A DIVISIBILITY of TWO AFFINE EXPRESSIONS - Problem S:E19.221

          In "Suplimentul ... " page 7

          English translation from Romanian, thanks to Miss Google:

           "Determine the natural numbers $n$ for which $2n-3$ divides $5n+2$."

 

ANSWER CiP:  $n=2$ and $n=11$

 Solution CiP

          If $2n-3 \mid 5n+2$, then $2n-3 \mid 5n+2-2(2n-3)$ $\Leftrightarrow $

(1)                             $2n-3 \mid n+8$.

From (1) we obtain $2n-3 \mid 2n-3-(n-11)$, so

(2)                            $2n-3 \mid n-11$.

If we subtract (1)-(2), we conclude $2n-3 \mid 19$ so $2n-3=1$ or $2n-3=19$.
We find the answer, both values checking the condition.

$\blacksquare$

           Remark If the statement required $n \in \mathbb{Z}$ instead of $n \in \mathbb{N}$ the answer will be

$n \in \{-8, 1, 2, 11\}$.

Alternative solution

 

          $2n-3 \mid 5n+2 \;\; \Leftrightarrow \;\; 5n+2=(2n-3) \cdot k$      for some $k \in \mathbb{Z}$

(we consider the case more general $n \in \mathbb{Z}$). The last equation is successively equivalent to

$2nk-5n-3k=2 \;\; \Leftrightarrow \;\;4nk-10n-6k=4\;\; \Leftrightarrow \;\;2n(2k-5)-3(2k-5)-15=4\;\; \Leftrightarrow \;\;(2k-5)(2n-3)=19$

       The numbers dividing 19 are $\pm1$ and $\pm 19$ so $2n-3 \in \{\pm 1, \pm 19 \}$. We get the answer again.

$\blacksquare$