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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 $$gmathbbZ = amathbbZ + bmathbbZ = nin mathbbZ .$$

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 $gmathbbZ$.

If $p in gmathbbZ$, 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 $x, y in mathbbZ$. 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, $sne 1$

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 $d mid a$, $dle a<a+b=p$. 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?

Edited:

I know it’s an old question, but since certain aspects of prior answers came across as arguably being like a second language, I figured some future readers who, like me, don’t have advanced math might benefit from a more basic approach to the OP’s requested “pointers” for “how to start a formal proof”:

For primes p > q , and positive integers m and n that are generally coprime but either or both of which can be 1, one can never have p = mq + nq because that would mean p/q = m + n . (Conversely, it would seem that any composite c would have at least one way of satisfying c = mq + nq).

I don’t mean any disrespect to the prior answers, but my aging brain cells prefer this kind of simplicity. :-)