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
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