# How to prove if there are exist positive integer solutions to these two variables inequalities system?

I would like to know how to prove if there are exist positive integer solutions (for $$m$$ and $$n$$) to: $$\begin{cases} 141n &- 143m &\leq -60\\ 143m &- 141n &\leq 138 \end{cases}$$

I need to prove it mathematically. Any comments are welcome. Thank you and regards, Tony.

• The idea of "if there are always integer solutions" is weird since this is a specific problem: Just find a solution! (or, show none exist). Commented Feb 13, 2020 at 2:13
• I need to prove that there is always exist integer solution (both $m$ and $n$) for those inequalities. Commented Feb 13, 2020 at 2:18
• Can you explain why you use the word "always"? It is like asking "Is there always a solution to $2x=4$?" Why use the word "always"? Either there is a solution, or there is not. Commented Feb 13, 2020 at 2:23
• I delete the word "always". Commented Feb 13, 2020 at 2:26
• If all you need to do is prove existence for one solution, plugging in an example works. Commented Feb 13, 2020 at 2:38

$$143 \cdot 71 - 141 \cdot 72 = 1$$

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If you multiply the second equation by $$-1$$, so the system becomes $$\begin{cases} 141n - 143m &\leq -60\\ 141n - 143m &\geq -138. \end{cases}$$ you see that all you have to consider is $$141n - 143m$$ for $$n, m \in \mathbb Z$$.

It might be easier to think of $$n$$ as fixed and $$m = n + k$$ as varying, so $$141n - 143m = 141n - 143n - 143k = -2n - 143k \in [-138, -60].$$ This is equivalent to $$-143k \in [-138 + 2n, -60 + 2n].$$

Let's define the interval $$I_n = [-138 + 2n, -60 + 2n]$$.

If $$n = 0$$, then does the condition $$-143k \in I_n = I_0 = [-138, -60]$$ have any solutions for $$k \in \mathbb{Z}$$? How about if $$n = -3$$?

Conversely, if $$k = 2$$, then is there any choice (or multiple choices, perhaps) of $$n$$ such that $$-143k \in I_n$$?

As food for thought, if $$-143k \in I_n$$ has a solution, is it unique?

EDIT: I didn't see the question was just looking for a start. I've removed some parts.

• sorry, actually i need to prove it mathematically. Commented Feb 13, 2020 at 3:38
• What do you mean? As far as I can tell, this is mathematical. If $m = n + k$ then $(n, k)$ is a solution if and only if $-143k \in I_n$. Are you unsure how to show there exists infinitely many choices of $(n, k)$ with this property? Commented Feb 13, 2020 at 3:50
• I mean you don't need to remove some parts before. I want to show that there exists at least one pair positive integer for $m$ and $n$. Commented Feb 13, 2020 at 4:35
• If that's the case, note if $n = -3$ then $-143 = -143 \times 1 \in I_{-3} = [-144, -66]$, that is if $k = 1$ and $n = -3$ then $-143k \in I_n$ is satisfied. This means that $n = -3$ and $m = n + k = -2$ is a solution. If you want other solutions, all you need to do is find $n$ such that $I_n$ contains an integral multiple of $143$. For instance, if $n = 200$ then $I_{200} = [262, 340]$ which contains $143 \times 2 = 286$. So $(n, k) = (200, -2)$, that is $(n, m) = (200, 198)$ is a solution. Commented Feb 13, 2020 at 5:23
• If you want to prove a solution exists without explicitly 'guessing' values of $n$, although you can make 'clever' guesses since you know you need to make sure $I_n$ contains a multiple of 143, one way is to note you can find $n$ such that $[-138 + 2n, -60 + 2n]$ contains $[0, 1]$. Since $[0 + 2n, 1 + 2n] = [2n, 2n + 1]$ and every number is either $2n$ or $2n + 1$, you can hit every multiple of 143. This shows that every $k \in \mathbb Z$ has an $n$ for which $-143k \in I_n$. (But there are multiple choices of $n$.) Commented Feb 13, 2020 at 5:28