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

 Some Reals Sum Two (Posted on 2008-09-01)
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.)
 solution checking | Comment 2 of 6 |
(In reply to solution (spoiler) by Daniel)

Nice work, Daniel, except for 2 quibbles:

QUIBBLE 1:

1/s<=s DOES NOT IMPLY that s^2 >= 1, because you have to reverse the inequality when multiplying by a negative number.

What it does imply is that s >= 1 or 0 > s >= -1

QUIBBLE 2:

s>=1 q>=1 q<=s DOES NOT IMPLY s=q=1

HOWEVER:

It does not alter the solution, however, because (as you pointed out) when rs = 1, p + q = 2, so q must be >= 1.  Thus r and s must be >= 1, which can only occur if r = s = 1.  I think this simplifies your solution (as well as making it correct).

But it does not affect the final solution.

 Posted by Steve Herman on 2008-09-01 14:06:10

 Search: Search body:
Forums (0)
Random Problem
Site Statistics
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox: