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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 ?

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pranav
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5 Answers5

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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).

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

Brian M. Scott
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  • Hi , @BrianMScott - I am not able to follow , exactly , would be grateful if you can expand upon your answer :) – pranav Feb 19 '15 at 02:25
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    @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. – Brian M. Scott Feb 19 '15 at 02:26
  • @pranav: You’ve proved that $Q_q$ must have a prime divisor larger than $q$, but by hypothesis $q$ is the largest prime ... – Brian M. Scott Feb 19 '15 at 02:27
  • But , @BrianMScott , in $Q_{n}$ -- n is not given to be prime -- why do you say that "by hypothesis $q$ is the largest prime" ? – pranav Feb 19 '15 at 02:29
  • @pranav: Read my answer again. I assumed, in order to get a contradiction, that there were only finitely many primes. In that case there must be a largest prime, which I’m calling $q$. In other words, the hypothesis for the remainder of the argument is that $q$ is the largest prime. – Brian M. Scott Feb 19 '15 at 02:31
  • Thanks @BrianMScott , I got it :) – pranav Feb 19 '15 at 02:33
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    @pranav: Whew! :-) You’re welcome. – Brian M. Scott Feb 19 '15 at 02:33
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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.

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

Yuval Filmus
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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.

Marc van Leeuwen
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