A property of relatively prime integers

Question

I need to prove the following fact:

1. If $(n_1, n_2,\cdots, n_m)=1$, then for any $n>n_1n_2\cdots n_m$, there are non-negative integers $k_1, k_2, \cdots, k_m$, such that $n=k_1n_1+k_2n_2+\cdots+k_mn_m$.

For the case of two integers, I used the following argument:

2. If $(a, b)=1$, then for any $n>ab$, there exist $k, l$ both non-negative, such that $n=ak+bl$.

Proof: Fix $n>ab$. Let $k_1,k_2\in \mathbb{Z}$ be such that $n=ak_1+bk_2$. If $k_1,k_2$ are already non-negative, we are done. Otherwise, assume $k_1>0$ and $k_2<0$, then from $ak_1+bk_2>ab$, we have $k_1>b$, write $k_1=br+q$, where $0\leq q<b$, then $ak_1+bk_2=aq+b(ar+k_2)>ab$, hence $b(ar+k_2)>a(b-q)>0$, inparticular $ar+k_2>0$, let $k=q$ and $l=ar+k_2$, we complete the proof.

My question is how to prove the general case, I failed to reduce statement 1 to 2, your hints are appreciated.

Answer 1

Let us argue by induction. The case $n = 2$ has been done above.

Consider $n_1,...,n_m$ as above, now with $m \geq 3$. Without loss of generality, we may assume that $2 \leq n_1 \leq n_2 \leq ... \leq n_m$, since if any $n_i$ is $1$ then the problem is trivial. The hypothesis implies $(n_1, (n_2...n_m)) = 1$. Let $d = (n_2,...,n_m)$. Consider $n > n_1...n_m= M$.

First, note that there is some $d > k \geq 0$ such that $n - kn_1$ is a multiple of $d$. This is because the numbers $n,n-n_1,n-2n_1,...,n-(d-1)n_1$ are $d$ different numbers, which leave $d$ different remainders when divided by $d$, because the difference of any two of them is of the form $jn_1$ with $j < d$, hence is not a multiple of $d$. One of these must leave remainder $0$, since there are only $d$ many possible remainders. The claim follows. Note that $k < d$, so $kn_1 < dn_1 \leq M$, therefore $n-kn_1 > 0$ is a positive multiple of $d$. Let $n - kn_1 = ld$.

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Suppose we can show that $l > \frac{n_2n_3...n_m}{d^{m-1}}$. Then, since $\left(\frac{n_2}d,...,\frac{n_m}d\right) = 1$, we get by the induction hypothesis that $l = \frac{\sum_{i=2}^m k_in_i}{d}$ for positive $k_i$, therefore $n = kn_1 + ld = kn_1 + \sum k_in_i$ is a non-negative combination of the $n_i$, hence we are done.

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To show that the condition holds, we go in steps.

• Note that $dl = n - kn_1$. Since $k <d$ we get $dl > n - dn_1$. Since $n > M$ we get $dl > M - dn_1$. Finally, since $d$ divides $n_2$ we have $d \leq n_2$ therefore $dl > M - n_2n_1$.

• Note that $n_1 \geq \frac{n_3...n_m}{n_3...n_m - 1}$ , since the left hand side is smaller than $2$ (it is equal to $2$ when $n_3...n_m = 2$ , but use the condition on $n_i$, and the gcd condition to derive a contradiction. Hence $n_3...n_m > 2$)

• Therefore, multiplying by $n_2$ on both sides , and noting that $M - n_2n_1 = n_2n_1(n_3...n_m - 1)$, we get $M - n_2n_1 > n_2n_3...n_m$. Of course, this implies $M - n_2n_1 > \frac{n_2n_3...n_m}{d^{m-2}}$.

Now the result follows by combining the bullet points.