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.
from (ii), 2*C^2 = B^2 + D^2
Let B = A + n, where n > 0
Then C = A + 2n
D = A + 50
Substituting gives
2*(A+2n)^2 = (A+n)^2 + (A+50)^2
Expanding and solving for A gives
A = (2500 - 7n^2) / (6n - 100)
The denominator is even, so the numerator must be even, so n must be even.
A is positive, so the numerator and the denominator must be of the same sign.
The numerator is negative if n > 18.9, and the denominator is negative if n < 16.6, and this is impossible, so they are not both negative.
The numerator is positive if n < 18.9, and the denominator is positive if n > 16.6, and we have already said that n must be even, so the only possible n is 18.
This leads to an integral A, as (2500 - 7*18^2)/(6*18-100) =
232/8 = 29.
So the only solution is the one already posted,
A = 29, n = 18 --> (29, 47, 65, 79)
Analytical solution (without quadratic equation)
