Suppose $a,b\in \mathbb{Z}$ are relatively prime and $c\in \mathbb{N}$ is a divisor of $a+b$. **Verify my proof** that gcd$(a,c)$=gcd$(b,c)=1$.

Question

I am looking for someone to verify my proof of the problem in the title. Of course, if you believe it to be wrong, you are welcome to shred it into a million pieces and if I made any logical errors please point them out. I'll write my solution out in the most coherent way I know how to.

I am also interested, even if the proof provided below is correct, is there a "simpler" (whatever you consider that to mean) way to approach this?

Proof

We have that $c$ divides $a+b$. This means $\exists \lambda \in \mathbb{Z}$, \begin{equation*} a+b = \lambda c. \end{equation*} Note that $a=\lambda c -b$ and $b = \lambda c - a$. This will be immediately useful.

Let $x \in \langle a \rangle + \langle c \rangle$. Then $x=sa+tc$ for some $s,t\in\mathbb{Z}$.

$sa+tc = s(\lambda c - b) + tc = (-s)b + (s\lambda + t)c \Rightarrow x \in \langle b \rangle + \langle c \rangle$. This means $\langle a \rangle + \langle c \rangle \subseteq \langle b \rangle + \langle c \rangle$. It is easy to verify that $\langle b \rangle + \langle c \rangle \subseteq \langle a \rangle + \langle c \rangle$ in the same way and hence $\langle a \rangle + \langle c \rangle = \langle b \rangle + \langle c \rangle$. Therefore we have gcd$(a,c)=$gcd$(b,c)$.

To show that $a$ is relatively prime to $c$ and $b$ is relatively prime to $c$ we suppose that there are some $f,g\in\mathbb{Z}$ with $f\neq \pm 1$ and $g\neq \pm 1$ such that $a=fc$ and $c=fa$, $b=gc$ and $c=gb$. Then \begin{equation*} a=fc=(fg)b \end{equation*} Now, since we have established that $f$ and $g$ are integers, $fg$ is an integer. Furthermore, $fg\neq \pm 1$. This is a contradiction, since $a$ and $b$ are relatively prime. Therefore we conclude that

\begin{equation*} \text{gcd}(a,c) = \text{gcd}(b,c) = 1. \end{equation*}

Answer 1

$\gcd(a,c)$ divides $a$ and divides $c$. And $c|a+b$ so $\gcd(a,c)|a+b$ so $\gcd(a,c)|b$ so $\gcd(a,c)$ is a common factor of $a$ and $b$. But $a$ and $b$ are relatively prime so $\gcd(a,c) =1$.

Same argument shows $\gcd(b,c)= 1$

Answer 2

It is Ok. But there is a shorter proof. Since $(a,b)=1$, there exist integers $p,q$ such that $pa+qb=1$. Now if $a+b = \lambda c$ then $pa+pb= p\lambda c= 1+(p-q)b $. Therefore $(c,b)=1$. Similarly. multiplying by $q$ instead of $p$ we get $(c, a)=1$.