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Is this right? And how to prove it ?

For $n \equiv 1,2 \bmod 4$ $$ \Bigg|\ \mathbb Z^3\cap\Big\{(a_1,a_2,a_3)\ \Big|\ a_1^2+a_2^2+a_3^2=n \Big\}\Bigg| \\ = \frac12\Bigg|\mathbb Z^3\cap\Big\{(a_1,a_2,a_3)\ \Big|\ |a_1a_2|+|a_2a_3|+|a_1a_3|=n \Big\}\Bigg| $$ Here $|\cdot| $denotes the cardinality of a set.

Denote the left hand side quantity by $r(n)$. In other words, $r(n)$ means the number of ways $n$ can be written as sum of three square of integers.

My experiment on Mathematica is here. In the picture, the first generating function is the generating function for the expression by sumf of three squares. The second generating function is half of the generating function of pairwise product absolute sum.

Notice that when $n \equiv 1,2 \bmod 4$ the coefficients of $x^n$ of two functions (the generating function of both sides respectively) are equal.

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    $\begingroup$ crossposted math.stackexchange.com/questions/4972062/… $\endgroup$
    – Will Jagy
    Commented Sep 16 at 17:19
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    $\begingroup$ For $n\equiv3\pmod8$, the ratio seems to be $1/4$. Of course, when $n\equiv7\pmod8$, the LHS vanishes $\endgroup$
    – Henri Cohen
    Commented Sep 16 at 17:36
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    $\begingroup$ I am wondering how you came across this observation. $\endgroup$
    – Iosif Pinelis
    Commented Sep 16 at 18:42
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    $\begingroup$ Very nice conjecture, I provide a proof below. $\endgroup$
    – GH from MO
    Commented Sep 16 at 18:46

1 Answer 1

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Let $G(n)$ denote the weighted number of $\mathrm{SL}_2(\mathbb{Z})$-equivalence classes of positive definite quadratic forms $$[a,b,c]:=ax^2+bxy+cy^2$$ with $2\mid b$ and $b^2-4ac=-4n$, where the classes represented by $[a,0,a]$ have weight $1/2$, the classes represented by $[a,a,a]$ have weight $1/3$, and all other classes have weight $1$.

Gauss (1801) proved the following formula for $n\equiv 1,2\pmod{4}$: $$\left|\left\{(a_1,a_2,a_3)\in\mathbb{Z}^3\,:\, a_1^2+a_2^2+a_3^2=n \right\}\right|=12G(n).$$ On the other hand, Mordell (1923) proved for every $n>0$ that $3G(n)$ counts the number of solutions of $a_1a_2+a_2a_3+a_3a_1=n$ in nonnegative integers if a weight $1/2$ is attached to the solutions with $a_1a_2a_3=0$. From here we obtain with a bit of combinatorics that $$\left|\left\{(a_1,a_2,a_3)\in\mathbb{Z}^3\,:\, |a_1a_2|+|a_2a_3|+|a_1a_3|=n \right\}\right|=24G(n).$$ Comparing the last two displays, the conjectured formula follows.

P.S. Mordell (1923) published his theorem in the American Journal of Mathematics (vol. 45, pp. 1-4), but he also included it in Chapter 30 of his book "Diophantine equations".

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    $\begingroup$ Can you give a detailed reference to Gauss (1801)? Is there an accessible English translation of that proof? $\endgroup$
    – Iosif Pinelis
    Commented Sep 16 at 20:00
  • $\begingroup$ @IosifPinelis I cannot give a reference off-hand. I guess it is in Disquisitiones. The usual formulation relates primitive lattice points to classes of primitive forms. As the two sides transform in the same way, the conversion is straightforward. Moreover, these days the condition $2\mid b$ is dropped for the forms, hence a slightly different class number is needed. The two kinds of class numbers are related by $H(-n)=h(-4n)$, and my memory tells that this one is in Dickson. There is a treatment with $h(-4n)$ as opposed to $H(-n)$ in Grosswald: Representations of integers as sums of squares. $\endgroup$
    – GH from MO
    Commented Sep 16 at 20:08
  • $\begingroup$ @IosifPinelis I looked up Dickson's book "History of the theory of numbers". He mentions on page 265 of volume 2, and also on page 109 of volume 3, that Kronecker (1860) used elliptic functions to prove Gauss's result in the form $r_3(n)=24F(n)-12G(n)$. Here $G(n)$ is as above, and $F(n)$ is the weighted number of classes of $[a,b,c]$ with $b$ even, $a$ or $c$ odd, $ac-b^2=n$. He also references the result that $F(n)=G(n)$ for $n\equiv 1,2\pmod{4}$, hence in fact $r_3(n)=12G(n)$ for such $n$. $\endgroup$
    – GH from MO
    Commented Sep 16 at 21:49
  • $\begingroup$ In my previous comment, $ac-b^2=n$ should be replaced by $b^2-4ac=-4n$. $\endgroup$
    – GH from MO
    Commented Sep 17 at 14:56

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