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- Thread starter al-mahed
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- #1

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- #2

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come on guys... this is an olympic problem, from the M.O. from Brazil

- #3

CRGreathouse

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sum(n=-2007,2007,issquare(Mod(n,2^2007))) = 670 (15ms in Pari)

Now if you actually want to solve it by hand, you'll need to look over quadratic reciprocity more carefully.

- #4

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you need to solve by hand

- #5

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Wikipedia, under "quadratic residue", says that a number is a residue mod 2^m (m any natural number) if and only if it is of the form (4^k)*(8n+1).

I counted 336 integers in [1,2007] that are of that form. I thought an equal number__might__ be in [-2007,-1], but that would give me 672, not 670.

I counted 336 integers in [1,2007] that are of that form. I thought an equal number

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- #6

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Is it at this point that quadratic reciprocity should be applied? But how if it applies to odd primes?

Something else (in my previous post): m any natural number > 2.

Something else (in my previous post): m any natural number > 2.

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