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

Home > Just Math
Some Reals Sum Two (Posted on 2008-09-01) Difficulty: 3 of 5
Determine all possible real quadruplet(s) (P, Q, R, S) with P ≤ Q ≤ R ≤ S that satisfy this system of equations:

P + Q*R*S = 2, and:

Q + R*S*P = 2, and:

R + P*Q*S = 2, and:

S + P*Q*R = 2.

See The Solution Submitted by K Sengupta    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts Unreal! | Comment 4 of 6 |
I suspect that the solution will always be integral for any set of equations

P + Q*R*S = W, and:
Q + R*S*P = X, and:
R + P*Q*S = Y, and:
S + P*Q*R = Z

where W, X, Y are Z are integral.

Can anybody prove or disprove this assertion?

My thinking (which does not constitute a proof) is that if P = a/b (relatively prime), and Q = c/d (relatively prime), and R and S are integers, then

1) R*S cannot be a multiple of d, or otherwise
   (a/b) + (c/d)*R*S is not integral

2) R*S cannot be a multiple of b, or otherwise
   (c/d) + (a/b)*R*S is not integral

3) S*a*c must be a multiple of b*d
4) R*a*c must be a multiple of b*d

Offhand, this feels inconsistent.

Help, anybody?





  Posted by Steve Herman on 2008-09-01 15:23:28
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 (5)
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