0
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ 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). $\endgroup$
    – Michael
    Commented Feb 13, 2020 at 2:13
  • $\begingroup$ I need to prove that there is always exist integer solution (both $m$ and $n$) for those inequalities. $\endgroup$
    – gu0hu1
    Commented Feb 13, 2020 at 2:18
  • $\begingroup$ 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. $\endgroup$
    – Michael
    Commented Feb 13, 2020 at 2:23
  • $\begingroup$ I delete the word "always". $\endgroup$
    – gu0hu1
    Commented Feb 13, 2020 at 2:26
  • 1
    $\begingroup$ If all you need to do is prove existence for one solution, plugging in an example works. $\endgroup$
    – Snacc
    Commented Feb 13, 2020 at 2:38

2 Answers 2

1
$\begingroup$

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

........................

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ sorry, actually i need to prove it mathematically. $\endgroup$
    – gu0hu1
    Commented Feb 13, 2020 at 3:38
  • $\begingroup$ 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? $\endgroup$
    – Riley
    Commented Feb 13, 2020 at 3:50
  • $\begingroup$ 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 $n$. $\endgroup$
    – gu0hu1
    Commented Feb 13, 2020 at 4:35
  • $\begingroup$ 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. $\endgroup$
    – Riley
    Commented Feb 13, 2020 at 5:23
  • $\begingroup$ 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 $n$.) $\endgroup$
    – Riley
    Commented Feb 13, 2020 at 5:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .