Consider three positive integers x< y< z in
Harmonic Sequence.
Determine all possible values of the positive integer constant S for which the equation 15x + Sy = 15z admits of valid solutions.
For any sequence of three positive harmonic integers, there will always be some values a, b, and k such that the sequence can be expressed as k/(a-b) > k/a > k/(a+b).
Subsituting that parameterization into 15x + Sy = 15z yields:
15k/(a+b) + S*k/a = 15k/(a-b).
Simplifying this equation yields:
S(a^2-b^2) = 30a*b
The trick to solving this problem is to square that equation:
S^2*(a^2-b^2)^2 = 900a^2*b^2
Rearranging this new equation yields:
S^2*(a^2+b^2)^2 = (2a*b)^2*(S^2 + 15^2)
Since the left hand side is a perfect square, then the right side must also be a perfect square, which means the possible values of S are limited to values which make sqrt(S^2+15^2) an integer.
There are only four of those values: 8, 20, 36, 112. A brute force search finds that all four of those values occur in solutions of the original equation, so the solution is S is a member of {8, 20, 36, 112}
I think I got it
