# Number Theory : Infinitude Of Primes - a different proof

I was doing some basic Number Theory problems and came across this problem :

• Show that the integer : $$Q_{n} = n ! + 1$$, where $$n$$ is a positive integer, has a prime divisor greater than $$n$$.

• Conclude that there are infinitely many primes.

My Solution (partial)

• We know that as $$Q_{n}$$ is $$\gt$$ $$1$$ $$\Rightarrow$$ $$Q_{n}$$ has a prime divisor $$p$$
• Let us assume , that $$p$$ $$\le$$ $$n$$
• If $$p \le n$$ then $$p \mid n!$$
• So , in the equality ; $$Q_{n} - n ! = 1$$ , $$p$$ divides the LHS $$\Rightarrow$$ it also divides $$1$$
• But that is not possible as no prime divides $$1$$
• Hence , we have achieved a contradiction and there exists a prime divisor $$>n$$

My Question :

I am not able to prove the infinitude of primes from this result , how can I do that ?

The proof implies that given any prime $\,\color{#c00}p\,$ there exists a larger prime (dividing $\,Q_{\large\color{#c00}p}$), therefore the set of primes is infinite.

Remark $\$ Because this way of proof is not by contradiction, it yields constructive information: iterating the above yields an algorithm to generate an infinite sequence of primes, viz.

\begin{align} &Q_1 = 1!+1 = 2\quad\ \text{has prime factor}\ \ \ \, 2 > 1\\ &Q_2 = 2!+1 = 3\quad\ \text{has prime factor}\ \ \ \, 3 > 2\\ &Q_3 = 3!+1 = 7\quad\ \text{has prime factor}\ \ \ \, 7 > 3\\ &Q_7 = 7!+1 = 71^2\ \text{has prime factor}\ \ 71 > 7\\ &\quad\ \ \ \vdots\qquad \qquad \qquad\qquad\ \ \vdots \end{align}\qquad This proof is a minor variation on Euclid's classical proof (which also was not by contradiction, despite many inaccurate historical claims to the contrary).

HINT: If there were only finitely many primes, there would be a largest prime; call it $q$. Now consider what you know about $Q_q$.

• Hi , @BrianMScott - I am not able to follow , exactly , would be grateful if you can expand upon your answer :) Feb 19, 2015 at 2:25
• @Michael: It’s correct, therefore it’s not flawed. Whether it’s needlessly complicated is a matter of opinion and taste; since I saw it before I saw the one that I suspect that you have in mind, I can’t really agree that it’s all that complicated. Feb 19, 2015 at 2:26
• @pranav: You’ve proved that $Q_q$ must have a prime divisor larger than $q$, but by hypothesis $q$ is the largest prime ... Feb 19, 2015 at 2:27
• But , @BrianMScott , in $Q_{n}$ -- n is not given to be prime -- why do you say that "by hypothesis $q$ is the largest prime" ? Feb 19, 2015 at 2:29
• @pranav: Whew! :-) You’re welcome. Feb 19, 2015 at 2:33

Let $P_n$ be a prime dividing $Q_n$. You've proved that $P_n>n$. The following is an infinite sequence of distinct primes: $$\left\{2,P_2,P_{P_2},P_{P_{P_2}},P_{P_{P_{P_2}}},P_{P_{P_{P_{P_{2}}}}},\ldots\right\}.$$ Therefore, there are infinitely many primes.

Hint: Suppose that there were finitely many primes. Let $n$ be the largest prime. Obtain a contradiction.

I'll not beat around the bush. For any given prime $p$ you can certainly find a positive integer $n$ such that $p$ is not greater than$~n$. So there cannot be a largest prime.