A < B < C < D are four positive integers such that:
(i) B is the arithmetic mean of A and C and:
(ii) C is the
rms of B and D, and:
(iii) D-A = 50
Determine all possible quadruplets satisfying the given conditions and prove that there are no others.
More generally, instead of D - A = 50, let D - A = K
Then
(1) 2*(A+2n)^2 = (A+n)^2 + (A+K)^2
Expanding and solving for A gives
(2) A = (K^2 - 7n^2) / (6n - 2K)
The denominator is even, so the numerator must be even, so n must be the same parity (odd or even) as K
A is positive, so the numerator and the denominator must be of the same sign.
The numerator is negative if n > K/sqrt(7) ~ K*.378, and the denominator is negative if n < K/3 ~ K*.333, and this is impossible, so they are not both negative.
The numerator is positive if n < K*.378, and the denominator is positive if n > K*.333, and so the only possible n is between K/3 and k/sqrt(7). Numerically, n must be between K*.333 and K*.378
For instance, instead of K = 50, let K = 176
Then n must be even and between 58.7 and 66.5.
The only possible n are 60, 62, 64, and 66
Substituting in (2) shows that there are actually 3 solutions if K = 176
n = 60 --> (722, 782, 842, 898)
n = 64 --> ( 72, 136, 200, 248)
n = 66 --> ( 11, 77, 143, 187)
Edited on November 6, 2014, 5:09 pm
General solution ( D-A = K )
