+118727 
+118727 A problems is being discussed here.
https://web2.0calc.com/questions/help_99455
This problem may be related to a problems I am having.
When I try to delete a response to a prior answer, that particular answer as well as all the other responses to it are deleted at the same time.
This may be why some of our guests answers have been deleteted. :/
+118727 Thanks Guest,
Ok lets look at this. I am redoing your working in the hope that all becomes clear to me. 
First a definition:
Prime triplet .
set of three prime numbers Which form of arithmetic sequence with common difference two is called a triplet prime .
Reference: https://artofproblemsolving.com/wiki/index.php/Prime_triplet
That defintiion appears to be nonesense because, this site https://en.wikipedia.org/wiki/Prime_triplet
says that 7,11 and 13 are prime triplets.
I assume prime triplets are any 3 consecutively prime numbers .... Is that correct? For now I will assume so.
\(\color{blue}{\text{We consider all positive prime triplets }P
We have
P, Q, R all consecutive primes
If P=2 then Q=3 and R=5 4+9+25=38 which is not prime so P is bigger than 2
So, let
\(P=3k_1+a\\ Q=3k_2+b\\ R=3k_3+c\\\)
where a,b and c are can equal 0,1 or 2
\(P^2+Q^2+R^2\\ =(3k_1+a)^2+(3k_2+b)^2+(3k_3+c)^2\\ =9k_1^2+6ak+a^2+9k_2^2+6bk+b^2+9k_3^2+6ck+c^2\\ =9k_1^2+9k_2^2+9k_3^2+6ak+6bk+6ck+a^2+b^2+c^2\\ =3(3k_1^2+3k_2^2+3k_3^2+2ak+2bk+2ck)+a^2+b^2+c^2\\ \)
Ok now please explain why this must be a multiple of 3 .....
