$a, b \in \mathbb{Z}^+$ whose sum is prime implied $\gcd(a,b) = 1$.

Question

I was hoping someone could let me know whether the logic in this proof is sound.

Let $a$ and $b$ be positive integers whose sum is a prime, $p$. Prove that $\gcd(a,b) = 1$.

Define $d = \gcd(a,b)$. By definition, $d \mid a$ and $d \mid b$, so $d \mid a + b$. That is, $d \mid p$. The only positive divisors of $p$, since it is prime, are $1$ and $p$. For a contradiction, suppose $d = p$. Then $p \mid a$ and $p \mid b$, so there exist integers $x$ and $y$ such that $p = ax$ and $p = by$. So $$p = ax = by$$ and $$ a + b = ax + by. $$ Since $a$ and $b$ are positive, and hence $p$ is positive, $ax$ and $by$ must be positive. Hence, $x = y = 1$ is the only possibility. Hence, $$p = a = b.$$ Hence, $$a + b = p + p = 2p$$ But $a + b = p$, so $$p = 2p,$$ which implies that $p = \frac{1}{2}$, which is absurd.

Answer 1

"For a contradiction, suppose d=p. Then p∣a and p∣b, so there exist integers x and y such that p=ax and p=by"

No! the exact opposite!

$p|a$ so there exist a positive integer, $x$ so that $a = px$ and $p|b$ means there is a positive integer $y$ so that $b=px$.

So..... $p = a+b = px+py = p(x+y)$ so $x + y = 1$. But $x \ge 1$ and $y \ge 1$ and so $x+y \ge 2$. Our contradiction.

Answer 2

The idea is correct, but as $p$ divides $a$ and $b$, I think you meant to write $$\exists x,y\in\mathbb{Z} \ \text{such that}\ a=px,\ b=py.$$ This gives then $$p=a+b=p(x+y)$$ which is a contradiction as $p,x,$ and $y$ are positive integers.