We consider all positive prime triplets p < q < r where also p 2+q 2 +r 2 is a prime. How many values are possible for p?
+312 a prime is a number that is not divisble by any number.
2*r+2*q+2*p can sometimes be a prime, according to you.
are you sure 2*r+2*q+2*p can be a prime? think about it (its a hint)
+9673 All primes except 2 is an odd number(Think about it. Why?). When p > 2, it must be an odd number.
Sum of the squares of 2 odd numbers is an even number. Therefore sum of the squares of 2, and 2 prime numbers is also an even number, not a prime number.
I tested a few combinations which involves p = 5 and I don't know why the results are always divisible by 3.
Tested a few combinations which involves p = 7 and I don't know why the results still are always divisible by 3.
I assume that all combinations involving p \(\geq\) 5 will always divisible by 3.
Therefore only 1 value is possible for p.
Why on Earth is everything divisble by 3? lol
+118680 Mmm I do not know either.
certainly p cannot equal 2.
Because then when you added the squares you would have even+ odd+odd = even = multiple of 2.
The trio 3,5,7 works.
9+25+49=83 which is prime.
I tried some trios with p=5 and like Max said te sum of the squares resulted in multiples of 3.
I'd be interesed if anyone could examine this more....
+2511 This may help...
... [A] prime triplet is a set of three prime numbers of the form (p, p + 2, p + 6) or (p, p + 4, p + 6). With the exceptions of (2, 3, 5) and (3, 5, 7), this is the closest possible grouping of three prime numbers, since one of every three sequential odd numbers is a multiple of three, and hence not prime. (Emphasis mine) Source: https://en.wikipedia.org/wiki/Prime_triplet
+312 The answer is 3
Proof:
P=3*n1+a1
Q=3*n2+a2
R=3*n3+a3
a1,a2,a3=1 OR 2
P2+Q2+R2=(3*n1+a1)2+(3*n2+a2)2+(3*n3+a3)2=(3n1)2+(3n2)2+(3n3)2+2*3*n1*a1+2*3*n2*a2+2*3*n3*a3+a12+a22+a32
If a1=a2=a3=1 OR 2 the formula is divisible by 3.
same applies if one of them is 1 and if 2 of them are 1.
meaning it will always be divisible. but something here doesnt make sense right? thats right, if one if them is 3 it works. meaning one of them has to be 3. we already know none of them can be 2, and 1 is not a prime, so P=3 (always).
+118680 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 .....
+33661 "I assume prime triplets are any 3 consecutively prime numbers .... Is that correct? For now I will assume so."
"3 consecutive prime numbers" is a necessary, but not sufficient condition, Melody. For example: 31, 37 and 41 are consecutive primes, but they don't have the form (p, p+2, p+6) or (p, p+4, p+6) that define prime triplets (as GingerAle pointed out above).
.
(3n1)2+(3n2)3+(3n3)2+2*3*n1*a1+2*3*n2*a2+2*3*n3*a3+a12+a222+a32=
[((3n1)2+(3n2)3+(3n3)2+(3n3)2+2*3*n1*a1+2*3*n2*a2+2*3*n3*a3)]+a12+a222+a32=
3*(3*n12+3*n22+3*n32+a3*n3*2+a1*n1*2+a2*n2*2)+a12+a222+a32.
remember, ak (k=1,2,3)=1 or 2.
there are 4 combinations for the values of the a's (3 1's, 2 1's, 1 1, or 3 2's).
all of the combinations are divisible by 3.
meaning, there is only one option- P=3.
You mentioned that ak can also be 0, but thats not true (unless nk=1, but that leads to the answer P=3)
why? because then the prime you chose will be divisible by 3.
+312 12+22+12 - divisible by 3
22+22+22 - divisible by 3
12+12+12 - divisible by 3
22+22+12 - divisible by 3
Did i convince you now?
No? Then you probably forgot ak CANNOT BE 0 UNLESS ONE OF THE NUMBERS IS 3. BUT THAT MEANS P=3 BECAUSE P IS THE SMALLEST AND 3 IS THE SMALLEST POSSIBLE PRIME (we cant use 2)
+118680 Hi Ehrlich,
No I am not convinced yet, I need to find the time to think about it a whole lot more.
Maybe then I will be convinced. :)
Troulbe is there is always a new question that attracts my attention and there is only so much time in the world :))
Thanks for trying so hard to convince me :) Maybe I will post more for you to comment on later :)
