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

[Max Alekseyev]

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.

Regards,
Max