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 | | From: | ha | | Subject: | problem about quadratic congruences | | Date: | Mon, 24 Jan 2005 07:37:50 GMT |
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 | Known that :
x^2 = a (mod m) is solvable iff x^2 a(mod 2^k) and x^2 =a (mod Pi ^ei) is
solvable.
If x^2 = a (mod m) is solvable, how many solutions does it have?
Thanks
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| | From: | William Elliot | | Subject: | Re: problem about quadratic congruences | | Date: | Mon, 24 Jan 2005 00:55:13 -0800 |
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| On Mon, 24 Jan 2005, ha wrote:
> Known that :
> x^2 = a (mod m) is solvable iff x^2 a(mod 2^k) and x^2 =a (mod Pi ^ei) is
> solvable.
>
What's x^2 a(mod 2^k)? Typo?
For what k? For k = 1, x^2 = a (mod 2) is always solvable.
x^2 = a (mod pi^ei) is gibberish.
(3.14159...)^(2.718... * sqr -1)
> If x^2 = a (mod m) is solvable, how many solutions does it have?
2 when m is prime.
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