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

Solution 1:

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 $$ (a^2+d^2)(b^2+c^2) = (ab+cd)^2+(ac-bd)^2.$$

Solution 2:

Disclaimer: this is only a partial answer, but something may be better than nothing. The approach is a brute force method, programming is in Pascal:

program china_math;

function twee(macht : integer) : integer;
{
  Compute 2^macht
}
var
  k : integer;
  h : integer;
begin
  h := 1;
  for k := 1 to macht do
    h := h shl 1;
  twee := h;
end;

procedure test(m : integer; blah : boolean);
{
  The problem
}
var
  i,j,k,a,b,c,d : integer;
  s,N,ii,jj,kk : integer;
  OK : boolean;
begin
  N := twee(m);
  s := 0;
  ii := (N div 2);
  for i := 1 to ii do
  begin 
    a := 2*i-1;
    jj := ((N div a) div 2)+1;
    for j := 1 to jj do
    begin 
      b := 2*j-1;
      kk := ((N-a*b) div 2)+1;
      if (N-a*b) > 0 then
      for k := 1 to kk do
      begin 
        c := 2*k-1;
        OK := (N-a*b) mod c = 0;
        if not OK then Continue;
        d := (N-a*b) div c;
        if blah then Writeln(a,'',b,'+',c,'',d,'=',N);
        s := s + a*c;
      end;
    end;
  end;
  if not blah then
  Writeln(m:6,s:12,' =?= ',twee(3*m-3));
end;

procedure doen;
var
  m : integer;
begin
  test(4,true);
  Writeln;
  for m := 1 to 11 do
  test(m,false);
end;

begin
  doen;
end.

a*b+c*d=2^m
1*1+1*15=16
1*1+3*5=16
1*1+5*3=16
1*1+15*1=16
1*3+1*13=16
1*3+13*1=16
1*5+1*11=16
1*5+11*1=16
1*7+1*9=16
1*7+3*3=16
1*7+9*1=16
1*9+1*7=16
1*9+7*1=16
1*11+1*5=16
1*11+5*1=16
1*13+1*3=16
1*13+3*1=16
1*15+1*1=16
3*1+1*13=16
3*1+13*1=16
3*3+1*7=16
3*3+7*1=16
3*5+1*1=16
5*1+1*11=16
5*1+11*1=16
5*3+1*1=16
7*1+1*9=16
7*1+3*3=16
7*1+9*1=16
9*1+1*7=16
9*1+7*1=16
11*1+1*5=16
11*1+5*1=16
13*1+1*3=16
13*1+3*1=16
15*1+1*1=16

     m         sum =?= 2^(3*m-3)
     1           1 =?= 1
     2           8 =?= 8
     3          64 =?= 64
     4         512 =?= 512
     5        4096 =?= 4096
     6       32768 =?= 32768
     7      262144 =?= 262144
     8     2097152 =?= 2097152
     9    16777216 =?= 16777216
    10   134217728 =?= 134217728
    11  1073741824 =?= 1073741824

Solution 3:

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{{\cal J}\pars{N} \equiv \sum_{\substack{ab +cd = 2^{m}\\[1mm] a,b,c,d\ \mbox{are odd}}}ac:\ {\Large ?}.\qquad\mbox{where}\ {\large N = 2^{m}}}$

\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 ?.