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 More Harmonic Integers (2) (Posted on 2007-03-28)
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.

 See The Solution Submitted by K Sengupta Rating: 2.5000 (2 votes)

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 I think I got it Comment 6 of 6 |

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}

 Posted by Brian Smith on 2007-03-30 13:10:34

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