Refined Bezout Lemma Fix positive integers $a,b,c$ such that $\mathrm{gcd}(a,b,c)=1$. Is it true that if $n$ is sufficiently large then there exist positive integers $x,y,z$ such that 
$$
ax+by+cz=n\,\,\,\text{ and }\,\,\,x+y+z \equiv 1\bmod{3}\,\,?
$$
 A: Sorry if I answer my own question, but I didn't think this question has a negative answer.
Countexample: Fix $(a,b,c)=(1,4,4)$. Then, for every positive integer $x,y,z$ such that $x+4y+4z=n$, we have
$$
x+y+z \equiv x+4y+4z \equiv n\bmod{3}.
$$
Therefore it is impossible to satisfy both constraints whenever $n$ is sufficiently large.
A: According to Bezout Lemma, there $\exists x_0,y_0,z_0 \in \mathbb{Z}$ such that
$$ax_0+by_0+cz_0=\gcd(a,b,c)=1 \tag{1}$$
Now


*

*If $3 \nmid (x_0+y_0+z_0)$ then, from FLT $$(x_0+y_0+z_0)^2 \equiv 1 \pmod{3}$$ and 
$$\color{red}{x=}x_0(x_0+y_0+z_0)+3^k$$
$$\color{red}{y=}y_0(x_0+y_0+z_0)+3^k$$
$$\color{red}{z=}z_0(x_0+y_0+z_0)+3^k$$
$$\color{red}{n=}x_0+y_0+z_0+a3^k+b3^k+c3^k$$
Effectively, we multiply $(1)$ by $x_0+y_0+z_0$ and then adding $a3^k+b3^k+c3^k$ to LHS and RHS. Then
$$x+y+z\equiv x_0(x_0+y_0+z_0)+3^k + y_0(x_0+y_0+z_0)+3^k+z_0(x_0+y_0+z_0)+3^k\equiv \\
x_0(x_0+y_0+z_0) + y_0(x_0+y_0+z_0)+z_0(x_0+y_0+z_0) \equiv (x_0+y_0+z_0)^2 \equiv 1 \pmod{3}$$

*If $3 \mid (x_0+y_0+z_0)$ or $$x_0+y_0+z_0 \equiv 0 \pmod{3} \tag{2}$$ then
$$\color{red}{x=}x_0+1+3^k$$
$$\color{red}{y=}y_0+3^k$$
$$\color{red}{z=}z_0+3^k$$
$$\color{red}{n=}1+a+a3^k+b3^k+c3^k$$
Effectively, we add $a+a3^k+b3^k+c3^k$ (for example) to the LHS and RHS of $(1)$ and from $(2)$
$$x+y+z \equiv x_0+1+3^k+y_0+3^k+z_0+3^k \equiv x_0+1+y_0+z_0 \equiv 1 \pmod{3}$$


By choosing large enough $k$, in both cases, we can make $x,y,z$ positive.
Update this proves the statement is true for some sufficiently large $n$, but not all.
