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.)
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
then we have

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
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}:

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

Please log in:
Remember me:
Sign up! | Forgot password

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

Copyright © 2002 - 2018 by Animus Pactum Consulting. All rights reserved. Privacy Information