How find this sum $\sum_{ab+cd=2^m}ac=?$ Find this sum (where $m$ is a fixed positive integer)
$$\sum_{\substack{ab+cd=2^m\\ a,b,c,d \text{ are odd}}}ac.$$
My idea: since $$ab+cd=2^m\Longrightarrow ab=2^m-cd$$
and $a,b,c,d$ is odd numbers,then
$$a=2a_{1}+1,b=2b_{1}+1,c=2c_{1}+1,d=2d_{1}+1$$
then I can't works,Thank you very  much
 A: $\newcommand{\+}{^{\dagger}}%
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$\ds{{\cal J}\pars{N} \equiv \sum_{\substack{ab +cd = 2^{m}\\[1mm] a,b,c,d\ \mbox{are odd}}}ac:\ {\Large ?}.\qquad\mbox{where}\ {\large N = 2^{m}}}$

\begin{align}
{\cal J}\pars{N}&=\sum_{\ell = 0}^{\infty}\pars{2\ell + 1}
\sum_{\ell' = 0}^{\infty}\pars{2\ell' + 1}
\delta_{\pars{2\ell + 1}b + \pars{2\ell' + 1}d,N}
\\[3mm]&=\sum_{\ell = 0}^{\infty}\sum_{\ell' = 0}^{\infty}
\pars{2\ell + 1}\pars{2\ell' + 1}
\int_{\verts{z} = 1}{1 \over
z^{-\pars{2\ell + 1}b -\pars{2\ell' + 1}d + N + 1}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\int_{\verts{z} = 1}
{{\cal F}\pars{z,b}{\cal F}\pars{z,d} \over z^{N + 1}}\,{\dd z \over 2\pi\ic}
\quad\mbox{where}\quad{\cal F}\pars{z,\mu}\equiv
\sum_{\ell = 0}^{\infty}\pars{2\ell + 1}z^{\pars{2\ell + 1}\mu}\tag{1}
\end{align}

Let's evaluate ${\cal F}\pars{z,\mu}$:
\begin{align}
{\cal F}\pars{z,\mu}&=
{z \over \mu}\,\totald{}{z}\sum_{\ell = 0}^{\infty}z^{\pars{2\ell + 1}\mu}
=
{z \over \mu}\,\totald{}{z}\pars{z^{\mu} \over 1 - z^{2\mu}}
=-\,{z \over \mu}\,\totald{}{z}\pars{{1 \over z^{\mu} - z^{-\mu}}}
\\[3mm]&={z \over \mu}\,
{\mu z^{\mu - 1} - \pars{-\mu}z^{-\mu - 1} \over \pars{z^{\mu} - z^{-\mu}}^{2}}
=
{z^{\mu} + z^{-\mu} \over \pars{z^{\mu} - z^{-\mu}}^{2}}\,,\quad
\mbox{Notice that}\
{\cal F}\pars{\expo{\ic\theta},\mu}=-\,\half\,{\cos\pars{\mu\theta} \over \sin^{2}\pars{\mu\theta}}
\end{align}

$$
{\cal J}\pars{N}
=
{1 \over \pars{N - 1}!}\,\lim_{z \to 0}\,\partiald[N - 1]{}{z}
\bracks{{z^{b} + z^{-b} \over \pars{z^{b} - z^{-b}}^{2}}\,
        {z^{d} + z^{-d} \over \pars{z^{d} - z^{-d}}^{2}}}\,,\qquad
N = 2^{m}
$$

Can you evaluate the derivatives and take the limit ?.
A: This is mainly a comment related to the answers of Han de Bruijn and Jack D'Aurizio, but it may be of interest.
Generalizing the OP's problem to ask for
$$\sum_{\substack{ab+cd=2N\\ a,b,c,d \text{ are odd}}}ac$$
and computing the cases $2N=6$, $10$, and $12$ (to complement Han de Bruijn's computations for $2N=2$, $4$, and $8$), I get the first six terms of A007331, which are coefficients of a modular form.
A: A simple idea is to fix both $a$ and $c$, then count the number of solution of $ab+cd=2^m$ with $b$ and $d$ both odd. If $a$ and $c$ are not coprime there is clearly no solution; in general, the number of solutions is given by the coefficient of $x^{2^m}$ in:
$$(x^a+x^{3a}+x^{5a}+\ldots)\cdot(x^c+x^{3c}+x^{5c}+\ldots)=\frac{x^{a+c}}{(1-x^{2a})(1-x^{2c})},$$
so your initial sum is just:
$$ S = [x^{2^m}]\sum_{a,c\text{ odd}}\frac{ax^a}{1-x^{2a}}\cdot \frac{cx^c}{1-x^{2c}}=[x^{2^m}]\left(\sum_{a\text{ odd}}\frac{ax^a}{1-x^{2a}}\right)^2,\tag{1}$$
or:
$$S = [x^{2^m}]\left(\sum_{n\text{ odd}}\sigma(n)\,x^n\right)^2,\tag{2}$$
where $\sigma(n)$ is just the sum of the divisors of $n$, that is a multiplicative function. $(2)$ gives:
$$ S = \sum_{n\text{ odd},\,n<2^m}\sigma(n)\cdot\sigma(2^m-n) = \sum_{n\text{ odd},\,n<2^m}\sigma(n2^m-n^2).\tag{3}$$
I do not know if $(3)$ can be further simplified, but since $\sigma(n)\geq n+1$, we have:
$$ S \geq \frac{1}{12}\cdot 2^{3m}+\frac{1}{6}\cdot 2^m,$$
with the correct magnitude being probably a bit bigger. By applying the Cauchy-Schwarz inequality to $(3)$, since the average order of $\sigma(n)^2$ is known, we can have an upper bound, too.

Update: The Han De Brujin conjecture 

$$ S = 2^{3m-3}$$

probably follows from the Eisenstein-series identity:
$$\sigma_3(n) = \frac{1}{5}\left\{6n\sigma(n)-\sigma(n) + 12\sum_{0<k<n}\sigma(k)\sigma(n-k)\right\}.$$
than can be found here or from the Ramanujan identity stated in the fourth-to-last line in here.

In facts, the identity $(1.14)$ and the Theorems $(4.1)$ and $(4.2)$
  in this paper of Hahn set the conjecture true. A simpler
  account is also given here by Pee Choon Toh through the identity
  $(1.13b)$ with $j=4$.


Now I wonder about the existence of a more elementary proof of the Han De Brujin conjecture, maybe through the Lagrange identity
$$ (a^2+d^2)(b^2+c^2) = (ab+cd)^2+(ac-bd)^2.$$
