Why are there usually an even number of representations as a sum of 11 squares 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.
 A: 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.
A: Throughout $N>0,$ and $N \equiv 3 \pmod 8.$ Let $I$ be the number of ordered triples $(a,d,e) \;\mbox{with} \; a,d,e \geq 0,$ such that
$$a^2+2 d^2+8 e^2=N.$$  I'll use a result of Gauss on sums of 3 squares to show that if there are 3 or more primes whose exponent in the prime factorization of $N$ is odd,  then $I$ is even.  As a consequence those $N$ for which $I$ is odd form a set of density 0;  in fact the number of such $N   < x$ for positive real $x$ is
 $$  O \left( \frac{x \; \log \log x}{\log x} \right). $$
      Let $ R = R(N)  $ be the number of triples  $ (a,b,c) \; \mbox{with} \; a,b,c > 0 $ and
$$ a^2+b^2+c^2=N,$$ and let $r(N)$ be the number of such
triples with the $ \gcd(a,b,c) = 1.$ Then $R$ is the sum of
the $r(N/k^2),$ the sum running over all $k>0$
for which $k^2 | N.$ Now in Disquisitiones, Gauss shows that  if
$N>3, \mbox{then} \; r(N)/3$ is the number of classes (under
proper equivalence) of positive primary binary forms of discriminant  $- 
N.$ (Or if you prefer, the number of
classes of invertible ideals in the quadratic order of discriminant $- 
N$). Now these classes form a group, and
Gauss uses genus theory to show that the order of this group is divisible by $2^{M-1}$ where $M$ is the
number of primes that divide $N.$ So if 3 or more primes have odd exponent in the prime factorization of $N,$
then all these primes divide $N/k^2,$  the corresponding group has order divisible by $2^{3-1}=4,$ so 4 divides
each $r(N/k^2),$ and 4 divides $R.$  $$  $$
     Now let $S=S(N)$ be the number of pairs  $(a,d) \; \mbox{with} \; a,d > 0$  and
$$a^2+2 d^2=N,$$ and $s(N)$ be the number of such pairs
with  $ \gcd(a,d) = 1.$ Then $S$ is the sum of the $s(N/ 
k^2).$ Using the fact that 
 $\mathbb{Z} \left[ \sqrt{-2} \right]$ is a UFD
we can calculate $s(N/k^2);$ it is zero when some prime $p \equiv 5,7 \pmod 8$  divides $N/k^2.$ When this doesn't happen there are 3 or more
primes  $q \equiv 1,3 \pmod 8$ dividing $N/k^2,$ so   4 divides each $s(N/ 
k^2)$ and
4 divides $S$ as well as $R.$ We conclude the proof by showing that $$2I=R+S.$$
Suppose $N \equiv 3 \pmod 8$ and $a^2+b^2+c^2=N,$ with $a,b,c>0.$ Of course $a,b, c$ are odd. If 
$b \equiv c \pmod 4,$ let $d=(b+c)/2$ and $e = | (b-c)/4 |.$ Otherwise let $d = | (b-c)/2 |$ 
and $e=(b+c)/4.$ Then $$a^2+2 d^2+8 e^2=a^2+b^2+c^2=N.$$ Furthermore $(a,b,c)$ and $(a,c,b)$ map to the same $(a,d,e).$ The fiber of the map $(a,b,c) \mapsto (a,d,e)$ has 1 element when $e=0$ and 2 elements otherwise. So $2I=R+S.$ $$ $$
     If $N = p q$ where $p$ and $q$ are primes congruent to 5 and 7 $ \pmod 8$
respectively, with $ (q | p ) = -1$ it can be
shown that $R \equiv 2 \pmod 4,$ so that $I$ is odd. This should
allow one to get a lower bound for
the number of $ N < x $ with $I$ odd that's a constant multiple of the upper
bound mentioned above. But
whether the number is asymptotic to a constant multiple of $x \; \log \log(x)/ 
\log (x)$ as Jagy's calculations suggest
isn't clear.
