Lagrange's four-square theorem asserts that each $n\in\mathbb N=\{0,1,2,\ldots\}$ is the sum of four integer squares. This nice result is actually very weak, for example, it is a consequence of the Gauss-Legendre theorem on sums of three squares. As $$4(w^2+x^2+y^2+z^2)=(2w)^2+(2x)^2+(2y)^2+(2z)^2,$$ Lagrange's four-square theorem is equivalent to that any positive integer not divisible by $4$ can be written as the sum of four squares.

Here I propose a new refinement of Lagrange's four-square theorem.

QUESTION 1. Is my following conjecture true?

**Big 1-3-5 Conjecture**. Any positive integer $n\not\equiv0\pmod8$ can be written as
$$w^2+\left(\frac{x(x+1)}2\right)^2+\left(\frac{y(3y+1)}2\right)^2+\left(\frac{z(5z+1)}2\right)^2,$$
where $w$ is a positive integer and $x,y,z$ are integers.

I have verified this for $n$ up to $2\times10^7$. For example, $28$ has a unique required representation: $$28=3^2+\left(\frac {2(2+1)}2\right)^2+\left(\frac{(-1)(3\times(-1)+1)}2\right)^2+\left(\frac{1(5\times1+1)}2\right)^2.$$ See also http://oeis.org/A306614 for related data and similar conjectures.

Note that the Big 1-3-5 Conjecture is different from my 1-3-5 conjecture published in this paper which states that any $n\in\mathbb N$ can be written as $x^2+y^2+z^2+w^2$ with $x,y,z,w\in\mathbb N$ such that $x+3y+5z$ is a square. We should also not confuse it with my little 1-3-5 conjecture (cf. http://oeis.org/A287616) which states that any $n\in\mathbb N$ can be written as $x(x+1)/2+y(3y+1)/2+z(5z+1)/2$ with $x,y,z\in\mathbb N$.

The standard algorithm to express $n\in\mathbb N$ as $w^2+x^2+y^2+z^2$ with explicit $w,x,y,z\in\mathbb N$ is as follows: Find $0\le x\le\sqrt n$, $0\le y\le\sqrt{n-x^2}$ and $0\le z\le\sqrt{n-x^2-y^2}$ with $n-x^2-y^2-z^2$ a square. Under the Big 1-3-5 Conjecture, we have a better algorithm: For $n\not\equiv0\pmod 8$, find \begin{gather}x\in\mathbb N\ \text{with}\ x(x+1)/2\le\sqrt n, \\y\in\mathbb Z\ \ \text{with}\ \frac{y(3y+1)}2\le\sqrt{n-\left(\frac{x(x+1)}2\right)^2} \\z\in\mathbb Z\ \ \text{with}\ \frac{z(5z+1)}2\le\sqrt{n-\left(\frac{x(x+1)}2\right)^2-\left(\frac{y(3y+1)}2\right)^2} \end{gather} such that $n-(x(x+1)/2)^2-(y(3y+1)/2)^2-(z(5z+1)/2)^2$ is a square.

QUESTION 2. Is there an algorithm to represent an arbitrary positive integer as the sum of four squares which is more efficient than the use of the Big 1-3-5 Conjecture?

The Big 1-3-5 Conjecture is much stronger than Lagrange's four-square theorem. Your comments or further check are welcome!