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## Topology Q+A Board

From: Bill Dubuque
Date: September 11, 2009
Subject: It's NOT Kummer's proof: Ribenboim's error

In reply to "Re: Re: Re: Re: Re: Re: Kummer's proof about there are infinitely many primes.", posted by Alireza on September 11, 2009:
>In reply to "Re: Re: Re: Re: Re: Kummer's proof about there are infinitely many primes.", posted by BD on September 11, 2009:
>>In reply to "Re: Re: Re: Re: Kummer's proof about there are infinitely many primes.", posted by Alireza on September 11, 2009:
>>>In reply to "Re: Re: Re: Kummer's proof about there are infinitely many primes.", posted by reader on September 11, 2009:
>>>>In reply to "Re: Re: Kummer's proof about there are infinitely many primes.", posted by Bill Dubuque on September 11, 2009:
>>>>>In reply to "Re: Kummer's proof about there are infinitely many primes.", posted by ... on September 10, 2009:
>>>>>>In reply to "Kummer's proof about there are infinitely many primes.", posted by Jen on September 10, 2009:
>>>>>>>
>>>>>>>Suppose that there exist only finitely many primes p1 < p2 < ... < pr. Let
>>>>>>N = p1.p2.....pr. The integer N-1, being a product of primes, has a prime divisor
>>>>>>pi in common with N; so, pi divides N - (N-1) =1.
>>>>>>>
>>>>>>Why is the N-1 odd?
>>>>>>
>>>>>>== Because p1 = 2 is even
>>>>>>
>>>>>> Why is N-1 composite?
>>>>>>
>>>>>>== Because N - 1 > pr
>>>>>
>>>>>Which is false if r=0 or r=1, p1=2. That's why the proof is usually
>>>>>written to say not "p1 < p2 <...." but rather "2 < 3 < p3 < ...",
>>>>>i.e. it must incorporate independent knowledge that there is at
>>>>>least one prime, and that 2 is not the only prime. That's one
>>>>>reason why the more uniform classical Euclid N+1 form is preferred.
>>>
>>>>One question: why N- 1 > p_r ???
>>>
>>>Continue with the following:
>>>...the integer N-1, being a product of primes, has a prime divisor p_i in common with N; so N = 0 (mod p_i) and N-1=0 (mod p_i) and from
>>>the properties of modulo we have N = N-1 (mod p_i) which is impossible since N-(N-1)=1 and there is no integer q to satisfy qp_i=1. A contradiction.
>>
>>Not only is that not relevant to the original question, but it
>>attains new heights for obfuscated proofs that d|N,N-1 => d|1
>
>That is Kummer's original proof! And he ultimately derives the absurd result stated above.

No, your obfuscated proof that N,N-1 are coprime isn't part of Kummer's proof.
Further, I had doubts that the original proof is due to Kummer and a quick
web search easily confirmed this. Namely, the attribution to Kummer appears
to have originated in Ribenboim's "Prime number records" books. However,
reading Kummer's original paper [1] shows that ist is quite misleading to
attribute to Kummer that minor N-1 variant of Euclid's N+1 proof. The point
of Kummer's paper is to present a proof based on Euler's phi function. Namely
phi(N) = 1 since any number between 2 and N-1 has a prime factor so is not
coprime to N. But since N = 2*3*... this contradicts phi(N) = (2-1)(3-1)... > 1,
using the well-known formula for phi. Kummer concludes the one page paper with
a remark that one may finish the proof in simpler way by noting that the
pair N-1, N+1 is a pair of integers coprime to N and this suffices to deduce
that phi(N) > 1 (note N>2 => N-1 != N+1 (mod N)). Thus Kummer is not promoting
the trivial N-1 variant but, rather, a proof based on phi(N)>1. This is not unlike
some of my proofs that use cardinality arguments to avoids units.