How to prove if there are exist positive integer solutions to these two variables inequalities system? { 141 n − 143 m ≤ − 60 143 m − 141 n ≤ 138

Question

How to prove if there are exist positive integer solutions to these two variables inequalities system?      {                            141          n                          −          143          m                          ≤          −          60                                      143          m                          −          141          n                          ≤          138

Jayvion Tyler · Accepted Answer

If you multiply the second equation by   −  1, so the system becomes      {                            141          n          −          143          m                          ≤          −          60                                      141          n          −          143          m                          ≥          −          138.                        you see that all you have to consider is   141  n  −  143  m for   n  ,  m  ∈  Z.It might be easier to think of   n as fixed and   m  =  n  +  k as varying, so  141  n  −  143  m  =  141  n  −  143  n  −  143  k  =  −  2  n  −  143  k  ∈  [  −  138  ,  −  60  ]  .This is equivalent to  −  143  k  ∈  [  −  138  +  2  n  ,  −  60  +  2  n  ]  .Let&#039;s define the interval       I    n    =  [  −  138  +  2  n  ,  −  60  +  2  n  ]If   n  =  0, then does the condition   −  143  k  ∈      I    n    =      I    0    =  [  −  138  ,  −  60  ] have any solutions for   k  ∈  Z? How about if   n  =  −  3?Conversely, if   k  =  2, then is there any choice (or multiple choices, perhaps) of   n such that   −  143  k  ∈      I    n  ?As food for thought, if   −  143  k  ∈      I    n   has a solution, is it unique?

Nickolas Taylor · Accepted Answer

If that&#039;s the case, note if   n  =  −  3 then   −  143  =  −  143  ×  1  ∈      I          −      3        =  [  −  144  ,  −  66  ], that is if   k  =  1 and   n  =  −  3 then   −  143  k  ∈      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  ×  2  =  286. So   (  n  ,  k  )  =  (  200  ,  −  2  ), that is   (  n  ,  m  )  =  (  200  ,  198  ) is a solution.

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