<s>[INST] Prove that $∀a, b, c ∈ \mathbb Z$ with $b \neq 0$, if $b|c$, then $\gcd(a, b) = \gcd(a + c, b)$. 
Prove that $∀a, b, c ∈ \mathbb Z$ with $b \neq 0$, if $b|c$, then $\gcd(a, b) = \gcd(a + c, b)$.

I've proved it in the following way:    
$\gcd(c,b) = b.$
Let $\gcd(a,b) = x$, $x|a$ and $x|b$
Since $b$ is the $\gcd$ of $c$,
if $x|b$ then $x|c$  
Thus, $\gcd(a+c, b) = x = \gcd(a, b)$.  
Was wondering if this is a solid proof. If not, is there anyway I can improve on this? Thanks for helping out
 [/INST] The conclusion $\gcd(a+c, b) = x = \gcd(a, b)$ is not that clear from your previous parts.
One way to show the claim $\gcd(a, b) = \gcd(a + c, b)$ is to split the task into two parts:
$\qquad\qquad\gcd(a, b)|\gcd(a + c, b)\qquad$ and $\qquad\gcd(a+c, b)|\gcd(a, b)$
from which equality follows.

We obtain
  \begin{align*}
\begin{array}{rlr}
\gcd(a,b)|b&\qquad\qquad(\text{by definition of }\gcd)&\qquad (1)\\
b|c&\qquad\qquad(\text{by assumption})\\
\Longrightarrow \gcd(a,b)|c&\qquad\qquad(\text{by  transitivity of }|)&\qquad(2)\\
\\
\gcd(a,b)|a&\qquad\qquad(\text{by definition of }\gcd)&\qquad(3)\\
\\
\Longrightarrow  \gcd(a,b)|(a+c)&\qquad\qquad(\text{by(2) and (3)})&\qquad(4)\\
\\
\color{blue}{\Longrightarrow  \gcd(a,b)|\gcd(a+c,b)}&\qquad\qquad(\text{by(1) and (4)})&\qquad\color{blue}{(\text{I})}\\
\end{array}
\end{align*}

We further obtain
\begin{align*}
\begin{array}{rlr}
\gcd(a+c,b)|b&\qquad\qquad(\text{by definition of }\gcd)&\qquad(5)\\
b|c&\qquad\qquad(\text{by assumption})\\
\Longrightarrow \gcd(a+c,b)|c&\qquad\qquad(\text{by  transitivity of }|)&\qquad(6)\\
\\
\gcd(a+c,b)|(a+c)&\qquad\qquad(\text{by definition of }\gcd)&\qquad(7)\\
\\
\Longrightarrow  \gcd(a+c,b)|a&\qquad\qquad(\text{by(6) and (7)})&\qquad(8)\\
\\
\color{blue}{\Longrightarrow  \gcd(a+c,b)|\gcd(a,b)}&\qquad\qquad(\text{by(5) and (8)})&\qquad\color{blue}{(\text{II})}\\
\end{array}
\end{align*}

We conclude from (I) and (II)
  \begin{align*}
\color{blue}{\gcd(a,b)=\gcd(a+c,b)}
\end{align*}

</s>