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Curious Real Additive Relationship II (Posted on 2008-05-28) Difficulty: 2 of 5
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?

See The Solution Submitted by K Sengupta    
Rating: 2.5000 (2 votes)

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Solution Aha! (spoiler) | Comment 2 of 5 |
I just recognized the process for iterating to find the square root of M.

Define a Series S(n)

Let S(n+1) = (S(n) + M/S(n))/2.
As long as S(1) <> 0, this converges very quickly to sqrt(M).

If s(1) <> sqrt(M), then |S(n+1) - sqrt(M)| < |S(n) - sqrt(M)| for n > 2. 

(If s(1) < 1 and M > 1, then this inequality does not hold for n = 1, but it holds thereafter.  For instance, if M = 16 and s(1) = .25, then the series is
1  0.25
2  32.125
3  16.31152724
4   8.646214322
5   5.248367593
6   4.148467273
7   4.002656708
8   4.000000882
and each term starting with the 3rd is between 4 and the term before it)

So what does this have to with our problem?

Well, if P*P > M, then
P*P > Q*Q > R*R > S*S > P*P, which is a contradiction

Similarly, if P*P < M, then
P*P < Q*Q < R*R < S*S < P*P, which is a contradiction

So, the only solution is the one I have already given,
P = Q = R = S = +/- sqrt(M)

And yes, I know I haven't done the algebra, but I could and it would work.

-------------------------------------------------------------
Or, to be more clear, if P*P <> M,
then Q*Q is between P*P and M,
 and R*R is between Q*Q and M,
 and S*S is between R*R and M,
 and P*P is between S*S and M,
but this is a contradiction, so P*P = M


Edited on May 28, 2008, 5:11 pm

Edited on May 29, 2008, 9:31 am
  Posted by Steve Herman on 2008-05-28 16:56:56

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