Proof Check: $\gcd(a+b,a-b) \leq 2$ if $\gcd(a,b)=1$

Question

Please peer review the following,

Proposition:

$\forall a,b\in \mathbb{Z}^{+}, \text{If gcd}(a,b)=1, \text{then} \text{ gcd}(a+b,a-b) \leq 2$.

Proof Attempt:

Assume $\gcd(a,b)=1$.

By way of contradiction, suppose $\text{gcd}(a+b,a-b) \geq 3$. Let $d=\text{gcd}(a+b,a-b)$.

Then $d \mid a+b$ and $d \mid a-b$. Say $dk = a+b$ and $dl = a-b$ for some integers $k,l$.

Hence, $d(k+l)=2a \\ d(k-l)=2b \tag{1}.$

Proceed by casework: either $d$ is a power of $2$ or it isn't.

$\underline{\text{Case }1}:$ Assume $d$ is a power of $2$. Since $d\geq3$, this means $d\geq4$. In other words, $d$ contains at least two factors of two. By (1) above, we can deduce that $2\mid a$ and $2\mid b$. We know $d$ must have at least two factors of two, so we can always cancel one of those factors of two with the 2 on the right-hand-side of both equalities in (1). Having at least one factor of two leftover from the factorization of $d$ implies $2\mid a$. By similar reasoning, we arrive at $2\mid b$. Hence gcd$(a,b)\geq2>1$. But this is a contradiction as we assumed gcd$(a,b)=1$.

$\underline{\text{Case }2}:$ Assume $d$ is not a power of $2$. Then $d$ contains a prime $p$ in its prime factorization such that $p\geq3$. Then we can say $p\mid d\mid 2a$ and similarly $p\mid d\mid 2b$. As $p$ is prime, we can use Euclid's lemma to say $p\mid 2a$ implies $p\mid a$ since $p\nmid 2$ because $p\geq3$. Analogous reasoning allows us to conclude $p\mid b$. But this means that gcd$(a,b)\geq p\geq3>1$. Again, this is a contradiction since we assumed gcd$(a,b)=1$

$\blacksquare$