I feel this is an intuitive result. If, for example, I was working with the prime number $11$, I could split it in the following ways: $\{1, 10\}$, $\{2, 9\}$, $\{3, 8\}$, $\{4, 7\}$, $\{5, 6\}$.
Then clearly there is no way that the $2$ numbers can have a $\gcd$ of anything other than $1$. However, I am sort of lost on how to start a formal proof for this. Any pointers would be much appreciated.
Let's show the contrapositive, because why not?
So we want to show that if $a,b>0$ and $\gcd(a,b) \neq 1$, then their sum is not prime.
Suppose that $\gcd(a,b) = d > 1$. Then $a = a'd$ and $b = b'd$ for some $a',b'$ natural numbers. But then $a + b = da' + db' = d(a' + b')$, and as each of $d,a',b' \geq 1$, we have that $a + b \geq 2d$, but is divisible by $d$. Thus it is not prime. $\diamondsuit$
Thus if the sum of positive integers is prime, then their gcd is $1$.
Let $c$ be the gcd. Then $c$ divides $a$ and $b$, hence it divides $a+b$, a prime number.
Let $d$ be their gcd. Then $d$ divides their sum $p$, so $d$ can be only 1 or $p$.
If $d = p$, then $p$ divides both $a$ and $b$. Since both of these are positive, they are each at least $p$, so their sum is at least $2p$.
I realize that this is a restatement of mixedmath's answer.
This can easily be generalized to show that this holds for the sum of $k$ positive integers for $k \ge 2$.
Here is a direct proof that is independent of the others:
Let $g = \gcd(a,b)$. Then we can see that $$g\mathbb{Z} = a\mathbb{Z} + b\mathbb{Z} = \{n\in \mathbb{Z} | ra + sb = n, \text{for } r,s \in \mathbb{Z} \}.$$
In other words, $g$ generates the set of all integer combinations of $a$ and $b$. (If you are not familiar with this, it is most likely presented as a theorem in your textbook.) Since we are $a+b=p$ (given) and $a+b$ is some integer combination of $a$ and $b$, then $p$ must be in $g\mathbb{Z}$.
If $p \in g\mathbb{Z}$, then $g$ must be equal to $1$ or $p$ (since $p$ is prime). But by the fact that $a$ and $b$ are both positive and less than $p$ (the sum of the two equals $p$), then $p$ can't possibly be the greatest common divisor, let alone a divisor of any kind. Therefore $g = 1$.
Suppose $a,b$ are positive integers whose sum is a prime, $p$. Then $a+b = p$. Also, because $a, b$ are both positive integers that add up to $p$, then $a,b < p$. This implies that $\gcd(a,p) = 1$ and $\gcd(b,p) = 1$. Thus we can write 1 as a linear combination of both $a$ and $p$ or $b$ and $p$. Let us do it the first way. $ax + py = 1$ for some $x, y \in \mathbb{Z}$. From our original assumption, $a+b = p$. Thus we can substitute. $ax + (a+b)y = 1$. This implies $ax + ay +by = 1$. Grouping like terms we get $a(x+y) + b(y) = 1$. Thus 1 can be written as a linear combination of $a$ and $b$. This implies that $\gcd(a,b) = 1$. We have shown what we wanted to show.
let prime p=a+b, a and b positive
Suppose gcd(a,b)=s, $s\ne 1$
Then s divides a and s divides b, so there are some m,n such that a=sm and b=sn
then $a+b=sm+sn=s(m+n)$, which shows that s is a factor of a+b
so s is a factor of p. Contradiction.
The way out for negatives is that, in the above example, 3 is a factor of p, since it is p itself. So there is no contradiction in that case.
Complete short proof:
Let $\gcd(a,b)=d$. Then, $d \mid a$ and $d \mid b$. Thus, $d \mid a+b=p$. Thus, $d=1$ or $d=p$. Since $d \mid a$, $d\le a<a+b=p$. Thus, $d=1$.
As $(a,b) \mid a$ and $(a,b) \mid b,\ \ (a,b) \mid (pa+qb)$ where a,b,p,q are integers.
If (a,b) is not prime, $pa+qb$ can not be prime.
So, if $pa+qb$ is prime, (a,b) must be.
But there exists, p,q of opposite parity such that $(a,b)=pa+qb$ (which is Bézout's Identity).
In that case, primality nature of $pa+qb$ will be dictated by that of $(a,b)$.
$$1=\gcd(p,a)=\gcd(a+b,a)=\gcd(a,b)$$
The last step follows from $\gcd(b+ma,a)=\gcd(a,b)$ for any integer $m$. The first step follows from $a<p$. Any particular reason someone voted to delete this answer?
