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Harmonic Integers (Posted on 2006-02-07) Difficulty: 3 of 5
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 a + b be equal to c?

  Submitted by K Sengupta    
Rating: 2.2000 (5 votes)
Solution: (Hide)
SOLUTION TO THE PROBLEM:

a^2 + b^2 can never be equal to c^2, in conformity with provisions of the problem.

Explanation:

By the problem, we are given three positive integers a < b < c which are in Harmonic Progression.

We shall prove that no integral solution exists for the equation given below:
a^P + b^P = c^P, where P is any positive integer.

PROOF:

Since a< b < c and a,b and c are in H.P., it follows that: b = 2ac/(a + c).

Now, a^P + b^P = c^P ..................(i)
or, a^P + (2ac/(a+c))^P = c^P
or, (c^P - a^P)*((a+c)^P) = (2ac)^P .........(ii)

Let 2^K ( K is a positive integer or zero) be the largest power of 2 dividing both a and c.
Accordingly, a = (2^K)*s and c = (2^K)*t, where t and s are positive integers and both t and s cannot be even for obvious reasons.........(iii)
From (ii), we obtain :

(t^P - s^P)*((s+t)^P = (2st)^P ..........(iv)

so that, LHS must be even implying inter-alia that s and t must possess the same parity, and so, both s and t must be odd in terms of (iii).

Consequently, LHS of (iv) must be divisible by 2^(P+1) so that,
LHS of (iv) = M*(2^(P+1)); for some integer M.

Hence,
Since, s and t possess different parities, it follows that R.H.S. of the above equation is even while the L.H.S. is odd.
Hence, M*(2^(P+1))=(2st)^P
or, 2*M = (st)^P, which is not feasible since both s and t must be odd.

Hence, no integral solution exists for the equation a^P + b^P = c^P.

Substituting P=2, we obtain our desired result.

------Q E D -------

ALTERNATE SOLUTION: ( Submitted By goFish)

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.

---------Q E D---------

A SECOND ALTERNATE SOLUTION: (Submitted By Mindrod with amendments as suggested by Eric and Richard)

" 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)

I then substituted these into the equation a^2 + b^2 = c^2.

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 c must be an integer while k must be a rational number 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. a^2 + b^2 can not be equal to c^2."

------Q E D ---------

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionThe ProofgoFish2006-02-08 20:16:09
re(3): I think I have itRichard2006-02-08 13:39:51
re(2): I think I have itMindrod2006-02-08 09:30:19
re: I think I have itEric2006-02-08 03:19:25
SolutionI think I have itMindrod2006-02-07 22:52:15
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