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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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 The Proof Comment 5 of 5 |

We recall that (a, b, c) is a pythagoran (a^2+b^2=c^2) triple if and only if

a = k( m^2 - n^2); b = k( 2 mn); c = k(m^2+n^2) with m>n and gcd(m,n)=1.

We assume that such (a,b,c) exists such that 1/a - 1/b = 1/b -1/c

This may be rewritten as

bc-ac=ac-ab or

b(c-a) = 2 a c and substituting we obtain

k (2mn)(k( m^2 + n^2 - m^2 +n^2)) = 2 k^2 (m^2-n^2)(m^2+n^2) or

mn ( 2 n^2 ) = (m^2-n^2)(m^2+n^2)

which implies that m| n and n|m: i.e n = m, contradicting both assumptions: that gcd(m,n)=1 AND m >n

Now suppose we let a = 2 mn and b = m^2-n^2 then bc - ac = ac - ab may be re-written as

(m^2-n^2-2mn)(m^2+n^2) = (b-a)c = a (c - b) = 2mn (2n^2)

which again implies that m| n and n|m: i.e n = m, contradicting both assumptions: that gcd(m,n)=1 AND m >n.

Hence (a,b,c) cannot be both a pythagoran triple and a harmonic progression.

 Posted by goFish on 2006-02-08 20:16:09

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