On aggregating two linear diophantine equations (Q1383381)

Given two diophantine equations \[ \sum_{j\in \mathbb{N}} a_{1j}x_j= b_1,\;\sum_{j\in \mathbb{N}}a_{2j}x_j= b_2, \quad x_j\geq 0\text{ for all }j\in \mathbb{N}= \{1,\dots, n\}, \tag{1} \] the problem is to find relatively prime integers \(t_1,t_2\), so that \[ \sum_{j\in \mathbb{N}} (t_1a_{1j}+ t_2a_{2j})x_j= t_1b_1+ t_2b_2 \tag{2} \] has the same solution set in nonnegative integers as (1). Using the notations \(w_j= c_1a_{2j} c_2a_{1j}\), \(j\in \mathbb{N}\), \(c_1,c_2\) integers not both equal, \(w_0= c_2b_1-c_1b_2\), \(x_0=1\), \(S(X)= S(X;c_1,c_2)= \sum_{j\in \mathbb{N}\cup \{0\}}w_jx_j\), \[ -\infty< L\leq\min_{X\in T}S(X), \qquad \max_{X\in T}S(X)\leq U<+\infty, \tag{3} \] \[ T= \{X: X=(x_1,\dots, x_n) \text{ satisfying }(2)\}, \] the main result is: the system (1) is equivalent to (2) if for arbitrary integers \(c_1,c_2\) not both equal the inequality \(t_1c_1+ t_2c_2> \max\{-L,U\}\) holds, where \(t_1,t_2\) are relatively prime integers and \(L,U\) any values satisfying (3). In general it is difficult to find \(\min S(X)\) and \(\max S(X)\), therefore in the following sections of the paper the main result is used to aggregate (1) under some additional conditions. In some cases the results are better than those given in the literature, i.e., the coefficients in (2) are smaller in the sense of absolute value.