Odd square as sum of $9$ distinct odd squares

Question

I'm interested in representing odd squares as sum of $9$ distinct odd squares.

So, let $n\in\mathbb{N}$ be odd and $x_1,\,x_2,\,...,\,x_9\in\mathbb{N}$ be odd and pairwise distinct. The question is, for which $n$ exists a solution for $$n^2=\sum_{j=1}^9x_j^2$$ I checked the possibilites up to $n=5555$ and in that range, for all $n\geq37$ a solution exists.

Furthermore, in that range besides $n=43$ we can even find solutions of the following more specific form with $x_1,\,x_2,\,x_3\geq 13$: $$n^2=1^2+3^2+5^2+7^2+9^2+11^2+\sum_{j=1}^3x_j^2\Longleftrightarrow n^2-286=\sum_{j=1}^3x_j^2$$ Solutions for $5555\geq n\geq 37$ can be found here: https://pastebin.com/raw/6B4MBaiw

However, that is obviously not a proof, and I'm wondering, whether it can be proven, that for all $n\geq 37$ at least one solution exists, and maybe even that for $n\geq 37$ with $n\neq 43$ the second equation has at least one solution?

If we don't need our odd squares to be distinct, the problem gets easy. (Or maybe it's still easy even with distinct odd squares and I just don't see it?)

We use a slight variation of the second equation.

Let $n\in\mathbb{N}$ be odd and $x_1,\,x_2,\,x_3\in\mathbb{N}$ be odd.

It is $$n^2-6\equiv 3 \mod 8$$ and thus Legendre's three-square theorem guarantees with $n\geq3$ a solution for $$n^2-6=\sum_{j=1}^3x_j^2\Longleftrightarrow n^2=1^2+1^2+1^2+1^2+1^2+1^2+\sum_{j=1}^3x_j^2$$

Note, that the theorem by itself does not guarantee $x_1,\,x_2,\,x_3$ to be odd, but you can only get to the form $8k+3$ with $3$ odd squares.

Answer 1

I tried to verify this by computer for n larger than 5555. My first attempt - adding nine odd squares and checking whether the sum is again an odd square - didn't work very well. I verified that every odd squares up to $1,000,000^2$ is the sum of nine distinct odd squares, and each solution started with $1^2 + 3^2 + 5^2 + 7^2 + 9^2 + 11^2$, so only three squares needed to be varied, and the 7th square was always less than $91^2$. The reason this search takes very long is that most of the sums are not squares. That search could have been improved a bit, but not to much.

To verify this for much larger n: Let t be a triangular number (0, 1, 3, 6, 10, 15, 21 etc.) then $8 \cdot t + 1$ is an odd square (1, 9, 25, 49, 81, 121, 169 etc.). So if s is the sum of eight triangular numbers, then $8 \cdot s + 8$ is the sum of eight odd squares.

$(2 \cdot s + 3)^2 - (2 \cdot s + 1)^2 = 8 \cdot s + 8$, so $(8 \cdot s + 8) + (2 \cdot s + 1)^2 = (2 \cdot s + 3)^2$; the left hand side is the sum of nine odd squares, and the right hand side is one odd square.

I verified by computer that every integer from 146 to 62,499,499,999 is the sum of 8 distinct triangular numbers, and therefore every odd square from 295 to $125,000,000,001^2$ is the product of nine odd primes.

But this search found solutions starting with 0 + 1 + 3 + 6 + 10 for all $s \ge 531$. So all integers from 511 to 62,499,999,979 are the sum of three distinct triangular numbers. We can add a fourth triangular number until the distance between two triangular numbers exceeds n = 62,499,999,979, which is at $n \cdot (n + 1)/2 \approx 1.95 \cdot 10^{21}$. We can add a fifth triangular number until the difference between two triangular numbers exceeds $1.95 \cdot 10^{21}$ at $1.9 \cdot 10^{42}$ and so on; each number up to about $1.37 \cdot 10^{336}$ is the sum of 8 triangular numbers.

