nt.number theory - Why are there usually an even number of representations as a sum of 11 squares

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.