Đặt d=ƯC(2n+1;\(\dfrac{n\left(n+1\right)}{2}\))\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2n+1⋮d\\\dfrac{n\left(n+1\right)}{2}⋮d\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}n\left(2n+1\right)⋮d\\\dfrac{4n\left(n+1\right)}{2}\end{matrix}\right.⋮d\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2n^2+n⋮d\\2n^2+2n⋮d\end{matrix}\right.\)\(\Leftrightarrow2n^2+2n-2n^2-n⋮d\Leftrightarrow n⋮d\Leftrightarrow2n⋮d\)
Mà 2n+1\(⋮d\)
Suy ra \(1⋮d\Leftrightarrow d=1\)
Vậy hai số 2n+1 và \(\dfrac{n\left(n+1\right)}{2}\) là hai số nguyên tố cùng nhau