Gọi $ƯCLN(3n+2;7n+1)=d$ (d là ước nguyên tố)
Ta có: $\left\{ {\matrix{{3n+2 \vdots d} \cr{7n+1 \vdots d} \cr} } \right.$
$=>\left\{ {\matrix{{7(3n+2) \vdots d} \cr{3(7n+1) \vdots d} \cr} } \right.$
$=>\left\{ {\matrix{{21n+14 \vdots d} \cr{21n+3 \vdots d} \cr} } \right.$
$=>(21n+14)-(21n+3) \vdots d$
$=>11 \vdots d$
$=>d∈\{±1;±11\}$
Do d là nguyên tố nên $d=±11$
$=>3n+2 \vdots 11$
$=>3n+11-9 \vdots 11$
$=>3n \not\vdots 11$
Do $ƯC(3;11)=1$
$=>n \not\vdots 11$
$=>$ Để p/s tối giản thì $n=11k$
