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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    
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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

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