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Lagrange’s Four-Square Theorem — a special case of the Fermat-Cauchy Polygonal Number Theorem (FCPNT) — states that every natural number can be written as the sum of the squares of at most four integers.

I’m trying to find a new proof of a conditional version of part of that theorem:

Theorem: If $n$ is a natural number such that $2n+1$ can be written as the sum of the squares of four integers, then $2n+1$ can be written as the sum of squares of four integers which sum to $1$. In other words, one can always find integers $t,u,v,w$ such that \begin{align} 2n+1 &= t^2+u^2+v^2+w^2, \\ 1 &= t+u+v+w. \end{align}

The non-conditional version of this has been proven many times in the past (cf. Pollock, Cauchy, Bradley, etc.); but as far as I know, it has always been proven by first assuming the Fermat-Gauss theorem [that every natural number is the sum of at most three triangular numbers].

I’m hoping to find an elementary proof — existing or new — of this conditional result that doesn’t rely on any other part of the FCPNT. Any references or hints would be appreciated.

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    $\begingroup$ How do you know that the proposition is true when it is not proved? Is it a conjecture? Do you have any computational arguments in support of your proposition? $\endgroup$ – SARTHAK GUPTA Jun 20 at 4:37
  • $\begingroup$ @SARTHAKGUPTA: It has been proven, many times. So, yes, I know it’s true. The problem is, it is always proved by assuming the Fermat-Gauss theorem, and I’m trying to prove it “from scratch”. $\endgroup$ – Kieren MacMillan Jun 20 at 13:28
  • $\begingroup$ @SARTHAKGUPTA: See, for example, the last page of Bradley’s article jstor.org/stable/3619945 $\endgroup$ – Kieren MacMillan Jun 20 at 13:48
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The problem is equivalent to determine whether the Fourier coefficient of $$ \Theta(q)=\sum_{ \begin{array}{cc} n_1,n_2,n_3,n_4\in\textbf{Z}\\ n_1+n_2+n_3+n_4=1 \end{array} }q^{n_1^2+n_2^2+n_3^2+n_4^2} $$ is always greater or equal to 1.

We can write equivalent $n_4=1-n_1-n_2-n_3$. Hence $$ \sum^{4}_{j=1}n_j^2= $$ $$ =1+2n_1+n_2^2+2n_2+2n_1n_2+2n_2^2+2n_3+2n_1n_3+2n_2n_3+2n_3^2= $$ $$ =(n_1+n_2+n_3)^2+2(n_1+n_2+n_3)+n_1^2+n_2^2+n_3^2+1 $$ Hence $$ \Theta(q)=\sum_{ \begin{array}{cc} n_1,n_2,n_3,t\in\textbf{Z}\\ n_1+n_2+n_3=t \end{array} }q^{(t+1)^2+n_1^2+n_2^2+n_3^2}= $$ $$ =\sum^{\infty}_{t=-\infty}q^{(t+1)^2}\sum_{ \begin{array}{cc} n_1,n_2,n_3\in\textbf{Z}\\ n_1+n_2+n_3=t \end{array} }q^{n_1^2+n_2^2+n_3^2}= $$ $$ =\sum^{\infty}_{t=-\infty}q^{(t+1)^2}\sum^{\infty}_{s=0}q^s\sum_{ \begin{array}{cc} n_1,n_2,n_3\in\textbf{Z}\\ n_1+n_2+n_3=t\\ n_1^2+n_2^2+n_3^2=s\\ \end{array} }1. $$ But the equations $$ n_1+n_2+n_3=t\textrm{, }n_1^2+n_2^2+n_3^2=s $$ have solutions $$ n_1=\frac{1}{2}\left(-n_3+t-\sqrt{-2n_3^2+2s+2n_3t-t^2}\right) $$ $$ n_2=\frac{1}{2}\left(-n_3+t+\sqrt{-2n_3^2+2s+2n_3t-t^2}\right) $$ and $$ n_1=\frac{1}{2}\left(-n_3+t+\sqrt{-2n_3^2+2s+2n_3t-t^2}\right) $$ $$ n_2=\frac{1}{2}\left(-n_3+t-\sqrt{-2n_3^2+2s+2n_3t-t^2}\right) $$ Hence if we set $$ A(t,s)=\sum_{ \begin{array}{cc} n_1+n_2+n_3=t\\ n_1^2+n_2^2+n_3^2=s\\ \end{array} }1, $$ then, if the discriminant of $n_1$ and $n_2$: $-2n_3^2+2s+2n_3t-t^2$ is zero possess one double root $(n_1=n_2)$. Otherwise if $-2n_3^2+2s+2n_3t-t^2=m^2$, $m\in\textbf{Z}^{*}$ two different. Hence if $s'=\left[\sqrt{s}\right]$ is the floor of the square root of $s\geq 0$, we get $$ A(t,s)=\sum_{ \begin{array}{cc} -s'\leq n_3\leq s'\\ -3n_3^2+2s+2n_3t-t^2=0 \end{array} }1+ $$ $$ +2\sum_{ \begin{array}{cc} -s'\leq n_3\leq s'\\ -3n_3^2+2s+2n_3t-t^2\neq 0 \end{array} }X_{\textbf{Z}}\left(\frac{1}{2}\left(-n_3+t+\sqrt{-3n_3^2+2s+2n_3t-t^2}\right)\right), $$ where $X_{\textbf{Z}}(n)$ is the characteristic function on integers.

