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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?
Submitted by K Sengupta
Rating: 2.2000 (5 votes)
SOLUTION TO THE PROBLEM:
a^2 + b^2 can never be equal to c^2, in conformity with provisions of the problem.
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.
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.
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.
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)
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 ---------
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