Re: Biquadratic Reciprocity Theorem

In article <1175201913.662713.100830@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
<Sunny123@xxxxxxxxxxxx> wrote:
hi

could someone please give a numerical example of how one would use the
Biquadratic Reciprocity Theorem as found at

http://mathworld.wolfram.com/BiquadraticReciprocityTheorem.html

any help would be very much appreciated.

Let us take two gaussian primes to play the role of sigma and pi. Say
we take sigma=11 and pi=1+2i, which are gaussian primes.


The number (N(pi)-1)/4 is

N(pi) = 1^2 + 2^2 = 5, so [N(pi)-1]/4 = 1
N(sigma) = 11^2 + 0^2 = 121, so [N(sigma)-1]/4 = 30.

Thus, Biquadratic Reciprocity says that
(pi/sigma)_4 * (sigma/pi)_4 = (-1)^{1*30}; that is, either pi is a
fourth power modulo sigma and sigma is a fourth power modulo pi, or
else neither is a fourth power modulo the other.

Moreover, pi is a square but not a fourth power modulo sigma if and
only if sigma is a square but not a fourth power modulo sigma (this
can be seen by noting that a square but not fourth power modulo sigma
will have biquadratic symbol equal to -1).

As to using them, suppose we want to figure out if 1+2i is a fourth
power modulo 11. Well, let us compute (11/1+2i)_4 instead.

Since 11 = (1+2i)(2-4i) + 1, we have that (11/1+2i)_4 = (1/1+2i)_4 =
1, so we have that (1+2i/11)_4=1 as well.

--
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"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes" by Bill Watterson)
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Arturo Magidin
magidin-at-member-ams-org

.