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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
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17+0 Answers

 #1
avatar+310 
+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
 #2
avatar
+2

 p^2+q^2 +r^2

sorry they are not multiple

Guest Mar 25, 2017
edited by Guest  Mar 25, 2017
 #3
avatar+310 
+4

Are you 100% sure its solveable?

 

i appreciate people that come up with their own questions, but it might be a very hard question, a question im afraid im not able to solve.

Ehrlich  Mar 25, 2017
 #4
avatar+6900 
+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
avatar+91247 
+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
avatar+833 
+1

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
 #7
avatar+91247 
0

Thanks Ginger,

I assumed that a prime triplet was just any three prime numbers.

I have not looked very well .... is a prime triplet any 3 consecutive prime numbers?

Melody  Mar 26, 2017
 #8
avatar+310 
+4

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+a22ā€‹+a32ā€‹

 

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).

Ehrlich  Mar 26, 2017
 #9
avatar+91247 
0

Thanks Guest,

Ok lets look at this.   I am redoing your working in the hope that all becomes clear to me.  indecision

 

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
avatar+26366 
+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
avatar
0

(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.

Guest Mar 26, 2017
 #12
avatar
0

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.

Guest Mar 26, 2017
 #15
avatar+310 
+6

12+22+12 - divisible by 3

22+22+2- divisible by 3

12+12+1- divisible by 3

22+22+1- divisible by 3

 

Did i convince you now?

No? Then you probably forgot aCANNOT 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
 #16
avatar+310 
+5

You added the numbers without squaring them. Thats the problem

Ehrlich  Mar 28, 2017
 #14
avatar+310 
+4

I already answered your question why cant i see my answer?

Ehrlich  Mar 27, 2017
 #17
avatar+91247 
+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
 #18
avatar+833 
+1

Ehrlich, why do you not post the source for this equation? The source should have more information related to the variables and how to derive the values 0, 1, and, 2 for the Ak variables. Iā€™m sure they are derived using a modulo system.

GingerAle  Mar 30, 2017
edited by GingerAle  Mar 31, 2017

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