All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
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    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
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
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information