In previous posts, we have learned about prime numbers, and we know from Euclid's Theorem that there exists an infinite number of primes. In this post, we continue to look at prime numbers. Leading mathematicians like Fermat, Euler, Gauss were all fascinated by prime numbers. There are many problems about primes that even the statements are simple, but they still remain unsolved even to this day.

Goldbach's conjecture

This is probably the most famous problem on prime numbers. The Goldbach's conjecture predicts that any even number greater than $2$ can be written as a sum of two prime numbers. For example,

- $4 = 2 + 2$
- $6 = 3 + 3$
- $8 = 3 + 5$
- $10 = 3 + 7 = 5 + 5$
- $12 = 5 + 7$
- $14 = 3 + 11 = 7 + 7$
- ...

The mathematician Goldbach stated this conjecture in a letter that he sent to the mathematician Euler in 1742. With current technology, mathematicians have used powerful computers to verify the conjecture and proved it correct up to numbers of billion billion's multitude. However, no one has ever been able to come up with a general proof for every even number. Some prizes of up to million dollars have occasionally been offered to a solution of this conjecture, but so far there have been no lucky winners.

Open problem.

Prove or disprove that any even number greater than $2$ can be written as a sum of two prime numbers.

Chebyshev's Theorem

Chebyshev's theorem is a very beautiful theorem asserting that for any number $n > 1$, there is always a prime number lying between $n$ and $2n$. For instance,

- between $2$ and $4$, there is a prime number $3$;
- between $3$ and $6$, there is a prime number $5$; and
- between $4$ and $8$, there are prime numbers $5$ and $7$, etc.

Chebyshev's Theorem.For any $n > 1$, there always exists a prime number $p$ such that $n < p < 2n$.

This result was first conjectured by Bertrand in 1845 and it was proved later by Chebyshev in 1850. The theorem is also widely known as Bertrand's postulate. This theorem has an elementary proof due to the famous mathematician Erdos. When he was 19 year old, Erdos came up with a simple elementary proof of the theorem. We will look at this proof in our future posts.

Chebyshev's theorem gives us the following corollary. Suppose that $p_i$ denotes the $i$th prime number. Then by Chebyshev's theorem, there is a prime number $p$ such that $p_i < p < 2 p_i$. It means that $p_i < p_{i+1} < 2p_i$, and we have the following

Theorem.If $p_i$ and $p_{i+1}$ are two consecutive prime numbers then $$\frac{p_{i+1}}{p_i} < 2.$$

We have observed that for any pair of consecutive primes $(p_i, p_{i+1})$ the fraction $\frac{p_{i+1}}{p_i}$ is bounded above by $2$. A similar question is raised as whether or not the difference $p_{i+1} - p_i$ is also bounded above. In other words, is there a constant $c$ such that for any pair of consecutive primes $(p_i, p_{i+1})$ it will hold that $p_{i+1} - p_i < c$? The answer to this question is negative.

We will show that $p_{i+1} - p_i$ can be as large as any number. To prove this, first, let us introduce a notation $n!$, it reads "$n$ factorial", its formula is as follows $$n! = 1 \times 2 \times 3 \times 4 \times \dots \times n.$$

Clearly, the number $100! + 2$ is divisible by $2$, the number $100! + 3$ is divisible by $3$, the number $100! + 4$ is divisible by $4$, ... So the numbers from $100! + 2$ to $100! + 100$ are all composite numbers. Thus, if $p_i$ is the prime number standing right before $100! + 2$, then the next prime number $p_{i+1}$ must be standing after $100! + 100$. We have $$p_i < 100! + 2 < \dots < 100! + 100 < p_{i+1}.$$

It means that we have found two consecutive prime numbers $p_i$ and $p_{i+1}$ such that $p_{i+1} - p_i \geq 100$. Of course, in the above arguments, we can use any number, instead of using the number $100$ we may just use the number 1 million and it still works. So $p_{i+1} - p_i$ can be as large as any number we want.

Twin prime numbers

If we write all the prime numbers in a sequence as $$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, \dots$$

we will see that there are quite a lot of consecutive primes that are also being the two consecutive odd numbers, such as $3$ and $5$, $5$ and $7$, $11$ and $13$, $17$ and $19$. These pairs of primes are called twin primes. To this day, mathematicians still don't know whether there are infinite or finite number of twin primes.

Open problem.

Is there an infinite number of prime pairs $(p_i, p_{i+1})$ such that $p_{i+1} - p_i = 2$.

Let us stop here for now, we will continue our story about prime numbers in the next post.

*Homework.*

1. For all $n > 0$, prove that if $2^n + 1$ is a prime number then $n$ must be of the form $n = 2^m$. It means that if $2^n + 1$ is a prime then it must be of the form $2^{2^m} + 1$.

2. Prove that there exists a polynomial $P(x)$ such that $P(1) = 2$, $P(2) = 3$, $P(3) = 5$, $P(4) = 7$, $P(5) = 11$,..., $P(100)=$ the $100$th prime number.

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