Gọi ƯCLN(2n + 1,3n + 2) = d
=> Ta có: \(\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}}\)=> \(\hept{\begin{cases}3.\left(2n+1\right)⋮d\\2.\left(3n+2\right)⋮d\end{cases}}\)=> \(\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}}\)
=> \(\left(6n+4\right)-\left(6n+3\right)⋮d\)
=> \(6n+4-6n-3⋮d\)
=> \(1⋮d\)
=> \(d=1\)
=> 2n + 1 ; 2n + 2 là 2 số nguyên tố cùng nhau