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Prove that the number of prime numbers is infinite

I want to solve it with this theorem :

For n, a positive integer, and integers i,j with 1≤i<j≤n, we know that gcd ($n!$ $\times$ $i+1$ , $n!$$\times$ $j+1$) $ =1$.

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The theorem has been proven.But I don't know how to use this theorem to prove my question.

user808266
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  • Relatively prime for infinitely many integers implies it, see this duplicate for the argument. So you are done. – Dietrich Burde Nov 06 '20 at 16:27
  • That produces an infinite number of relatively prime numbers. If there were only a finite number of primes you would eventually run out of such pairs. If $n \ge $ the largest prime you've got then $\gcd(n!+1, n!\cdot 2+1)=1$ but neither are divisible by any of the primes you have so far so there must be more primes somewhere. – fleablood Nov 06 '20 at 17:08

2 Answers2

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Let us use a proof by contradiction. Let us assume $P$ is finite, let $n=\max(P)$, you can see that $n!+1$ is prime. In fact, you can see that each prime number can't divide $n!+1$ (because they divide $n!$, if one would divide $n!+1$, it would divide $n!+1-n!=1$.

So $n!+1$ is prime and $n!+1>n $, so here you have your contradiction

NHL
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$\gcd(n!+1, n!\times 2 + 1)=1$ but neither $n!+1$ nor $n!\times 2 + 1$ is divisible by any prime less than or equal to $n$.

so there must always be two primes larger than any $n$-- one that divides $n!+1$ and another to divide $n!\times 2 + 1$.

ANd as $n$ is unbounded so are the primes larger than $n$.

.....

But you didn't really need the awkward theorem.

Its enough to point out that $n!+1$ is not divisible by any prime less than or equal to $n$

......

But even that is overkill. The classic way, the way Euclid did it, is that if you have any finite list of primes, $A$, then $1+\prod_{p\in A}p $ is not divisible by any prime in the list so there must be primes not on the list so no finite list is complete.

Sometimes the classics are best.

fleablood
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