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Strange Equation (Posted on 2013-03-11) Difficulty: 3 of 5
Solve the equation
(a^2,b^2)+(a,bc)+(b,ac)+(c,ab)=199. in positive integers.
(Here (x,y) denotes the greatest common divisor of x and y.)

No Solution Yet Submitted by Danish Ahmed Khan    
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re: Equation | Comment 9 of 10 |
(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 2013-03-21 01:15:39

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