# Proof that there are infinitely many prime numbers

I answered a question to prove that there are infinitely many prime numbers, but I'm not sure if my attempt is right. Can somebody help me to check if my attempt is right? I would like, if I am wrong, not a answer, but just a hint, thank you to whoever helps me!

Question: Prove that the sequence of prime numbers is infinite using the sequence $\{R_n\}_{n\geq1}$ given by $R_n = n! + 1$.

My attempt:

Suppose that the sequence of prime numbers is finite, so the sequence $R_n$ is finite, therefore, $R_{n+1}$ is not a prime number, in other words, $R_{n+1} = t*k$, $t,k \in \mathbb{Z}^+$, but $R_{n+1} = (n+1)! + 1$ by definition of the sequence $R_n$.

Note that $R_{n+1} = (n+1)n(n-1)! + 1$ and $6|n(n+1)$, so $R_{n+1} = 6v(n-1)! + 1, v \in \mathbb{Z}^+$

The only integer number that divide 1 is himself, therefore, $1|R_{n+1}$, because $1|6v(n-1)!$ and $1|1 \Longrightarrow 1|6v(n-1)! + 1$.

Furthermore, $R_{n+1}|R_{n+1}$, therefore, $R_{n+1}$ is a prime number, this is a absurd, because $R_{n+1}$ is not a prime number by hypothesis, therefore, $R_{n+1}$ is a prime number and then the sequence of prime numbers is infinite.

Suppose that the sequence of prime numbers is finite, so the sequence $R_n$ is finite

is wrong. Also, you then say that $R_{n+1}$ is not a prime number, but you did not determine what the value of $n$ is.

Furthermore, you then proved that $R_{n+1}$ is divided by $1$ and itself, and concluded that that means $R_{n+1}$ is a prime number. This is nonsense, since you did not prove that there is no other number that divides $R_{n+1}$. For example, look at this proof that $12$ is prime:

We can see that $1|12$ because $12=12\cdot 1$ and we also know that $12|12$ because $12=1\cdot 12$, therefore, because $1$ and $12$ both divide $12$, $12$ must be prime.

I believe you meant to say that if the sequence of prime numbers is finite, then the values of $n$ for which $R_n$ is prime must also be finite.

From here, you can simply take $n$ to be the product of all the primes (only finitely many by assumption), and then following the same argument as in Euclid's proof, there must be some prime that divides $R_n$ but it can't be on the list by construction. (Note that $R_n \equiv 1 \mod{p}$ if $p \mid n$).

$\textbf{Hint}$:

Let $p_n$ be a prime divisor of $R_n=n!+1$, and show that $p_n$ must be larger than n.

Then conclude from this that there are infinitely many primes.

• but I need to consider $p_n$ like the largest prime number and there are finite prime numbers too, right? – George Aug 24 '15 at 1:08
• @George You can let $p_n$ be any prime divisor of $n!+1$, and then show that if $p_n\le n$ that you get a contradiction. This will show that for every $n$, you can find a prime larger than $n$. – user84413 Aug 24 '15 at 15:57
• Sorry for being slow to answer @user84413 . I understood thank you! – George Sep 2 '15 at 22:10
• @George You're welcome -- I'm glad that made sense. – user84413 Sep 2 '15 at 22:25