I wish to prove that if $a+b$ and $a-b$ are coprime, then $a$ and $b$ are coprime.
I want to make sure that my proof holds, so any reasonable insight concerning this problem would be appreciated.
Suppose $a+b$ and $a-b$ are coprime where $a$ and $b$ are integers. Then for some $n \in \mathbb{Z}$ where $n = \pm 1$, we have $n|(a+b)$ and $n|(a-b)$, which implies that
$$a+b = nk \tag{1}$$
$$a-b = nj \tag{2}$$
for some $k,j \in \mathbb{Z}$.
Thus, equation $(1)$ implies that $b = nk-a$, from which equation $(2)$ becomes $a = nj + b = nj + (nk-a)$. Adding $a$ to both sides of the equation and factoring the $n$, we have the equation
$$2a = n(k+j)$$
But if we divide both sides by $2$, we have
$$a = n \cdot \frac{k+j}{2}$$
Since we are assuming that $a \in \mathbb{Z}$, it follows that $n \cdot \frac{k+j}{2} \in \mathbb{Z}$ (in particular, $\frac{k+j}{2} \in \mathbb{Z}$). Therefore, $n| a$. A similar argument can be made to show that $n|b$.
Updated Portion
Now assume there is another integer $d$ that divides $a$ and $b$. Consequently, we have the following equations
$$a = dk$$
$$b = dj$$
for some $k,j \in \mathbb{N}$.
If we add and subtract the two equations, we have
$$a+b = d(k+j)$$
$$a-b = d(k-j)$$
so $d|(a+b)$ and $d|(a-b)$. But $a+b$ and $a-b$ are coprime, so we must have $d=n$. Therefore, $n =\pm 1$ is the only integer that divides $a$ and $b$.
Hence $a$ and $b$ are coprime under the assumption that $a+b$ and $a-b$ are coprime.
$\blacksquare$
Best Answer
A simpler approach: $\gcd(a,b)=(a,b)=1$ if and only if there are integers $j,k\in\Bbb Z$ with $aj+bk=1$. Thus, since $(a+b,a-b)=1$, there are integers $j,k\in\Bbb Z$ so that $(a+b)j+(a-b)k=1$, but this is the same as $(j+k)a+(j-k)b=1$. Since $j+k,j-k\in\Bbb Z$, we have that $(a,b)=1$.