Gọi \(d\) là \(UCLN\left(n+1;3n+4\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n+1⋮d\\3n+4⋮d\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3n+3⋮d\\3n+4⋮d\end{matrix}\right.\)
\(\Rightarrow3n+4-3n-3⋮d\)
\(\Rightarrow1⋮d\Leftrightarrow d=1\)
Nên \(\left(n+1;3n+4\right)=1\left(đpcm\right)\)