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)