# 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 $p<100000$.

#### Solutions Collecting From Web of "A condition for being a prime: $\;\forall m,n\in\mathbb Z^+\!:\,p=m+n\implies \gcd(m,n)=1$"

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.

If $d \mid p$ and $d<p$, then $1 = \gcd(d, p-d) = \gcd(d, p) = d$, so $p$ is prime.

for p = 1 obviously wrong
(for all positive integers m, n with m+n=p (of course there are no ones, doesn’t matter) there is gcd(m,n)=1, but 1=p is not prime)