Parity of number "r" of representations of "n" as x^2+5y^2.

[Joshua Zucker]

Hi Zak,
I suspect it has something to do with writing
(x+isqrt(5)y)(x-isqrt(5)y) and so on.
Basically you're looking at factorizations of n over the integers
adjoin isqrt(5), though it is slightly more complicated than that.
And for the same reason that most numbers have an even number of
factors (except the squares), most numbers have an even number of
representations here as well.

Those (regular) factorizations are like (but again, slightly more
complicated than) writing n = (x+y)(x-y) ... [The added complication
here is that these factorizations only give n as a product of two
numbers with the same parity.]

--Joshua Zucker


On Sun, Apr 27, 2008 at 10:27 AM, zak seidov <zakseidov at yahoo.com> wrote:
> Dear seqfans,
>
>  i observe that for large numbers n (say n>~10^6)
>  with sufficiently large r (say r>~20)
>  of representations of n as x^2+5y^2 (x,y >=0),
>  the even r's are much more frequent that odd r's.
>
>  Is there any theory behind this?
>  Thanks, zak
>
>
>
>
>
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