TITLE: Any two positive integers are co-prime if their sum is a prime number
QUESTION [14 upvotes]: Any two positive integers are co-prime if their sum is a prime number
Is this a trivial thing to say? I have a proof, but I don't want to bore anyone if it's just a trivial matter-of-fact thing about primes and co-primes.

REPLY [3 votes]: Yes, it is trivial. It is a consequence of $\rm\:(a,n\!-\!a)\, =\, (a,n),\:$ in the special case when the gcd $=1.\:$ Namely, in your case,  $\rm\ a+b \,=\, n\:$  and $\rm\ 0< a,b\:\Rightarrow\: a,b<n,\:$ so $\rm\:(a,n) =1\: $ when $\rm\:n\:$ is prime, therefore $\rm\,\ (a,b)\, =\, (a,n\!-\!a) \,=\, (a,n) \,=\, 1.\ $  
Generally $\rm\: (a,n\!+\!ak)\, =\, (a,n)\: $ since if $\rm\: d\mid a\ $ then $\rm\: d\mid n\!+\!ak\!\iff\! d\mid n.\, $ So $\rm\:a,n\!+\!ak\:$ and $\rm\:a,n\:$ have the same set $\rm\,D\,$ of common divisors $\rm\,d,\,$ so the same greatest common divisor $\rm(= max\ D).\:$ Notice that this is precisely the inductive (descent) step in the Euclidean algorithm for the gcd. Therefore your result also follows by applying Euclid's algorithm to compute the gcd $\rm\, (a,a\!+\!b)$