Determining the number of ways a number can be written as sum of three squares

Question

I was going through Erich Friedman's "What's Special About This Number?" and there some numbers are classified based on the number of ways we can write them as sum of squares. I want to prove the following claim by Friedman:

129 is the smallest number that can be written as the sum of 3 squares in 4 ways.

Indeed, as given in Wikipedia, $$11^2+2^2+2^2 = 10^2+5^2+2^2 = 8^2+8^2+1^2 = 8^2+7^2+4^2 = 129$$ So what remains to prove is that this is the smallest such number.

Is it possible to write a proof for this fact using some insights along with brute force/cases? How can we solve this problem using only brute-force?

Also, since I know the proof of Legendre's three-square theorem. I am also curious to know:

How can we determine the number of ways we can write a non-negative integer which satisfies Legendre's three-square theorem as sum of three squares?

Edit1: Related discussions on MathOverflow:

• Is there a simple way to compute the number of ways to write a positive integer as the sum of three squares?: Note that this is not answer of my question since $r_k(n)$ counts the number of representations of $n$ by $k$ squares, allowing zeros and distinguishing signs and order.
• Efficient computation of integer representation as sum of three squares

Edit2: Related discussions on ComputerScience.SE

• Listing integers as the sum of three squares $m=x^2+y^2+z^2$

Edit3: Related discussions on Mathematics.SE

• When is a rational number a sum of three squares?

• Why can't this number be written as a sum of three squares of rationals?

• Sum of one, two, and three squares

Answer 1

For the system of equations.

$$x_1^2+x_2^2+x_3^2=x_4^2+x_5^2+x_6^2=x_7^2+x_8^2+x_9^2=x_{10}^2+x_{11}^2+x_{12}^2$$

Solutions can be parameterized.

$$x_1=a(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$

$$x_2=b(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$

$$x_3=c(y^2+z^2+n^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$

$$x_4=(ay^2-2byn+az^2-2czn-an^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$

$$x_5=(bz^2-2cyz-by^2-2ayn+bn^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$

$$x_6=(cy^2-2byz-cz^2-2azn+cn^2)(p^2+s^2+n^2)(k^2+t^2+n^2)$$

$$x_7=(ap^2-2bpn+as^2-2csn-an^2)(y^2+z^2+n^2)(k^2+t^2+n^2)$$

$$x_8=(bs^2-2cps-bp^2-2apn+bn^2)(y^2+z^2+n^2)(k^2+t^2+n^2)$$

$$x_9=(cp^2-2bps-cs^2-2asn+cn^2)(y^2+z^2+n^2)(k^2+t^2+n^2)$$

$$x_{10}=(ak^2-2bkn+at^2-2ctn-an^2)(y^2+z^2+n^2)(p^2+s^2+n^2)$$

$$x_{11}=(bt^2-2ckt-bk^2-2akn+bn^2)(y^2+z^2+n^2)(p^2+s^2+n^2)$$

$$x_{12}=(ck^2-2bkt-ct^2-2atn+cn^2)(y^2+z^2+n^2)(p^2+s^2+n^2)$$

It is interesting that such triples can be too much. The formula can be increased to any number. That is the same to write not only for 4 partitions, but for any number. The main thing that all the variables were not identical to each other.

Answer 2

You can try something like:

check=new Array();
for (i=1;i<12;i++)
for (j=i;j<12;j++)
for (k=j;k<12;k++) {
n=i*i+j*j+k*k;
if (!check[n]) check[n]=0;
check[n]++;
}
for (a=0;a<check.length;a++)
if (check[a]>=4) console.log(a);

which proves $129$ is the smallest such number, followed by:

$$134=11^2+3^2+2^2=10^2+5^2+3^2=9^2+7^2+2^2=7^2+7^2+6^2$$

You can determine the number of representations by $3$ squares by subtracting squares from $N$, and using $\sum_\limits{x\le \sqrt{N}} r_2(N-x^2)$, where $r_2$ is the sum of (two) squares function.

Answer 3

Legendre gave the following answer, which he could not prove (Gauss gave the first proof of the 3-squares theorem). For the sake of simplicity I will only cover the case of numbers $c \equiv 5 \bmod 8$. In this case consider the equivalence classes of forms with discriminant $-4c$; like Legendre, we will consider equivalence with respect to the action of GL$_2({\mathbb Z})$. Let $P$ denote the form classes that represent primes $p \equiv 1 \bmod 4$, and $Q$ those that represent primes $q \equiv 3 \bmod 4$. Then the number of classes $P$ is equal to that of classes $Q$, and to the number of ways in which $c$ can be written as a sum of three squares.

If $c = 29$, the classes $P$ are $x^2 + 29y^2$ and $5x^2 + 2xy + 6y^2$, and these correspond to the two distinct ways of writing $29$ as a sum of three squares: $29 = 0^2 + 2^2 + 5^2 = 2^2 + 3^2 + 4^2$.

Legendre conjectured similar formulas for other values of $c$, which were later proved by Gauss, but using classes with respect to the action of SL$_2({\mathbb Z})$.