Two other methods to prove the infinite number of prime numbers

We already know that there are infinite prime numbers. At that time, we used the most common proof method:

Suppose there is the largest prime number P, then we can construct a new number 2*3*5*7 *... * p + 1 (all prime numbers multiply by 1 ). Obviously, this number cannot be divisible by any prime number (all prime numbers except it all have more than 1), which indicates that we have found a larger prime number.

Here, we will provide two new proofs from cut-the-knot.

Use the Fermat number to prove the infinite number of prime numbers
The Fermat number refers to the number of 2 ^ (2 ^ n) + 1. We record 2 ^ (2 ^ n) + 1 as F (N ), where n can take all natural numbers. Obviously, all the Fermat numbers are odd. In a moment, we will see that any two Fermat numbers are mutually unique. That is to say, each prime factor of each Fermat number is different from that of other Fermat numbers. This shows that the number of prime numbers is infinite.
Theorem 1: F (0) * F (1) * F (2) *... * F (n-1) = f (N)-2, n> = 1
Proof: mathematical induction. F (0) = 3 and F (1) = 5, then k = 1 is obviously true. If K = N is true, when k = n + 1:
F (0) * F (1) * F (2) *... * F (N)
= (F (0) * F (1) * F (2) *... * F (n-1) * F (N)
= (F (N)-2) * F (N)
= (2 ^ (2 ^ N)-1) * (2 ^ (2 ^ n) + 1)
= 2 ^ (2 ^ (n + 1)-1
= F (n + 1)-2

Theorem 2: For any two unequal natural numbers N and M, F (N) and F (m) are mutually unique.
Proof: Assume that t is divisible by F (N), F (M), and m <n at the same time. According to the Theorem 1, there are:
F (n) = f (0) * F (1) * F (2) *... * F (m) *... * F (n-1)-2
This means t can be divisible.
F (0) * F (1) * F (2) *... * F (m) *... * F (n-1)-f (n) = 2
Note that 2 only has two factors: 1 and 2. As mentioned above, the number of Fermat is an odd number, so it is impossible to be divisible by 2. In this way, t can only be 1, which proves that two numbers are mutually unique.

Use *-set to prove the infinite number of prime numbers
*-The set is a positive integer set {A1, A2,... an}, enabling Ai-AJ to divide all unequal I and j.
Theorem 1: For all N> = 2, there is a *-set with the size of N.
Proof: mathematical induction. {1, 2} is obviously a *-Set of 2. Suppose {A1, A2,... an} is a *-set. Define B0 as A1 * A2 *... * an (that is, the product of all AI ). For all positive integers K that cannot exceed n, so BK = b0 + AK, then {B0, B1, B2 ,..., BN} is a *-set of N + 1.

Theorem 2: Suppose {A1, A2,... an} is a *-set. For all the positive integers I that do not exceed n, define Fi = 2 ^ ai + 1, then F1, F2,..., FN mutual element.
Proof: Obviously, FI is an odd number. If FK and FM (FK> FM) can be divisible by the same prime number P, P can only be an odd number. P can be divisible by FK-FM, that is, 2 ^ am * (2 ^ (AK-Am)-1 ). Since P is an odd number, it can only be divided by 2 ^ (AK-Am)-1.
If s excludes T, 2 ^ S-1 excludes 2 ^ t-1. Therefore, according to the definition of *-set, 2 ^ (AK-Am)-1 divisible by 2 ^ ak-1. Then P can remove 2 ^ ak-1. However, P can also divide 2 ^ aK + 1, so we can conclude that P can divide 2, which is in conflict with P as an odd number.

Theorem: There are infinite numbers of prime numbers
Proof: According to the latencies 1 and 2, for any large N, there is a set of N in size, and there are at least N different prime factors in the number pair. This shows that the number of prime numbers can be any number.