Therefore every odd square $n^2$ for $n \le 2.74 \cdot 10^{336}$ is the sum of nine odd squares.

Answer 2

Here are some large-$n$ results from order-of-magnitude arguments.

Theorem 1. There is an ineffective constant $N_3$ such that all $n\geq N_3$ with $n\equiv 3\pmod{8}$ satisfy $n=a^2+b^2+c^2$ for some distinct odd integers $a,b,c>11.$

Theorem 2. There is an effective constant $N_4$ such that all $n\geq N_4$ with $n\equiv 4\pmod{8}$ satisfy $n=a^2+b^2+c^2+d^2$ for some distinct odd integers $a,b,c,d>9.$

"Effective" means that in principle you could work out a value by going through all required proofs and replacing all "there exists $c$..." by explicit values. Non-effective constants come from non-constructive arguments, in this case by results affected by Siegel zeroes. Unfortunately ineffective constants aren't directly useful.

For $k\in\{2,3,4\},$ define $r_k(n)$ to be the number of tuples $a\in\mathbb Z^k$ such that $n=\sum_{j=1}^k a_j^2.$ Define $r_2'(n)$ to be the number of pairs $(a,b)\in\mathbb Z^2$ such that $a^2+2b^2=n.$ Denote the divisor function by $d(n).$

Lemma 3.

(a) $r_2(n)\leq 4d(n)$

(b) $r_2'(n)\leq 2d(n)$

(c) $d(n)=n^{o(1)}$

(d) $r_3(n)\geq n^{1/2+o(1)}$ for $n\equiv 3\pmod{8},$ but the implicit constants are ineffective

(e) $r_4(n)=8\sum_{m|n\\4\not|m}m$

(a) follows from "Jacobi's two-square theorem", see Wikipedia's article, "k=2" section.

(b) can be shown by a similar argument as (a), using prime factorization in $\mathbb Z[\sqrt{-2}]$ instead of $\mathbb Z[\sqrt{-1}].$ The factor of $2$ comes from the fact that there are only two units instead of four.

For (c) see Wikipedia.

(d) needs a bit of work to deal with the fact that $n$ might not be square free. let $n=m^2 N$ with $N$ square-free. Then Hirschhorn and Sellers "On representations of a number as a sum of three squares" equation (3) implies $r_3(n)\geq r_3(N) m$ - the factor for each prime $p$ in their equation (3) is $\geq p^{\lambda_p}$ where $p^{\lambda_p}$ is the biggest power of $p$ dividing $m.$ And in our case all prime factors are odd. We can then apply the equation $r_3(N)=24h(-N),$ on Wikipedia's article, "k=3" section where $h(-N)$ is the class number of $\mathbb Q(\sqrt{-N}).$ The estimate $h(-N)=N^{1/2+o(1)}$ is given by the Brauer–Siegel theorem. This gives $r_3(n)/n^{1/2}\geq N^{o(1)}$ where the $o(1)$ tends to zero as $N\to \infty.$ That's also $n^{o(1)}$ as $n\to\infty.$

(e) is "Jacobi's four-square theorem", see Wikipedia's article, "k=4" section.$\square$

Proof of Theorem 1.

By Lemma 3(a,b,c,d), for sufficiently large $n$ we have $r_3(8n+3) > 6\sum_{m\in\{1,3,5,7,9,11\}}r_2(8n+3-m^2) + 6r_2'(8n+3).$ So there is some representation $8n+3=a^2+b^2+c^2$ even after excluding representations with $\{a,b,c\}\cap\{1,3,5,7,9,11\}\neq\emptyset$ or $a=b$ or $b=c$ or $a=c.$ (The factors of $6$ comes from having three options $a,b,c,$ plus an extra choice of sign.) The condition $a^2+b^2+c^2\equiv 3\pmod{8}$ forces $a,b,c$ to be odd.

Proof of Theorem 2.

