# Discussion on twin prime conjecture

I understand where I am wrong in my previous post. Also, I am very thankful to all members, who answered and showed my errors in post. Now, I would like to know the proof for the following.

"The difference of any two prime numbers $p$ and $q$ is expressible as $2k$; and the probability of finding primes $p$ at $n + k$ and $q$ at $n - k$ decreases as $n$ increases."

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@Zev Chonoles! Thank you for proper editing of my post. – mahi Dec 21 '11 at 15:31
I understood the intuition behind your question, you are looking for some generalization of the twin-prime conjecture, let me try to answer it – Iyengar Dec 25 '11 at 10:52
I just now re-edited and re-tagged your question asking for reference requests too, so that proficient users can give you some good references @mahi – Iyengar Dec 25 '11 at 11:28

You are referring to the generalized proof of " Twin-Prime Conjecture ", by using the Prime Number Theorem ( PNT ).

Prime number theorem states that the density of primes is ruled by the law $$\pi(n) \sim \large\frac{n}{log(n)}$$ We can state this in a more precise form using Riemann's Li function, as $$\pi(n)=Li(n)+O(ne^{-a\sqrt{Log(n)}})$$ for some constant $a$.

we can compute that the probability of finding twin primes ( $a-b=2$ , but you are looking at $2k$ which will be an implication ) $a$ at $n + 1$ and $b$ at $n -1$ is about $\large\frac{2}{\rm{Log(n)}}$ , but you are writing about the case for some $k$ but here in this case I mentioned above its $k=1$ and the number of twin primes in the interval $n$ is about $\large\frac{2n}{\rm{Log(n)}^2}$.

So its clearly evident to say from the above things that

• The difference of two prime numbers $a$ and $b$ is an even number $2k$.
• The probability of finding primes $a$ at $n + k$ and $b$ at $n - k$ decreases as $n$ increases.

The prime number theorem states that the number of primes less than $n$ is asymptotic to $1/\rm{Log(n)}$. So if we choose a random integer $m$ from the interval $[1,n]$, then the probability that $m$ is prime is asymptotic to $1/\rm{Log(n)}$

So its crystal clear that one can apply the same thing to your conditions by adjusting the values and as the $n$ is in the denominator of probability its evident that they are inversely proportional.

I end here, and I seriously advice you to completely go through this and this.

But I find there are some more beautiful articles about this, but it takes some time for me to fish them out. I surely re-edit once if I find any of such things.

Thank you,

Yours truly,

Iyengar.

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Is there any problem with my answer ? , it annoys the person who answers the question, if he doesn't hear any response even after taking strains to answer ( may be positive or negative response ) . – Iyengar Dec 26 '11 at 17:26

There are several problems here

• The difference between an odd prime $p$ and 2 is not an even number
• If $n$ and $k$ have the same parity and $n>k+2$ then neither $n+k$ nor $n-k$ are prime
• You cannot talk about the probability of finding primes in this sense: a given number is either prime or not. You might turn it into a probabilistic statement if you had a meaningful way of choosing $n$ and $k$ with some kind of distribution. Or you could take the route of the Bateman-Horn conjecture
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You're assuming a certain interpretation of probabilities; see e.g. this discussion. – joriki Dec 21 '11 at 15:38
@joriki The linked to discussion does not invalidate Henry's remark that one needs a probability distribution. If you told me "I have picked a natural number $n$; at what odds are you willing to bet that it is prime", I would say "at no odds, until I know how you picked it". – Alex B. Dec 25 '11 at 14:14
@Alex: I agree you'll want to know how I picked the number, but I don't agree this implies that one needs a probability distribution. All you need to know is that there's no reason to expect it to tend to be or not to be a prime. For instance, if we agree to go out on the street, ask the first person we meet for a big natural number $k$ and choose $n$ uniformly between $2^{k+1000}$ and $2^{k+1001}$, it would be rational to bet on the basis that $n$ is prime with probability $\approx1/\log n$ without knowing the probability distribution with which people on the street pick big natural numbers. – joriki Dec 25 '11 at 17:07
The main statement in Henry's post I was reacting to was "You cannot talk about the probability of finding primes in this sense: a given number is either prime or not", which I took to claim that it doesn't make sense to talk about the probabilities of individual numbers being prime. Your valid point that you'll want to know how I picked the number isn't in conflict with my point that (as long as you know that the number wasn't specifically picked to be or not to be a prime) it does make sense under some interpretations of probabilities to talk about its individual probability of being prime. – joriki Dec 25 '11 at 17:07
@joriki The question whether I can assume that people on the street pick numbers from a given interval uniformly is one for psychologists, not for mathematicians. There are many possible ways of picking numbers that do not intentionally favour primes but which will influence the probability of such a chosen number being prime. I don't think that as stated, this is a well-defined mathematical claim. Yes, you can talk about the probability of a number being prime, provided you have a concrete random process that chooses numbers, i.e. a probability distribution. – Alex B. Dec 26 '11 at 2:53