# $\gcd(p_1+p_2,p_1+p_3,p_2+p_3)\leq 2$ if $p_1<p_2<p_3$ are primes [closed]

Conjecture about greatest common divisors and primes:

$\gcd(p_1+p_2,p_1+p_3,p_2+p_3)\leq 2$ if $p_1,p_2,p_3$ are different primes.

Tested for all $p_1,p_2,p_3<500$.

I'm pretty sure that someone can find an easy proof of this, but I have no clue myself.

## closed as off-topic by Namaste, GNUSupporter 8964民主女神 地下教會, Claude Leibovici, eranreches, polfosolDec 27 '17 at 14:39

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Let $d=(p_1+p_2,p_1+p_3,p_2+p_3)$. Then since it divides all three numbers, it also divides $(p_1+p_2)-(p_1+p_3)+(p_2+p_3)=2p_2$.
However $d$ has to be coprime with $p_2$, because otherwise $p_2$ would divide all three numbers, including for example $p_2+p_3$, which means $p_2 \mid p_3$, which is impossible.
So we are left with $d$ divides $2$ and that is what we wanted.