Calculate the gcd of $21 n+4$ and $14 n+3$.
We apply the Euclidean algorithm.

$$
\begin{aligned}
21 n+4 & =(14 n+3) \cdot 1+7 n+1 \\
14 n+3 & =2 \cdot(7 n+1)+1 \\
7 n+1 & =1 \cdot(7 n+1)+0
\end{aligned}
$$

Therefore, this gcd is 1, and this holds for all $n \in \mathbb{N}$.