number theory - gcd as positive linear combination

Good evening,

I have a question concerning the euclidean algorithm.

One knows that for $a_1 , \ldots , a_n \in \mathbb{N} $ and $k\in \mathbb{N} $ there exist some $\lambda_i \in \mathbb{Z}$ such that :

$$\gcd(a_1, \ldots, a_n) = \frac{1}{k}\sum_{i=1}^n \lambda_i a_i$$

Here is my question: can one find a $m_0 \in \mathbb{N}$ that for every $m \geq m_0$ there are scalars $\mu_i \in \mathbb{N}$ such that:

$$\gcd(a_1, \ldots , a_n) = \frac{1}{m}\sum_{i=1}^n \mu_i a_i$$

Unfortunately I have only very rudimentary knowledge about number theory ...

With best regards
Mat

Answer

Let's say that $\gcd(a_1,\ldots,a_n)=d$ and $d=\displaystyle{\sum_{i=1}^{n}\lambda_ia_i}$ for some $\lambda_i\in \mathbb Z$.
Suppose that $s_i\in \mathbb N$ are sufficiently large such that $r_i=\lambda_i+s_ia_1a_2\ldots a_{i-1}a_{i+1}\ldots a_n>\dfrac{a_1}{d}|\lambda_i|$ and $\displaystyle{\sum_{i=1}^{n}r_ia_i}=m_0d$ for some $m_0\in \mathbb N$.
For all $r=0,1,\ldots,\dfrac{a_1}{d}-1$ we have $$(m_0+r)d=\displaystyle{\sum_{i=1}^{n}(r_i+r\lambda_i)a_i}$$ and $r_i+r\lambda_i>0$ for all $i.$
For $r\geq\dfrac{a_1}{d}$ if $r=q\dfrac{a_1}{d}+s$ with $q \in \mathbb N, \ s\in\left\{0,1,\ldots,\dfrac{a_1}{d}-1\right\}$ we have $$(m_0+r)d=(r_1+s\lambda_1+q)a_1+\displaystyle{\sum_{i=2}^{n}(r_i+s\lambda_i)a_i}$$
and $r_1+s\lambda_1+q>0, \ r_i+s\lambda_i>0$ for all $i\geq2.$
Therefore for every $m\geq m_o,\ \ md=\displaystyle{\sum_{i=1}^{n}\mu_i a_i}\Rightarrow d=\displaystyle{\frac{1}{m}\sum_{i=1}^{n}\mu_i a_i}$ for some $\mu_i\in \mathbb N$.