Sufficient conditions for solvability of linear Diophantine equations and Frobenius numbers

Abstract: A linear Diophantine equation $\sum_{i=1}^{n}a_ix_i=b$ is considered, where $a_1,a_2,...,a_n$ are coprime natural numbers, $b$ is an non-negative integer, $x_i$ $(i=1,2,...,n)$ are non-negative integers. It is proved that if $\left[\frac{b}{M}\right]\geq \left[n-\frac{\sum_{i=1}^{n}a_i+r}{M}\right]$, then this equation is solvable (here $M$ is the least common multiple of the numbers $a_1,a_2,...,a_n$; $r$ is the remainder of $b$ modulo $M$). With the aid of this result, it is shown that the considered equation is solvable if either $b\geq nM-\sum_{i=1}^{n}a_i$ or $b\geq (n-1)M$. The case where $\sum_{i=1}^{n}a_i\geq (n-2)M+2$ is considered closer. It is proved that, in that case, the equation is solvable if $b>(n-1)M-\sum_{i=1}^{n}a_i$. If, in addition, $r>(n-1)M-\sum_{i=1}^{n}a_i$, then it has $\frac{M^{n-1}}{a_1a_2...a_n}C^{n-1}_{b'+n-1}$ integer non-negative solutions; it is also shown that if $b\geq M$ and $r\leq (n-1)M-\sum_{i=1}^{n}a_i$, then the number of solutions of the equation is greater than or equal to $\frac{M^{n-1}}{a_1a_2...a_n}C^{n-1}_{b'+n-2}$; here $b'=\left[\frac{b}{M}\right]$ (and $C^{k}_{m}$ denotes the binomial coefficient ${\begin{pmatrix}m\k\\end{pmatrix}}$). Moreover, it is proved that if the numbers $a_1,a_2,...,a_n$ are coprime, then $Frob(a_1,a_2,...,a_n)\leq c$, where $c$ is the smallest among the numbers $a_ia_j-a_i-a_j$ with $(a_i,a_j)=1$ ($1\leq i< j\leq n$).