this summation is an integer of value at least 1. Thus we can immediately restrict 1<=x<=14
Now due to symmetry we can restrict z>=y. Finally, if we consider when y>=x then we have
x^2+xy/(x,y)+xz/(x,z)+yz/(y,z)>=x^2+y+2z
so now using the two following inequalities we can create a finite space of {x,y,z} triplets to test
x^2+y+2z<=199 and 1<=x<=14
doing so, I was only able to find the following possible solutions for {x,y,z}:
{1,2,98}
{1,6,96}
{1,8,38}
{1,18,90}
{1,22,88}
{1,36,54}
{1,66,66}
{2,3,63}
{2,39,39}
{3,2,46}
{3,10,40}
{4,9,21}
{5,4,22}
{5,6,24}
{7,10,10}
{9,2,10}
However, none of these are valid values for x,y,z as we require that gcd(y,z) divide x which all of these fail to satisfy. This requirement can be seen from that fact that gcd(y,z)y and gcd(y,z)z and thus gcd(y,z)a and gcd(y,z)b and thus gcd(y,z)gcd(a,b) or gcd(y,z)x.
Thus there can be no solution to this equation.

Posted by Daniel
on 20130321 01:15:39 