Prove $\{12a+25b:a,b\in\mathbb{Z}\}=\mathbb{Z}$ 
Prove $\{12a+25b:a,b\in\mathbb{Z}\}=\mathbb{Z}$. 

My approach 
Proof: Since, the only prime factors of $12$ are $2$, and $3$ and the only prime factors of $25$ are $5$, the $\gcd(12,25)=1$. Thus, by Bezout's lemma, 1=12x+25y for some $x,y\in\mathbb{Z}$. Now, for all $c\in\mathbb{Z}$, 
  \begin{align*}
      c(12x+25y)&=1(c)\\
      12(xc)+25(yc)&=c\\
  \end{align*}
    Thus, by closure of the set of integers under integer addition and multiplication,$xc,yc\in\mathbb{Z}$. Thus, $12(xc)+25(yc)=12a+25b=c$ for some $a,b\in\mathbb{Z}$, namely, $a=cx$ and $b=cy$. Therefore,  $12a+25b=\mathbb{Z}$, since $c$ is any integer.
This was wrong but can you help show me why?
The reason it is wrong is because this only shows its a subset  of the integers not equal to them. I don't quite see that but would appreciate a full proof. 
 A: In general your proof is correct, but there are some minor mistakes. When you want to prove that two sets $A$ and $B$ are the same you have to prove that $A\subseteq B$ and $B\subseteq A$. 
Let $X=\{12 a + 25 b : a, b \in \mathbb{Z}\}$. So, if $x\in X$, $x=12a+25b$ for some $a, b\in \mathbb{Z}$, then "by closure of the set of integers under integer addition and multiplication" you have $x\in \mathbb{Z}$, therefore $X\subseteq \mathbb{Z}$. On the other hand, suppose $c\in \mathbb{Z}$, we want to show that $\mathbb{Z}\subseteq X$. In order to do that we have to prove that $c\in X$. Since $12(-2)+25(1)=1$, multiplying by $c$ last equality gives us $c(12(-2)+25(1))=c$, then $12(-2c)+25(c)=c$. So if we put $a=-2c$ and $b=c$ we get $12a+25b=c$ and thus $c\in X$. This proves $\mathbb{Z}\subseteq X$. Hence $X=\{12 a + 25 b : a, b \in \mathbb{Z}\}=\mathbb{Z}$.
A: The core of your proof is fine, except for maybe some trivialities to tie the whole thing together. Given $c \in \mathbb Z$, you show that there are $a, b \in \mathbb Z$ such that $12 a + 25 b = c$. Hence $c \in \{ 12 a + 25 b : a, b \in \mathbb Z \}$, so $\mathbb Z \subseteq \{ 12 a + 25 b : a, b \in \mathbb Z\}$. The reverse inclusion holds trivially, hence both sets are equal.
