Gọi \(d=ƯCLN\left(3n+1;4n+2\right)\left(d\in N\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}3n+1⋮d\\4n+1⋮d\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12n+4⋮d\\12n+3⋮d\end{matrix}\right.\)
\(\Leftrightarrow d=1\)
Vì \(d\in N;1⋮d\Leftrightarrow d=1\)
\(\LeftrightarrowƯCLN\left(3n+1;4n+1\right)=1\)
\(\Leftrightarrow3n+1;4n+1\) nguyên tố cùng nhau với mọi \(n\in N\)
Gọi \(d\) = ƯCLN (3n + 1: 4n + 2).
\(\Leftrightarrow\left\{{}\begin{matrix}3n+1⋮d\\4n+1⋮d\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4\left(3n+1\right)⋮d\\3\left(4n+1\right)⋮d\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}12n+4⋮d\\12n+3⋮d\end{matrix}\right.\)
\(\Rightarrow\left(12n+4\right)-\left(12n+3\right)⋮d\)
\(\Leftrightarrow12n+4-12n-3⋮d\)
\(\Leftrightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(3n+1\) và \(4n+1\) nguyên tố cùng nhau với mọi \(n\in N\)* .
