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 Strange Equation (Posted on 2013-03-11)
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 No Rating

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 solution | Comment 1 of 10
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

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