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1 Oct 2009 10:47:32 IST
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is a-b always a factor of a^n-b^n .........if yes prove
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1 Oct 2009 10:55:40 IST
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n belongs to natural nos na?
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1 Oct 2009 10:56:31 IST
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yes
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1 Oct 2009 12:07:12 IST
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TIll we get the solution...Here's something interesting.
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Everyone's Stupid. |
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1 Oct 2009 12:54:17 IST
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we have, a = b mod (a-b) an = bn mod (a-b) an - bn = 0 mod(a-b) hence (a-b) will always be a factor of (an - bn ) for every real n .
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all's well that end's well, but if it dosen't, then it is not the end . . . |
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1 Oct 2009 12:58:24 IST
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i ddidnt understand ur proof akki
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1 Oct 2009 12:59:53 IST
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but here is the best proof given by nishant sir....x=y is always a root of x^n-y^n..so (x-y)(polynomial)....amazing naaa??????????
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1 Oct 2009 13:09:36 IST
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But see.. a^n - b^n = (a-b)(something ...) We can apply 'reverse-engineering' here .. cant we ? 
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Everyone's Stupid. |
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1 Oct 2009 13:10:41 IST
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hey guys isnt it enough if we have only proof by induction........
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