Sum of Twin Primes is Divisible by 12
\[p, \; p+2\]
, both greater than 3, are twin primes. Then their sum is divisible 12.\[p+(p+2)=2p+2=2(p+1)\]
\[p\]
is odd so \[p+1\]
is even hence \[2(p+1)\]
is divisible by 4.Working modulus 3, and noticing that since
\[p \gt 3\]
we must have \[p \equiv 1, \; 2 )mod \; 3)\]
.\[p \equiv 1 \; (mod \; p) \rightarrow p+2 \equiv 0 \; (mod \; 3)\]
so \[p+2\]
is composite contradicting that \[p+2\]
is prime.\[p \equiv 2 \; (mod \; p) \rightarrow p+2 \equiv 1 \; (mod \; 3)\]
so \[p+p+2 \equiv 2+1 \; (mod \; 3)\]
.Hence
\[2p_2\]
is divisible by 4 and 3 so is divisible by 12. 
