Good evening,
I have a question concerning the euclidean algorithm.
One knows that for a1,…,an∈N and k∈N there exist some λi∈Z such that :
gcd
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.
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