Counting Triple Primes - How Many Are There?

[saadsarfraz]

Homework Statement

Here's the problem. We define the triple primes as triples of natural numbers (n,n+2,n+4) for which all three entries are prime. How many triple primes are there? (Hint:mod 3.) (By way of contrast, it is not yet known whether the twin primes-that is, pairs (n,n+2) with both entries prime-form an infinite collection.)

Homework Equations

see above

The Attempt at a Solution

i'm stuck in this problem, don't know where to begin.

 

[Dick]

You should first ask yourself if there are ANY triple primes. Pay attention to the mod 3 clue.

 

[saadsarfraz]

i do know that they exist put n=3 and u get (3,5,7). I still don't know how mod 3 would work in this case.

 

[Dick]

Oh, yeah. That one. Notice that one of those primes is divisible by three. Write down any 3 consecutive odd numbers and notice that one of them is divisible by 3. Can you give the reason why using a 'mod 3' argument? Then ask yourself how many primes are divisible by 3?

 

[saadsarfraz]

so if i write (3,5,7) with 3$$\equiv0$$ mod 3 and 5$$\equiv2$$ mod 3 and 7$$\equiv1$$ mod 3 how does that help? and there are no primes which are divisible by 3 unless it would not be a prime number.

 

[Dick]

Just do the general case. Take n,n+2,n+4. n mod 3 is either 0 (in which case you are already done),1 or 2. Consider the other two cases and figure out what n+2 and n+4 are.

 

[saadsarfraz]

n+2 is going to be mod 5 with 0,1,2,3,4. and n+4 is going to be mod 7 with 0,1,2,3,4,5,6. right?

 

[saadsarfraz]

btw how does n mod 3 = 0 would prove that there is only one triple prime.

 

[jgens]

I think what Dick is trying to get you to do is to show that given any natural number n: n, n +2, or n + 4 one of them must be divisible by 3; hence, (3, 5, 7) must be the only set of triple primes.

 

[saadsarfraz]

i still don't understand the procedure of how i would show that any natural number n,n+2,n+4 must be divisible by 3.

 

[saadsarfraz]

so for any n, either n, n+2 or n+4 must be divisible by 3 which means either of them should be mod 3?

 

[NoMoreExams]

Yes, you know that if you have k sequential numbers, one of them will be divisible by k. What's n + 4 mod 3?

 

[saadsarfraz]

n+4 mod 3 would be 0,1,2.

 

[NoMoreExams]

I was going to say n + 1 mod 3

 

[saadsarfraz]

n+1 mod 3 would also give me 0,1,2

 

[NoMoreExams]

Obviously but the thing I was hoping that you would notice is that you have n, n + 2, n + 4. If we do them mod 3 we get n, n + 1, n + 2 which is THREE sequential numbers so one of them MUST be divisible by 3.

 

[saadsarfraz]

i still don't understand, what do you mean by a three sequential number?

 

[NoMoreExams]

By 3 sequential numbers I mean 3 numbers in order, 1, 2 ,3 or 24424, 24425, 24426 or ... n, n + 1, n + 2

 

[saadsarfraz]

im sorry cause I am fairly new to this concept but i have one more thing to ask. i still do not understand how n,n+2,n+4 mod 3 would give you n,n+1,n+2.

 

[NoMoreExams]

Do you agree that n mod 3 is n mod 3?
Do you agree that n + 2 mod 3 is n + 2 mod 3?
Do you agree that n + 4 mod 3 is n + 1 mod 3?

 

[saadsarfraz]

i don't understand how n+4 mod 3 would give me n+1 mod 3

 

[NoMoreExams]

doesn't (a + b) mod n = a mod n + b mod n?

 

[Dick]

Case I: n=0 mod 3. Done. n is divisible by 3.
Case II: n=1 mod 3. So n+2=? mod 3 and n+4=? mod 3. Fill in the blanks and make a conclusion.
Case III: n=2 mod 3. So n+2=? mod 3 and n+4=? mod 3. Fill in the blanks and make a conclusion.
That's all of the possibilities.

 

[saadsarfraz]

oh so, for n+4 it would be like (n+4)mod3 = n mod3 +4 mod3, and 4 is congruent to 1 therefore (n+4)mod3 = n mod3 + 1mod3 = (n+1)mod3

 

[saadsarfraz]

btw thanks soo much for your help and effort.