We knew that any natural number can be expressed as a sum of at most three triangular numbers, which Gauss wrote as $n=\Delta+\Delta+\Delta$. The number $4$ for the polynomial that we coined the four square theorem.
With the help of Shnirelman theorem we can assert that $\sigma(\Delta^{(2)})>0$, but cannot be $1$, where $\Delta$ in this case represent the set of all triangular numbers, and $\sigma(S)$ be the density of $S$. In addition, $\Delta^{(2)}=\left\{a+b : a,b\in\Delta\right\}\cup \Delta$
Is there any type of quardratic polynomial that used only $2$ sets of it to cover $\mathbb N$. In other word, Is there any $\mathscr P=\left\{an^2+bn+c\right\}$ that $\sigma(\mathscr P^{(2)})=1$?
Otherwise, Three would be the smallest number for any quardratic polynomial to additively form $\mathbb N$.
Write $$ 4a ( a n^2 + bn +c ) = (2an+b)^2 + (4ac-b^2) $$ Shift every positive integer by adding $2(4ac-b^2),$ then ask if a positive density of those might be represented. The answer is no, we now have (with $n,m$) $$ (2an+b)^2 + (2am+b)^2 = W, $$ where the original target number $4aT$ is shifted to $W = 4aT + 2(4ac-b^2).$ We continue to ignore congruences.., and the exact expression for $W$ is not critical. Taking $X= 2an+b$ and $Y = 2am+b$ we have constructed $$ X^2 + Y^2 = W $$ The collection of $W$ that can be so represented has density zero.
Proof is in the last chapter of W. LeVeque, Topics in Number Theory. I have a Dover reprint, two volumes in one. This is volume 2, section 7-5, pages 257-263. For your purpose, a proof of density zero is good enough; I suspect this can be done just by writing out prime factorizations, as a number fails to be represented when the exponent of any prime $q \equiv 3 \pmod 4$ is odd. I think I will look in Kac's little book on statistical independence in number theory. Same guy as the Erdos-Kac Theorem