co.combinatorics - Variations on a theme of O'Bryant, Cooper and Eichhorn concerning power series over $mathbb Z/2mathbb Z$

(Part 1)--My argument uses the following curious fact about ideals in $Z[i]$ and $Z[sqrt{-2}].$
Suppose $n=8m+1$. Let $I=I(n)$ and $J=J(n)$ be the number of ideals of norm $n$ in $Z[i]$ and $Z[sqrt{-2}]$. Then $Iequiv J$ (4) except when $m$ is odd triangular, in which case $Iequiv J+2$ (4).

As a corollary we find that in $Z/2[[x]]$, $fg^2-fg$ is $x+x^3+x^{15}+x^{21}+cdots$, the exponents being the odd triangular numbers. (I wonder if this is previously observed, and if it's
related to congruence relations for modular forms). To prove the corollary it suffices to
show: Let $I_1=I_1(m)$ be the number of solutions of $m=t+2s$ and $J_1=J_1(m)$ be the number of solutions of $m=t+s$ with $t$ triangular, $s$ a square. Then when $m$ is odd triangular, $I_1-J_1$ is
odd; otherwise it is even.

We compare $I_1(m)$ with $I(n)$. Suppose $m=t+2s$. Then $n=(8t+1)+16s$ and $8t+1=x^2$, $16s=y^2$ with
$x$ odd and $x$, $y$ in $N$. $x+iy$ and $x-iy$ generate ideals of norm $n$ in $Z[i]$. These 2 ideals are
distinct except when $m$ is triangular and $s=y=0$. Using the fact that $Z[i]$ is a PID we find
that every ideal of norm $n$ comes from a decomposition $m=t+2s$, and that $I=2I_1$, except when
$m$ is triangular in which case $I=2I_1-1$.

Suppose $m=t+s$. Then $n=(8t+1)+8s$, and $8t+1=x^2$, $4s=y^2$ with $x$ odd and $x$, $y$ in $N$. $x+ysqrt{-2}$ and $x-ysqrt{-2}$ generate ideals of norm $n$ in $Z[sqrt{-2}]$. As above, we find
that $J=2J_1$, except when $m$ is triangular in which case $J$ is $2J_1-1$. Combining the
result of this paragraph and the last with the curious fact we get the corollary.

One now derives the answer to my question. Let $R=Z/2[[x^2]]$. As R-module, $A$ is the direct sum of $R$ and $xR.$ Let $pr$ be the $R$-linear map $Ato R$ which is id on $R$ and 0 on $xR$.
Since $fg^2-fg=x+x^3+x^{15}+cdots$, $pr(fg^2)=pr(fg)$. Since $pr$ is R linear and $1/g^2$ is in $R$.
$pr(f)=pr(f/g)$. So for even $m$, the coefficient of $x^m$ in $f/g$ is the coefficient of $x^m$ in $f$,
answering my question. (In his answer Wadim saw the projection argument but missed the
implications)---To be continued