Refer to the earlier problem.

Can you find all possible non zero real quadruplet(s) (P, Q, R, S) that satisfy the following system of simultaneous equations?

2*Q = P + 19/P, and:

2*R = Q + 19/Q, and:

2*S = R + 19/R, and:

2*P = S + 19/S.

Bonus Question:

In the problem given above, if the number 19 was replaced throughout by a real constant M > 1, what would have been the possible non zero real quadruplet(s) (P, Q, R, S) that satisfy the given set of equations, in terms of M?

if P≥0 P+19/P ≥ 2√19=> 
Q ≥19, P ≥√19, R ≥√19, S≥√19
if P<0 (-p-19/p)≥2√19 => p+19/p≤-2√19
Q≤-√19,P≤-√19,R≤-√19,S≤-√19

Add all the 4 equations
P+Q+R+S=19(1/P+1/Q+1/R+1/S)
(P-19/P)+(Q-19/Q)+(R-19/R)+(S-19/S)=0
Case1: (P-19/P)≥0,(Q-19/Q)≥0,(R-19/R)≥0,(S-19/S)≥0
So, P=Q=R=S=√19
Case2: (P-19/P)≤0,(Q-19/Q)≤0,(R-19/R)≤0,(S-19/S)≤0
So, P=Q=R=S=-√19

For any given M, P=Q=R=S=+/-√M is the solution

Posted by Praneeth on 2008-05-30 07:12:49