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

[Max Alekseyev]

On Mon, Apr 28, 2008 at 11:31 AM, Max Alekseyev <maxale at gmail.com> wrote:
> 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?
>
>  The number of such representations is equal to the number of solutions
>  to the congruence:
>  z^2 == -20 (mod 4n)
>  such that 0 <= z < 2n, or simply the half of the number of all
>  solutions for n>5.
>
>  Let n=2^k*5^s*m where m is coprime to 2*5. Then the original
>  congruence is split into three (or two if s=0):
>  z^2 == -20 (mod 2^(k+2))
>  z^2 == -20 (mod 5^s)
>  z^2 == -20 (mod m)
>
>  It is clear that for k>=2 the first congruence has no solutions. For
>  smaller k it has an unique solution modulo 2^(k+1):
>  k=0: z == 0 (mod 2)
>  k=1: z == 2 (mod 4)
>
>  The second congruence (that present only if s>0) is solvable only for
>  s=1: z==0 (mod 5).
>
>  The third congruence, if solvable, has exactly 2^t solutions modulo m
>  where t is the number of distinct prime factors of m.
>
>  Therefore, the total number of solutions, if non-zero, is 2^t and the
>  number of representations of n>5 is 2^(t-1).
>
>  To have an odd number of representations, we must have t=1, meaning
>  that n=p^u, n=2p^u, n=5p^u, or n=10p^u where p is prime (with
>  (-20/p)=1) - these numbers are rare.

Actually, this analysis was about "primitive" representations with (x,y)=1.
If q(n) is the number of such representations then the total number of
representations r(n) is
r(n) = sum_{d^2|n} q(n/d^2)
and this number (if non-zero) is odd if and only if n = p^u,
n=2p^(2u+1)m^2, n=5p^(2u+1)m^2, or n=10p^(2u+1)m^2 where p is prime
and (m,10p)=1.

Regards,
Max