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Re: A quicker way?
Posted:
Feb 4, 2013 11:03 AM
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Le lundi 4 février 2013 16:36:14 UTC+1, quasi a écrit :
> quasi wrote:
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> >luttgma@gmail.com wrote:
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> >
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> >>Let n^2 = N + a^2, where n, N and a are integers.
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> >>Knowing N, it is of course possible to find n by trying
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> >> a = 1, 2, 3, 4, etc...
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> >>But doesn't exist a quicker way ?
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> >
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> >if you rewrite the equation in the form
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> >
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> > (n - a)(n + a) = N
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> >
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> >then for each pair of integers u,v with u*v = N, you can
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> >solve the equations
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> >
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> > n - a = u
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> > n + a = v
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> >
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> >for n and a.
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> Also, since n - a and n + a have the same parity (both even
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> or both odd), you only need to consider pairs of integers u,v
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> with u*v = N for which u,v are both even or both odd.
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>
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> quasi
Thank you, but then you must find u and v, which doesn't seem quicker than trying a=1,2,3, etc.
Marcel
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