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Topic: Radical power equality (Read 1133 times) |
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Michael Dagg
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Radical power equality
« on: Feb 4th, 2008, 7:20pm » |
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Let positive integers a, n be given. Show that
there is a positive integer b for which
(sqrt(a) - sqrt(a - 1))^n = sqrt(b) - sqrt(b - 1).
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Regards,
Michael Dagg
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Eigenray
wu::riddles Moderator
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Re: Radical power equality
« Reply #1 on: Feb 4th, 2008, 8:04pm » |
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First, if n=2, we take b=(2a-1)2. Now by replacing a with b, and n by n/2, repeatedly, we can assume n=2k+1 is odd In this case,
| hidden: | [ a - (a-1)]n = ca - d(a-1), where
c =  C(2k+1,2i+1) ai(a-1)k-i,
d = C(2k+1,2i) ai(a-1)k-i.
But we have also [a + (a-1)]n = ca + d(a-1), and multiplying these two equations gives
1 = ([a - (a-1)][a + (a-1)])n = [ca - d(a-1)][ca + d(a-1)] = c2a - d2(a-1),
and therefore we take b = c2a:
(a - (a-1))n = (c2a) - (d2(a-1)) = (c2a) - (c2a-1). |
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Michael Dagg
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Re: Radical power equality
« Reply #2 on: Feb 5th, 2008, 3:42pm » |
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Clever (and economical) ! 
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Michael Dagg
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