By this way we get if $r_0(t,2s)$ is the number of representations of $2s$ in the form $$ 3x^2-2xt+t^2 $$ and $r_1(t,2s)$ the number of representations of $2s$ in the form $$ 3x^2-2xt+m^2+t^2, $$ then $r(t,2s)=r_0(t,2s)+2r_1(t,2s)=A(t,s)$ is the Fourier coefficient of $$ \Theta(q)=\sum^{\infty}_{t=-\infty}\sum^{\infty}_{s=0}A(t,s)q^{(t+1)^2+s}\textrm{, }|q|<1 $$ If $s-t=p$, then $$ 2s=3x^2-2xt+m^2+t^2\Leftrightarrow 3x^2-2tx+t^2-2t+m^2=2p.\tag 1 $$ We want to show that the above last equation have always integer solutions for every even non negative integer $p\geq0$ and for $p$ odd positive none. This will enable us to conclude, that the odd Fourier coefficients of $\Theta(q)$ are always occur and are non zero iff $p$ is even. For $p$ odd we have no representations. For this we define: $$ \phi(q)=\sum^{\infty}_{n,t=-\infty}q^{3n^2-2tn+t^2-2t} $$ and assume the transformation $n\rightarrow an+bt$, $t\rightarrow c n+d t$, where $a=-1$, $b=1$, $c=1$, $d=-2$. Then $ad-bc=1$ and $$ \phi(q)=2\sum^{\infty}_{n,t=-\infty}q^{n^2-2n+8 t^2+4t}. $$ Hence $$ \phi(q)=2q^{-1}\left(\sum^{\infty}_{n=-\infty}q^{n^2-2n+1}\right)\left(\sum^{\infty}_{t=-\infty}q^{8t^2+4t}\right)= $$ $$ =2q^{-1}\theta_3(q)\sum^{\infty}_{n=-\infty}q^{8n^2+4n}, $$ where $\theta_3(q)=\sum^{\infty}_{n=-\infty}q^{n^2}$, $|q|<1$. Hence (1) have analog:
$$ \phi(q)\theta_3(q)=\sum_{n,k,m\in\textbf{Z}}q^{3n^2-2nk+k^2-k+m^2}=2q^{-1}\theta_3(q)^2\sum^{\infty}_{n=-\infty}q^{8n^2+4n}= $$ $$ =2q^{-1}\theta_3(q)^2\phi_e(q^2), $$ where $\phi_e(q)=\sum^{\infty}_{n=-\infty}q^{4n^2+2n}$. But $$ \theta_2(q)=q^{1/4}\sum^{\infty}_{n=-\infty}q^{n^2+n}=q^{1/4}\sum^{\infty}_{n=-\infty}q^{4n^2+2n}+\sum^{\infty}_{n=-\infty}q^{(2n+1+1/2)^2} $$ and $$ \sum^{\infty}_{n=-\infty}q^{(2n+1+1/2)^2}=\sum^{\infty}_{n=-\infty}q^{(2n+3/2-2)^2}= $$ $$ \sum^{\infty}_{n=-\infty}q^{(2n-1/2)^2}=q^{1/4}\sum^{\infty}_{n=-\infty}q^{4n^2-2n}. $$ Hence $$ \theta_2(q)=2q^{1/4}\sum^{\infty}_{n=-\infty}q^{4n^2+2n} $$ and therefore $$ \phi(q)\theta_3(q)=q^{-3/2}\theta_3(q)^2\theta_2(q^2).\tag 2 $$

Hence is equivalent to show that $$ x^2+y^2+2z^2+2z=2p+1,\tag 3 $$ have always solutions when $p$ is even non negative integer and none when $p$ positive odd. But indeed: $2z^2+2z\equiv 0(4)$ and when $p$ is odd we have no solutions since no integer of the form $4n+3$ have representation as a sum of two squares. Hence we left with case $p$ even.

The function $f(z)=\theta_3(q)^2\theta_2(q^2)$, $q=e(z)=e^{2\pi i z}$, $Im(z)>0$ is the associated theta function of (3) and is a modular form of weight $3/2$ in $\Gamma(8)$, i.e. it holds $$ f\left(\frac{az+b}{cz+d}\right)=\epsilon_{c,d}(cz+d)^{3/2}f(z), $$ when $Im(z)>0$ and $a,b,c,d$ integers with $ad-bc=1$, $a,d\equiv 1(8)$, $c,b\equiv 0(8)$. The function $\epsilon_{c,d}$ is such that $\epsilon_{c,d}^4=1$. But I don't have much experience to go further.

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