For any representation $a^2+b^2+c^2+d^2=n$ with $n\equiv 4\pmod{8},$ we either have $a,b,c,d$ all even, or $a,b,c,d$ all odd. Write $r_4(n)=r^{\mathrm{even}}_4(n)+r^{\mathrm{odd}}_4(n)$ where the first term is the number of representations with $a,b,c,d,$ even, and the second term is the number with $a,b,c,d$ odd. By dividing each of $a,b,c,d$ by two we have $r^{\mathrm{even}}_4(n)=r_4(n/4).$ Lemma 3(e) implies $r_4(n)=3r_4(n/4).$ Note $r_4(n/4)\geq n/4.$ Therefore $r^{\mathrm{odd}}_4(n)$ is at least $n/2.$

By considering the "$a$" term of a representation $a^2+b^2+c^2=n$ we have $r_3(n)=\sum_{a^2 \leq n}r_2(n-a^2)$ (with $a$ allowed to be negative). Combining this with Lemma 3(a,c) gives $r_3(n)\leq n^{1/2+o(1)}.$ So there are $\leq n^{1/2+o(1)}$ representations $a^2+b^2+c^2+d^2=n$ with $\min\{a,b,c,d\}\leq 9.$

Using Lemma 3(a,c) again, there are at most $6\sum_{2m^2\leq n}r_2(n-2m^2)\leq n^{1/2+o(1)}$ representations $a^2+b^2+c^2+d^2=n$ with $a,b,c,d$ not all distinct. Here $m$ is allowed to be negative. The factor of $6$ comes from the six choices of a two element subset of four.

For sufficiently large $n\equiv 4\pmod 8,$ the quantity $n/2$ is larger than the sum of these $n^{1/2+o(1)}$ terms. Since $r^{\mathrm{odd}}_4(n)\geq n/2,$ there is a representation $a^2+b^2+c^2+d^2=n$ with $a,b,c,d$ distinct and odd and greater than $9.$

Answer 3

Note: This is not a complete answer, but a result that may be used to progress.

There is a result of Oliverio [1] generalizing Pythagorean triples to multiple dimensions.

Let $S=(a_1,a_2,\cdots,a_{n-2})$, where $a_i$ are integers, and let $T$ be the number of odd integers in $S$. Then iff $T≢2 \pmod{4}$, there exist integers $a_{n-1}$ and $a_n$ such that

$$ a_1^2+a_2^2+ \cdots +a_{n-1}^2=a_n^2 \tag{1} $$

The OP asks (reframed using the notation used here):

For $n=10$, which $a_{n}$ does Eqn. $(1)$ have a solution in integers $a_1, a_2 \cdots a_{9}$, all odd and pairwise distinct?

If we choose odd integers $a_1, a_2, \cdots a_8$, with the added condition required by the OP that they be pairwise distinct, then we have $$T = 8 (\equiv 0 \pmod{4} ≢ 2 \pmod{4}).$$

Therefore, by Oliverio's result, there exist integers $a_{9}, a_{10}$ satisfying the equation

$$a_1^2+a_2^2+\cdots+a_8^2 =a_{10}^2 - a_{9}^2 \tag{2}$$

$a_9, a_{10}$ can be found by factoring the LHS in the Eqn. $(2)$.

The added condition is that $a_9$ is odd.

The LHS is the eight squares representation from Degen's Eight-Square identity.

Therefore, for reversing the process, if we have an integer of the form $A \cdot B$, with $A = a_{10} - a_9$ and $B = a_{10} + a_9$ and $a_9$ odd, then we can find 8-square representations of $A, B$ and use that to get $a_1, a_2 \cdots a_{8}$ such that they are odd.

May be suitable for a computer procedure.

References:

[1]: Oliverio, P. "Self-Generating Pythagorean Quadruples and N-tuples." Fib. Quart. 34, 98-101, 1996. URL: https://www.fq.math.ca/Scanned/34-2/oliverio.pdf