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I am interested in understand the proof of infinitely many primes. It seems like quite an easy proof, ( I know there are many but I am referring to the proof that goes as follows);

" Suppose there are only finitely many primes, lets say n of them, we can denote them as p1,p2,..,pn. Now we construct a new number P=(p1)(p2)..(pn)+1. Clearly P is larger than any of the primes, so it does not equal any one of them. Since p1,p2,,pn constitute all primes, P cannot be a prime. Thus it must be divisible by at least one of our finitely many primes, say pn. But when we divide P by Pn we get a remainder of 1 which is a contradiction. So our original assumption must be false."

I understand a lot of what it is saying, I am just not understanding where the +1 ties in. I see that if we assume P is not prime, and then cannot divide without a remainder there is a contradiction, but what if that +1 was not in the statement question originally?

I hope what I am asking makes sense to you guys,

Thanks a lot.

(Source for proof Hans Riesel, Prime Numbers and Computer Methods for Factorization, Birkhaeuser, 1985, ISBN 0-8176-3291-3.)

Bill Dubuque
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  • The +1 is what guarantees that $P$ is not divisible by any of our primes. Thus, there must be new primes which divide it that we didn't list already. – Alex G. Jun 09 '14 at 21:07
  • Adding $1$ to the product of the primes is what makes it coprime to all factors (all primes). It's a special case of $,\gcd(n,n+1)= 1,$ for $,n = $ product of all primes. – Bill Dubuque Jun 09 '14 at 21:09
  • Without the $+1$, $P$ would be divisible by every known prime, rather than indivisible by them. – user2357112 Jun 09 '14 at 21:21
  • You can use $-1$ instead of $+1$ unless you have the single prime $2$ – Mark Bennet Jun 09 '14 at 21:36
  • As I just noted in an answer this evening, this widespread way of proving the infinitude of primes is inferior to the one that Euclid wrote and is frequently incorrectly reported to be what Euclid wrote. Some of the best mathematicians are guilty of this error. http://math.stackexchange.com/questions/842187/proof-of-infinite-primes-clarification/842225#842225 – Michael Hardy Jun 21 '14 at 04:29

2 Answers2

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The key idea is that we can construct an integer coprime to any finite set of integers by taking their product, then adding $\:\!{\bf\color{#c00}1}.\,$ So iterating this process generates an infinite sequence of (pair) $\rm\color{#0af}{ coprimes}$ $\ 2,\,3,\,7,43,\, \ldots,\,$ via $\ \color{#0a0}{f_{n}} = \,\color{#c00}{\bf 1} + f_1\cdots f_{n-1},\,$ i.e. $\,\color{#0af}{{\rm gcd}(f_n,f_k) = 1}\,$ if $\,n>k\,$ since any common divisor of $\,\color{#0a0}{f_n},\,\color{#c0e}{f_k}\,$ divides $\, \color{#c00}{\bf 1} = \color{#0a0}{f_n} - f_1\cdots \color{#c0e}{f_k}\cdots f_{n-1}.$

Any infinite sequence $\,f_n > 1 \,$ of coprimes yields an infinite sequence of distinct primes $\, p_n $ obtained by choosing $\,p_n$ to be any prime factor of $\,f_n,\,$ e.g. the least factor $> 1.$

Remark $ $ A shorter variant of Euclid's proof arises by noting that iterating the map $\, n\mapsto n^2\!+n$ generates integers with an unbounded number of prime factors, because $\,n(n+1)\,$ includes all prime factors of $\,n\,$ plus some (new!) prime factor of $\,n+1,\,$ since $\,n\,$ and $\,n+1\,$ are coprime.

Bill Dubuque
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If the +1 were not there, then P would be composite, and we wouldn't know whether there was another prime beyond pn. This way, we're guaranteed that there is one:

With the +1, either

  1. P is composite, the product of primes other than the pi in the list (since it's definitely not divisible by any prime in the list) - contradicting our assumption about the pi; or
  2. P is itself a prime; but it's not in the list - again a contradiction.

Without the +1, P is definitely composite, and the product of primes in the list - which doesn't let us conclude anything about primes that may not be in the list.

Matt Gutting
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