Curious Real Additive Relationship II (Posted on 2008-05-28)

	Submitted by K Sengupta
Rating: 2.5000 (2 votes)
Solution:	(Hide)
(P, Q, R, S) = (Ã19, Ã19, Ã19, Ã19), (-Ã19, -Ã19, -Ã19, -Ã19) are the only possible solutions to the given problem. For an explanation, refer to the solution submitted by Praneeth at this location. Solution To The Bonus Question: P + M/P = (ÃP – ÃM/ÃP)2 + 2ÃM or, P + M/P ³ 2ÃM , whenever P is positive, with equality being satisfied whenever P= ÃM. Thus, 2*Q ³ 2*ÃM, giving: Q ³ ÃM. In a similar manner, R ³ ÃM, S ³ ÃM and: P ³ ÃM ……..(i) If P is negative, then it follows that: -P – M/P ³ 2*ÃM or: P + M/P ² -2*ÃM. Thus, Q ² -ÃM, R ² -ÃM, S ² -ÃM and: P ² -ÃM………(ii) Now, if each of P, Q, R and S is > ÃM, then each of P-M/P, Q-M/Q, R-M/R and S-M/S is >0. Similarly, if each of P, Q, R and S is < -ÃM, then each of P-M/P, Q-M/Q, R-M/R and S-M/S is < 0 Also, if (P, Q, R, S) = (ÃM, ÃM, ÃM, ÃM), or:(-ÃM, - ÃM, -ÃM, -ÃM), then: P-M/P = Q-M/Q = R-M/R = S-M/S = 0 Now, summing over the 4 given equations and rearranging, we have: (P- M/P) + (Q – M/Q) + (R – M/R) + (S-M/S) = 0 Now, in terms of (i) and (ii), the lhs of the above equation is always nonzero, unless: P = Q = R = S = ÃM, or: P= Q= R= S = -ÃM. Consequently, (P, Q, R, S) = (ÃM, ÃM, ÃM, ÃM), (-ÃM, -ÃM, -ÃM, -ÃM) are the only possible solutions to the bonus question.

	Subject	Author	Date
	re: Solution	K Sengupta	2008-07-28 11:57:58
	Solution	Praneeth	2008-05-30 07:12:49
	More generally ...	Steve Herman	2008-05-28 17:01:29
	Aha! (spoiler)	Steve Herman	2008-05-28 16:56:56
	An obvious start (spoiler)	Steve Herman	2008-05-28 13:56:14