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why (4*i+7) in Z[i] not a prime?norm(4*i+7) =65, 65mod4=1

asked 2013-12-05 18:28:51 -0600

cjsh gravatar image

updated 2013-12-06 03:39:58 -0600

kcrisman gravatar image
K.<a> = NumberField(x^2 +1);
(4*a+7).norm()
65
mod (65,4)
1

factor(4*a+7)
(-3*a - 2) * (a - 2)

(4*a+7).multiplicative_order();
+Infinity

K.elements_of_norm(65);factor(-8*a + 1);K.prime_factors(4*a -7);

-8*a - 1, 4*a - 7, 4*a + 7, -8*a + 1]
(a - 2) * (3*a - 2)
[Fractional ideal (3*a - 2), Fractional ideal (-a - 2)]
[Fractional ideal (a - 2), Fractional ideal (3*a - 2)]
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answered 2013-12-06 03:45:07 -0600

kcrisman gravatar image

updated 2013-12-06 03:45:39 -0600

This is because 65 is not prime. Note that $1^2+2^5$ and $3^2+2^2=13$ are both primes (of the form $4n+1$, so the factors you find are indeed (Gaussian) primes.

You may find the Wikipedia article helpful.

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thank you very nuch!

cjsh gravatar imagecjsh ( 2013-12-06 19:23:25 -0600 )edit

Great! In order to let future users know that was the correct answer, please accept it (there should be a check mark of some kind below the thumbs up/down symbols).

kcrisman gravatar imagekcrisman ( 2013-12-07 03:12:06 -0600 )edit

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Asked: 2013-12-05 18:28:51 -0600

Seen: 96 times

Last updated: Dec 06 '13