A condition for being a prime: $\;\forall m,n\in\mathbb Z^+\!:\,p=m+n\implies \gcd(m,n)=1$
If $\;p=m+n$ where $p\in\mathbb P$, then $m,n$ are coprime, of course. But what about the converse?
Conjecture:
$p$ is prime if $\;\forall m,n\in\mathbb Z^+\!:\,p=m+n\implies \gcd(m,n)=1$
Tested (and verified) for all $p<100000$.
Solution 1:
It is true. Suppose $p\geqslant 2$ is not prime. Then we can write $p=xy$ with $x,y\geqslant 2$. Then we find $p=m+n$, with $m=x$ and $n=x(y-1)$. Those are obviously not coprime.
Solution 2:
If $d \mid p$ and $d<p$, then $1 = \gcd(d, p-d) = \gcd(d, p) = d$, so $p$ is prime.