The number $1680$ has the factorization $2^4\times3\times5\times7$. Rather to my surprise, I found that this number has certain mysterious connection with Lagrange's four-square theorem.
QUESTION: Can we write each $n\in\mathbb N=\{0,1,2,\ldots\}$ as $x^2+y^2+z^2+w^2$ with $x,y,z,w\in\mathbb N$ such that $x^4+1680y^3z$ is a square?
My computation suggests that it might have a positive answer. Of course, if $n\in\mathbb N$ is not of the form $4^k(8l+7)$ with $k,l\in\mathbb N$ then by the Gauss-Legendre theorem there are $x,y,z\in\mathbb N$ such that $n=x^2+y^2+0^2+z^2$ with $x^4+1680y^3\times0$ a square. If $$n=x^2+y^2+z^2+w^2\ \ \text{with}\ x^4+1680y^3z=m^2,$$ then $$4n=(2x)^2+(2y)^2+(2z)^2+(2w)^2\ \ \text{with}\ (2x)^4+1680(2y)^3(2z)=(4m)^2.$$ So it suffices to consider the question only for $n\equiv 7\pmod 8$.
For $n\in\mathbb N$ let $a(n)$ denote the number of ways to write $8n+7$ as $x^2+y^2+z^2+w^2$ $(x,y,z,w\in\mathbb N)$ with $x^4+1680y^3z$ a square. Then $a(n)=1$ for $$n=0,\ 11,\ 244,\ 289,\ 664,\ 749,\ 983,\ 1228,\ 1819,\ 2503,\ 2506.$$ For example, \begin{gather*}8\times0+7 = 1^2 + 1^2 + 1^2 + 2^2\ \text{with}\ 1^4 + 1680\times1^3\times1 = 41^2,\\ 8\times11 + 7 = 6^2 + 3^2 + 1^2 + 7^2\ \text{with}\ 6^4 + 1680\times3^3\times1 = 216^2,\\8\times244 + 7 = 13^2 + 13^2 + 39^2 + 10^2\ \text{with}\ 13^4 + 1680\times13^3\times39 = 11999^2, \\ 8\times1228 + 7 =35^2 + 10^2 + 91^2 + 15^2\ \text{with}\ 35^4 + 1680\times10^3\times91 = 12425^2. \end{gather*} The values $a(n)$ with $0\le n\le 10^4$ are available from http://oeis.org/A280831.
Moreover, I conjecture that for nonzero integers $a$ and $b$ with $\gcd(a,b)$ squarefree, every $n\in\mathbb N$ can be written as $x^2+y^2+z^2+w^2\ (x,y,z,w\in\mathbb N)$ with $ax^4+by^3z$ a square if and only if $(a,b)$ is among the following ordered pairs $$(1,1),\ (1,15),\ (1,20),\ (1,36),\ (1,60),\ (1,1680),\ (9,260).$$
I'm really puzzled by my question and have no explanation for the curious phenomenon. Any helpful ideas?
