# Positive integer solutions of linear equations under the constraint of Frobenius number

Let $$a,b,c$$ be three pairwise coprime positive integers, and $$\Gamma=\langle a,b,c\rangle$$ be the corresponding numerical semigroup. Consider the linear equations:

$$n_1a=m_{12}b+m_{13}c$$

$$n_2b=m_{21}a+m_{23}c$$

$$n_3c=m_{31}a+m_{32}b$$

Where $$n_i,m_{ij}$$ are positive integers. Also, let $$n_1=\min\{x\in\mathbb{Z}_{>0}|xa\in\langle b,c\rangle\}$$, i.e, we choose the smallest $$n_1$$ that satisfies the equation, and $$n_2$$ and $$n_3$$ are chosen similarly.

Then it's not hard to prove that $$n_1=m_{21}+m_{31}$$, $$n_2=m_{12}+m_{32}$$, and $$n_3=m_{13}+m_{23}$$. For the prove see http://hera.ugr.es/doi/15773139.pdf.

My question is, does there exist $$a,b,c$$ such that $$n_1a, $$n_2b, $$n_3cwhere $$F$$ is the Frobenius number?

Edit: thanks for the example, the answer is positive. Furthermore, does there exist $$a,b,c$$ such that $$\max\{n_1a,n_2b,n_3c\}

• The formula for the Frobenius number should be $F(\langle a,b \rangle)=ab-a-b$. – Francesco Nov 8 at 11:03

Let $$a=30$$, $$b=31$$ and $$c=37$$. Then
$$\begin{array}{ll} 7a=2b+4c=210 \\ 7b=6a+c=217 \\ 5c=a+5b=185 \\ \end{array}$$
and the Frobenius number of $$\langle a,b,c \rangle$$ is 267.