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Topic: Arithmetic Powers (Read 611 times) |
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K Sengupta
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Arithmetic Powers
« on: Feb 19th, 2006, 10:57pm » |
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Without referring to Fermat's Last Theorem, prove that, in general, it is not feasible to determine three positive integers in Arithmetic Progression with the Kth power of the largest integer being equal to the sum of Kth powers of the remaining integers , where K is a positive integer greater than 2.
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TenaliRaman
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Re: Arithmetic Powers
« Reply #1 on: Feb 20th, 2006, 1:25am » |
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Assume that we can actually find three such numbers in arithmetic progression. Let these numbers be (n-d),n,(n+d). WLOG, assume (n,d) = 1.
So,
(n-d)^k + n^k = (n+d)^k
n^k
= (n+d)^k - (n-d)^k
= 2d((n+d)^(k-1)+(n+d)^(k-2)(n-d)+...+(n-d)^(k-1)) .. (*)
d<n => d|n (contradiction)
d>=n => RHS(*) > L.H.S(*)
-- AI
P.S -> Hmm, not given this much time, so there could be something totally bizarre going above that i havent noticed.
[edit]formula corrected  [/edit]
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| « Last Edit: Feb 20th, 2006, 2:12am by TenaliRaman » |
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Barukh
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Re: Arithmetic Powers
« Reply #2 on: Feb 20th, 2006, 1:47am » |
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TenaliRaman, one problem that I see with your argument is the formula (*). It doesn't look right.
Also, your argument covers also the case k = 2, where it's not correct (3, 4, 5). The problem is that (n,d) = 1 and d|n are contradictory unless d = 1.
Your argument for the case n >= d looks right.
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| « Last Edit: Feb 20th, 2006, 1:52am by Barukh » |
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TenaliRaman
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Re: Arithmetic Powers
« Reply #3 on: Feb 20th, 2006, 2:10am » |
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on Feb 20th, 2006, 1:47am, Barukh wrote:TenaliRaman, one problem that I see with your argument is the formula (*). It doesn't look right.
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Thanks. Corrected it now.
Quote:| Also, your argument covers also the case k = 2, where it's not correct (3, 4, 5). The problem is that (n,d) = 1 and d|n are contradictory unless d = 1. |
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Yes, forgot to take care of that case.
Quote:| Your argument for the case n >= d looks right. |
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Yes but that said, the case for n<d seems a bit hard now.
-- AI
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Obob
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Re: Arithmetic Powers
« Reply #4 on: Feb 20th, 2006, 11:10pm » |
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I could be missing something, but it appears to me that in the case n<d there is nothing to prove since it is assumed the three numbers are positive.
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Barukh
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Re: Arithmetic Powers
« Reply #5 on: Feb 21st, 2006, 12:07am » |
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on Feb 20th, 2006, 11:10pm, Obob wrote:| I could be missing something, but it appears to me that in the case n<d there is nothing to prove since it is assumed the three numbers are positive. |
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Hmm... I think the intention was to consider three numbers n, n+d, n+2d and then assume d < n.
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| « Last Edit: Feb 21st, 2006, 12:12am by Barukh » |
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Eigenray
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Re: Arithmetic Powers
« Reply #6 on: Feb 21st, 2006, 4:52pm » |
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The only way to have d|nk with (n,d)=1 is to have d=1; there's no need to distinguish d<n or d>n.
Now, it's easy to see there are no solutions if n<2. So assume n>2. Then from (-1)k+0=1 mod n, we have that k must be even. Thus we may expand| hidden: | nk = (n+1)k - (n-1)k = 2( (kC1)nk-1 + (kC3)nk-3 + . . . + (kC3)n3 + (kC1)n ).
Since n is therefore even, and k>4, we conclude 2n3 | 2kn, so k>n2. But then
(n-1)k = (n+1)k - nk > k nk-1 > nk+1, |
which is absurd. Nice problem; had me stuck for a while. See also the harmonic case.
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TenaliRaman
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Re: Arithmetic Powers
« Reply #7 on: Feb 23rd, 2006, 10:44pm » |
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on Feb 21st, 2006, 4:52pm, Eigenray wrote:| The only way to have d|nk with (n,d)=1 is to have d=1; there's no need to distinguish d<n or d>n. |
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Jah! Indeed!
-- AI
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