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?

Guest Mar 25, 2017

#1**+4 **

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)

Ehrlich
Mar 25, 2017

#4**+2 **

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

MaxWong
Mar 26, 2017

#5**+1 **

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....

Melody
Mar 26, 2017

#6**0 **

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

GingerAle
Mar 26, 2017

#8**+4 **

The answer is 3

Proof:

P=3*n_{1}+a_{1}

Q=3*n_{2}+a_{2}

R=3*n_{3}+a_{3}

a1,a2,a3=1 OR 2

P^{2}+Q^{2}+R^{2}=(3*n_{1}+a_{1})^{2}+(3*n_{2}+a_{2})^{2}+(3*n_{3}+a_{3})^{2}=(3n_{1})^{2}+(3n_{2})^{2}+(3n_{3})^{2}+2*3*n_{1}*a_{1}+2*3*n_{2}*a_{2+}2*3*n_{3}*a_{3}+a_{1}^{2}+a_{2}^{2}ā+a_{3}^{2}ā

If a_{1}=a_{2}=a_{3}=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).

Ehrlich
Mar 26, 2017

#9**0 **

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 .....**

Melody
Mar 26, 2017

#10**+1 **

"*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).

.

Alan
Mar 26, 2017

#11**0 **

(3n_{1})^{2}+(3n_{2})^{3}+(3n_{3})^{2}+2*3*n_{1}*a_{1}+2*3*n_{2}*a_{2}+2*3*n3*a_{3}+a_{1}^{2}+a2_{2}^{2}+a_{3}^{2}=

[((3n_{1})^{2}+(3n_{2})^{3}+(3n_{3})^{2}+(3n_{3})2+2*3*n_{1}*a_{1}+2*3*n_{2}*a_{2}+2*3*n3*a_{3})]+a_{1}^{2}+a2_{2}^{2}+a_{3}^{2}=

3*(3*n_{1}^{2}+3*n_{2}^{2+}3*n_{3}^{2}+a_{3*}n_{3}*2+a_{1*}n_{1}*2+a_{2}*n_{2}*2)+a_{1}^{2}+a2_{2}^{2}+a_{3}^{2}.

remember, a_{k} (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.

Guest Mar 26, 2017

#12**0 **

You mentioned that a_{k} can also be 0, but thats not true (unless n_{k}=1, but that leads to the answer P=3)

why? because then the prime you chose will be divisible by 3.

Guest Mar 26, 2017

#15**+6 **

1^{2}+2^{2}+1^{2} - divisible by 3

2^{2}+2^{2}+2^{2 }- divisible by 3

1^{2}+1^{2}+1^{2 }- divisible by 3

2^{2}+2^{2}+1^{2 }- divisible by 3

Did i convince you now?

No? Then you probably forgot a_{k }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)

Ehrlich
Mar 28, 2017

#17**+1 **

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 :)

Melody
Mar 28, 2017