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 Harmonic Integers (Posted on 2006-02-07)
Given that, a, b, and c are all positive integers so that a < b < c, and 1/a, 1/b, and 1/c are in arithmetic progression, can a² + b² be equal to c²?

 See The Solution Submitted by K Sengupta Rating: 2.2000 (5 votes)

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 I think I have it | Comment 1 of 5

I started by setting 1/b + k = 1/c, and 1/a +2k = 1/c, then rewrote these in terms of a, and b to get:

a = c/(1-2ck)

b=c/(1-ck)

then substituted these into the equation a2 + b2 = c2.

After I reorganized the equation I ended up with:

4(ck)4 - 6(ck)3 + 8(ck)2 - 1 = 0.

Solving for (ck) gave me four solutions.  Two were complex, and the other two were real, but irrational.  Since both c and k must be integers to fit the requirements of the problem, and since the product of two integers can not be irrational nor complex, then the answer to the question posed by K Sengupta must be:

No.   a2 + b2 can not be equal to c2.

Edited on February 8, 2006, 9:19 am
 Posted by Mindrod on 2006-02-07 22:52:15

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