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        riddles >> medium >> HARMONIC POWERS?
        
(Message started by: K_Sengupta on Nov 21^st, 2005, 11:56pm)

Title: HARMONIC POWERS?
Post by K_Sengupta on Nov 21^st, 2005, 11:56pm

Prove either by using algebraic techniques or utilising number theoretic methods ( but strictly in accordance with High School Standard) that it is not feasible to determine THREE POSITIVE INTEGERS IN HARMONIC PROGRESSION with the seventeenth power of the largest integer being equal to the sum of seventeenth powers of the remaining integers .

Title: Re: HARMONIC POWERS?
Post by JocK on Nov 22^nd, 2005, 3:49am

Do you mean to say "with the seventeenth power of the largest integer being equal to the sum of seventeenth powers of the remaining integers"...?

I'm sure these days they tell secondary school pupils about the fact that Fermat's last theorem has finally been proven to be correct... ;D

Title: Re: HARMONIC POWERS?
Post by Eigenray on Nov 22^nd, 2005, 1:41pm

What Fermat's last theorem won't tell you is that the result still holds when 17 is replaced by 2, or even 1.

Suppose m,k,n are in harmonic progression; this means
k=2mn/(m+n).
Suppose
n^p = m^p+k^p.
[hide]Rewrite this as
(n^p-m^p)(m+n)^p = (2mn)^p,
and let 2^r be the largest power of 2 dividing both m and n. Thus m=2^rx, n=2^ry, with x,y not both even. Cancelling a factor of 2^2rp, we have
(y^p-x^p)(x+y)^p = (2xy)^p.
Since the RHS is even, x and y must have the same parity. But then the LHS is divisible by 2^1+p, while the RHS isn't.[/hide] Contradiction.

Title: Re: HARMONIC POWERS?
Post by JocK on Nov 22^nd, 2005, 1:58pm

on 11/22/05 at 13:41:05, Eigenray wrote:

What Fermat's last theorem won't tell you is that the result still holds when 17 is replaced by 2, or even 1.

Nice.

Makes an elegant variation to the above problem: "show that three integers can not be in arithmetic progression and in harmonic progression".

Title: Re: HARMONIC POWERS?
Post by Eigenray on Nov 22^nd, 2005, 2:38pm

on 11/22/05 at 13:58:32, JocK wrote:

Makes an elegant variation to the above problem: "show that three integers can not be in arithmetic progression and in harmonic progression".

Well, that's a different problem altogether.
If m, k, n, is in both AP and HP, then m=k=n follows from the AM-HM inequality [(m+n)/2=2mn/(m+n) becomes (m-n)²=0].

On the other hand, replacing 17 by 1, the result (that n=m+k is impossible) holds in a domain iff 2 is not a square in its field of fractions.

Title: Re: HARMONIC POWERS?
Post by JocK on Nov 23^rd, 2005, 3:36am

on 11/22/05 at 14:38:22, Eigenray wrote:

Well, that's a different problem altogether.

Of course... you're right!

I mixed up the linear (power 1) result with arithmetic averaging. And yes, the arithmetic-harmonic inequality is trivial and even applies to real numbers (as long as they are non-identical).

Title: Re: HARMONIC POWERS?
Post by K Sengupta on May 27^th, 2006, 12:09am

I furnish an alternative methodology culminating in a resolution to the generalised version corresponding to the problem under reference.

Alternate Method:

Let x, y and z are in H.P. with x<y<z, where x, y and z are positive integers and P is a positive whole number.
then, x^P + y^P = {xy/(2x-y)}^P;
Or,{(2x-y)^P}= (xy)^P/( x^P + y^P) .
For 2x<y, LHS is negative for odd P, while RHS is positive for odd P
For 2x=y, 0=(xy)^P/( x^P + y^P) , which does not admit of positive integral solution for x and y, for positive integral P.
For 2x>y, LHS is always greater than 1, while RHS is less than one for x>1.
For 2x>y and x=1, we obtain y<2,or x=y=1. This contradicts x<y.
Hence, if possible, 2x<y and P is even, so that P=2Q(say), giving:
{x^(2Q) + y^(2Q)}{(4*x^2 – 4*xy + y^2)^Q} = (xy)^(2Q)
But, y>x. Accordingly, 4*x^2 – 4*xy + y^2 <y^2, so that:
(xy)^(2Q) > (y^2Q) {x^(2Q) + y^(2Q)}
Or, x^(2Q) > {x^(2Q) + y^(2Q)}.
This is a contradiction.

Consequently, it is not feasible to determine THREE POSITIVE INTEGERS IN HARMONIC PROGRESSION with the Pth power of the largest integer being equal to the sum of Pth powers of the remaining integers , where P can correspond to any given positive integer.

Title: Re: HARMONIC POWERS?
Post by K Sengupta on May 27^th, 2006, 12:21am

I wish to thank Eigenray especially for his solution to "Arithmetic Powers"{Reference: (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_medium;action=display;num=1140418676)} which inspired me to derive the alternative methodology in terms of the previous post.
It would be needless to mention here that it was again Eigenray who originally derived the solution to "Harmonic Powers".

Title: Re: HARMONIC POWERS?
Post by K Sengupta on May 27^th, 2006, 6:41am

Quote:

K Sengupta wrote "I furnish an alternative methodology culminating in a resolution to the generalised version corresponding to the problem under reference."

I have effected some minor amendments in the abovementioned post. Any inconvenience caused due to the foregoing is sincerely regretted.