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I'm studying the theorems used in the paper which explains how the sieve of Atkin works, but I cannot understand a point.

For example, in the paper linked above, theorem 6.2 on page 1028 says that if $n$ is prime then the cardinality of the set which contains all the norm-$n$ ideals in $\mathbf Z[(-1+\sqrt{-3})/2]$ is 2. I don't understand why, and I am not able to relate this result to the quadratic form $3x^2+y^2=n$ used in the proof.

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1 Answer 1

The main thing is that the norm of $s + t \omega$ is $s^2 + s t + t^2,$ which is a binary form that represents exactly the same numbers as $3x^2 + y^2.$

It is always true that, for an integer $k,$ the form $s^2 + s t + k t^2$ represents a superset of the numbers represented by $x^2 + (4k-1)y^2.$ For instance, with $k=2,$ the form $x^2 + 7 y^2$ does not represent any numbers $2\pmod 4,$ otherwise it and $s^2 + s t + 2 t^2$ agree.

With $k=-1,$ it turns out that $x^2 - 5 y^2$ and $s^2 + s t - t^2$ represent exactly the same integers.

Take $s^2 + s t + k t^2$ with $s = x - y, \; t = 2 y.$ You get $$ (x-y)^2 + (x-y)(2y) + k (2y)^2 = x^2 - 2 x y + y^2 + 2 x y - 2 y^2 + 4 k y^2 = x^2 + (4k-1) y^2.$$

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Thank you very much! And why must $n$ be in $1+6\mathbf{Z}$? –  zar Dec 17 '11 at 11:39
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@zar $u^2 + 3 v^2$ does not represent any number $m \equiv 2 \pmod 3.$ So the primes are 3 itself and all $p \equiv 1 \pmod 3.$ A number $n \equiv 4 \pmod 6$ is even. So you might as well talk about $p \equiv 1 \pmod 6.$ –  Will Jagy Dec 17 '11 at 22:03
    
Ok, thank you. Regarding the paper, I think that I understand theorem 6.1. Can you give me some hints also for theorem 6.3? –  zar Dec 17 '11 at 23:07
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The two classes of discriminant 12 are $x^2 - 3 y^2$ and $3 x^2 - y^2.$ The first one represents (positive) primes that are $1 \pmod{12}.$ The second one represents (positive) primes that are $11 \pmod{12}.$ If any number $n$ is represented by one of these forms, then $-n$ is represented by the other. –  Will Jagy Dec 18 '11 at 3:50
    
(Is there an error on page 1029? Should $\gamma$ be equal to $\sqrt{-3}$ instead of $\sqrt3$?) –  zar Dec 18 '11 at 19:01

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