The question was:
- Find $n$ where $GCD(a_{n}, 14) = 7$ where $n$ is natural if you knew that $a_{n} = n + 3$
The book solved it by saying that means $a_{n}$ is a multiple of $7$ but not $14$ so $a_{n} = 7a$ and by putting it in the equation
$GCD(7a, 14) = 7$ means $GCD(a,2) = 1$ which means $a = 2k + 1$ and by putting it in $a_{n}$ we find that n = 14a + 4$
The first, I understand that because if it was a multiple of 14 then we would take $14$ as the $GCD$ not $7$
What I didn't understand is that what putting $a_{n} = 7a$ has anything to do with not being a multiple of 14? 14 can be written as $7(2)$ which means $a_{n}$ can be clearly a multiple of 14 too.
The proof is incorrect / incomplete since it gives only a necessary condition for $n$ to be a solution. To fix it you either need to check that the candidate $n$ works or else verify that all inferences are actually bidirectional (see here for more on the insufficiency of unidirectional inferences). Let's do the proof with bidirectional inferences, which will answer your question along the way.
First, $7$ is a common divisor of $\,n+3\,$ and $\,14,\,$ thus a divisor of $\,n+3,\,$ i.e. $\,\color{#0a0}{7\mid n+3},\,$ Using this and $\, 7x = (7y,7z)\color{#c00}\iff x = (y,z)\,$ by $\rm\color{#c00}{GDL}$ = gcd distributive law we have
$$\begin{align} 7 = (n+3,14) &\iff \ \ \color{#0a0}{7\mid n+3}\ \ \ \ \ \&\ \ \ 7 = (n+3,\ \ 14)\\[.1em] &\iff n=4+7k\ \ \ \&\ \ \ 7 = (7+7k,\:\!14)\\[.1em] &\color{#c00}\iff n=4+7k\ \ \ \&\ \ \ 1 = (1+k,\ 2)\ \ \ \rm by\ \color{#c00}{GDL}\\[.1em] &\iff n=4+7k\ \ \ \&\ \ \ k = 2j\\[.1em] &\iff n = 4+7(2j) \end{align}\qquad$$