TITLE: Why are there usually an even number of representations as a sum of 11 squares
QUESTION [20 upvotes]: Question: Why do so few $n\equiv 3 \bmod 8$ have an odd number of representations in the form $$n=x_0^2 + x_1^2 + \dots + x_{10}^2$$ with $x_i \geq 0$?
Note that $x_i\geq 0$ spoils the symmetry enough that this is not the usual "number of representations as a sum of squares" question. The question may be ill-posed, too: I do not know that there are few such $n$, but computations suggest that their counting function is $\sim cx/\log x$.
The question I'd like to answer is this: which $n$ have an odd number of representations of the form $$n=x_0^2+2x_1^2+4x_2^2+ \dots = \sum_{i=0}^\infty 2^i x_i^2,$$ where $x_i \geq 0$? Does the set of such $n$ have 0 density? The motivation for this question is that it is nontrivially the same as this question.
If $n$ is even and has an odd number of reps, then (nice exercise) $n$ has the form $2k^2$. If $n\equiv 1\pmod4$ and has an odd number of reps, then (challenging but elementary) $n$ has a special sort of factorization, and there are very few such $n$.
The question I'm asking here is for $n\equiv 3 \bmod 8$. Reducing modulo 8 reveals that $x_2$ must be even. If it isn't a multiple of 4, then we can pair off the two reps:
  $$(x_0,x_1,x_2,x_3,x_4,x_5,\dots) \leftrightarrow (x_0,x_1,2x_4,x_3,x_2/2,x_5,\dots).$$
If $x_2$ is a multiple of 4 but not 8, then we can make a similar pairing with $x_2$ and $x_5$, and so on. This only leaves the situation when $x_2$ and $x_4,x_5,\dots$ are all zero.
This leaves us at: Which $n\equiv 3\bmod 8$ have an odd number of representations in the form:
 $$n = x_0^2 + 2x_1^2 + 8 x_3^2 \; , \; \; \mbox{with} \; \; x_0, x_1, x_3 \geq 0 \;?$$
Some play with the parities of binomial coefficients reduces this to the question I led with.

REPLY [4 votes]: This is not an answer to your stated question, but a relevant remark.  Suppose that $a(x)$ is a unipotent formal power series in the formal power series ring $(\mathbb{Z}/p)[[x]]$.  (Here unipotent just means constant term 1, so you could say topologically or adically unipotient.)  Then obviously $a(x)^n$ is well-defined for any integer $n$.  What is somewhat less obvious, but not hard and an interesting general principle, is that $a(x)^d$ is well-defined for any $p$-adic integer $d$.  Namely, if $d$ has digits $\ldots d_2d_1d_0$, then 
$$a(x)^d := a(x)a(x^{d_1p})a(x^{d_2p^2})\cdots$$
It's not hard to check that this formula is (a) correct when $d$ is an integer using the Frobenius map, (b) convergent when $a(x)$ unipotent, and (c) continuous in $d$ and $a$.  Therefore (d) it satisfies $a^{d+e} = a^d a^e$ and $(ab)^d = a^db^d$.
You use this formula twice in your question.  You use it with $d=-1$ when you call your power series "nontrivially the same" as your other question.  You use it again with $d=11$ with the phrase "some play with parities of binomial coefficients".  Of course in both cases you're using the power series from your paper,
$$a(x) = 1+x+x^4+x^9 + x^{16} + \cdots \in (\mathbb{Z}/2)[[x]].$$
I hadn't thought of this style of $p$-adic exponentiation and I think that it's cute, and it could be a unifying principle for some of what you are doing.