Use the properties of gcd that (m,n) = (n,m) and (m,np) = (m,n)*(m,p).
Let (a,b) = d, (a,c) = e, (b,c) = f.
Then
(a^2,b^2) = d^2
(a,bc) = (a,b)*(a,c) = de
(b,ac) = (b,a)*(b,c) = df
(c,ab) = (c,a)*(c,b) = ef
and the equation becomes d^2 + de + df + ef = 199
Writing the last 3 terms as (e+d)(f+d)-d^2 reduces the equation to
(e+d)(f+d) = 199
e,f,d are each >=1 so the sum of two of them is >=2.
But 199 is prime which requires one of the left-hand factors to equal 1. This contradiction shows there are no solutions to the original equation.
Frankly I find that result anticlimactic. Maybe I've made an error.
|
Posted by xdog
on 2013-03-14 15:27:43 |