Q: Could you too prove one of the most important results in Number Theory and Mathematics: The infinitude of prime numbers??…

A: I would be so glad to post an original proof!

Euclid’s Theorem

Euclid’s Theoremis a fundamental statement in number theory that asserts that there are infinitely many prime numbers.## Euclid’s Proof

Euclid offered a proof published in his work

Elements(Book IX, Proposition 20),^{[1]}which is paraphrased here.^{[2]}Consider any finite list of prime numbers

p_{1},p_{2}, …,p_{n}. It will be shown that at least one additional prime number not in this list exists. LetPbe the product of all the prime numbers in the list:P=p_{1}p_{2}…p_{n}. Letq=P+ 1. Thenqis either prime or not:

If

qis prime, then there is at least one more prime than is in the list.If

qis not prime, then some prime factorpdividesq. If this factorpwere on our list, then it would divideP(sincePis the product of every number on the list); butpdividesP+ 1 =q. IfpdividesPandq,thenpwould have to divide the difference^{[3]}of the two numbers, which is (P+ 1) −Por just 1. Since no prime number divides 1,pcannot be on the list. This means that at least one more prime number exists beyond those in the list.

Euler’s ProofAnother proof, by the Swiss mathematician Leonhard Euler, relies on the fundamental theorem of arithmetic: that every integer has a unique prime factorization. If

Pis the set of all prime numbers, Euler wrote that:

The first equality is given by the formula for a geometric series in each term of the product. The second equality is a special case of the Euler product formula for the Riemann zeta function. To show this, distribute the product over the sum:

in the result, every product of primes appears exactly once and so by the fundamental theorem of arithmetic the sum is equal to the sum over all integers.

The sum on the right is the harmonic series, which diverges. Thus the product on the left must also diverge. Since each term of the product is finite, the number of terms must be infinite; therefore, there is an infinite number of primes.

Erdős’s ProofPaul Erdős gave a third proof that also relies on the fundamental theorem of arithmetic. First note that every integer

ncan be uniquely written as

where

rissquare-free, or not divisible by any square numbers (lets^{2}be the largest square number that dividesnand then letr=n/s^{2}). Now suppose that there are only finitely many prime numbers and call the number of prime numbersk. As each of the prime numbers factorizes any squarefree number at most once, by the fundamental theorem of arithmetic, there are only2^{k}square-free numbers (see Combination#Number ofk-combinations for allk).Now fix a positive integer

Nand consider the integers between 1 andN. Each of these numbers can be written asrs^{2}whereris square-free ands^{2}is a square, like this:

- ( 1×1, 2×1, 3×1, 1×4, 5×1, 6×1, 7×1, 2×4, 1×9, 10×1, …)
There are

Ndifferent numbers in the list. Each of them is made by multiplying a squarefree number, by a square number that isNor less. There are floor(√N) such square numbers. Then, we form all the possible products of all squares less thanNmultiplied by all squarefrees everywhere. There are exactly 2^{k}floor(√N) such numbers, all different, and they include all the numbers in our list and maybe more. Therefore, 2^{k}floor(√N) ≥N. But, since this inequality does not hold forNsufficiently large, there must be infinitely many primes.

Proof Using the Irrationality of πRepresenting the Leibniz formula for π as an Euler product gives

^{[9]}

The numerators of this product are the odd prime numbers, and each denominator is the multiple of four nearest to the numerator.

If there were finitely many primes this formula would show that π is a rational number whose denominator is the product of all multiples of 4 that are one more or less than a prime number, contradicting the fact that π is irrational.

Source: WikipediA