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Topic: HARMONIC POWERS? (Read 746 times) |
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K Sengupta
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HARMONIC POWERS?
« on: Nov 21st, 2005, 11:56pm » |
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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 .
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| « Last Edit: Dec 1st, 2005, 12:15am by K Sengupta » |
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JocK
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Re: HARMONIC POWERS?
« Reply #1 on: Nov 22nd, 2005, 3:49am » |
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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... 
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| « Last Edit: Nov 22nd, 2005, 3:51am by JocK » |
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Eigenray
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Re: HARMONIC POWERS?
« Reply #2 on: Nov 22nd, 2005, 1:41pm » |
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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
np = mp+kp.
Rewrite this as
(np-mp)(m+n)p = (2mn)p,
and let 2r be the largest power of 2 dividing both m and n. Thus m=2rx, n=2ry, with x,y not both even. Cancelling a factor of 22rp, we have
(yp-xp)(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 21+p, while the RHS isn't. Contradiction.
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JocK
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Re: HARMONIC POWERS?
« Reply #3 on: Nov 22nd, 2005, 1:58pm » |
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on Nov 22nd, 2005, 1:41pm, 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.
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Nice.
Makes an elegant variation to the above problem: "show that three integers can not be in arithmetic progression and in harmonic progression".
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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Eigenray
wu::riddles Moderator
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Re: HARMONIC POWERS?
« Reply #4 on: Nov 22nd, 2005, 2:38pm » |
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on Nov 22nd, 2005, 1:58pm, JocK wrote:
Makes an elegant variation to the above problem: "show that three integers can not be in arithmetic progression and in harmonic progression". |
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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.
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JocK
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Re: HARMONIC POWERS?
« Reply #5 on: Nov 23rd, 2005, 3:36am » |
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on Nov 22nd, 2005, 2:38pm, Eigenray wrote:
Well, that's a different problem altogether. |
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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).
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solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.
xy - y = x5 - y4 - y3 = 20; x>0, y>0.
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K Sengupta
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Posts: 371
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Re: HARMONIC POWERS?
« Reply #6 on: May 27th, 2006, 12:09am » |
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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.
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| « Last Edit: May 27th, 2006, 6:35am by K Sengupta » |
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K Sengupta
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Re: HARMONIC POWERS?
« Reply #8 on: May 27th, 2006, 6:41am » |
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Quote:| K Sengupta wrote "I furnish an alternative methodology culminating in a resolution to the generalised version corresponding to the problem under reference." |
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I have effected some minor amendments in the abovementioned post. Any inconvenience caused due to the foregoing is sincerely regretted.
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