How to prove 'infinite primes' kind of problem

[l-1j-cho]

Other than the Eucledean method
(1+p1p2p3...)and so on

other than this, how do we prove that there are infinite prime numbers?
in algebreic way (excluding analyses)

 

[HallsofIvy]

The Euclidean proof is so neat and simple, I cannot imagine why you would want a different proof.

 

[l-1j-cho]

indeed

but I was just wondering if there is any other technique to prove there are infinite primes so it can be applied to prove twin prime conjecture, mersenne prime conjecture, or sophie-germain conjecture.

 

[Bill Simpson]

There are many many proofs of the infinitude of the primes

http://primes.utm.edu/notes/proofs/infinite/

And about 1/4 of the way down through this
http://en.wikipedia.org/wiki/Prime_number

and many more, but I hesitate to claim an infinite number of them.

However thus far nobody no one has seen how to use these to prove any of the conjectures you mention.

On a brighter and prime related note, there is now a non-crank proposed proof of Goldbach that has been submitted. It would be interesting to see if that could be explained in a form that those not in the "bright student" group could understand.

 

[RamaWolf]

One proof due to the German mathematician Kummer:

Assumption:
Let P_{r}:={2,3,5,7,...,p_{r}} the finite set of all primes

form the prime factorial p_{r}#:=2*3*5*...*p_{r}
and the number q:=-1+ p_{r}#

q is not in P_{r}, that means, q is composite, with prime
factors from P_{r}; let p_{k} be one of those factors

Now: p_{k} divides both p_{r}# and q, hence it
must divide their difference: but this difference is 1
and p_{k} does NOT divide 1.

Therefore our assumption must be false

 

[disregardthat]

The above proof is essentially equivalent to euclids proof.

 

[RamaWolf]

The above proof is essentially equivalent to euclids proof.

I think not.

Recall the original version of Euclids proof:

Let Q_{r}:={q_{1}, q_{2},...,q_{r}} be the set a r distinct
prime numbers, and set q:=1+q_{1}*...*q_{r}

(Note: the r primes must NOT be the first r primes 2,3,..,p_{r} !)

Then q is NOT divisible by any of the q_{i} and must therefore be an new
prime or at least a composite number formed with additional primes.

Therefore: from any Q_{r} you can construct a Q_{r+1}, and that proves the infinitude of the primes

Now compare this to the Kummer proof I cited.

 

[RamaWolf]

The following proff is due to Polya

Let e:=2^{k} and F_{k}:=1+2^{e} be the Fermat numbers

with F_{0}:=3, F_{1}:=5, F_{3}:=17. F_{4}:=257, F_{5}:=65537, F_{6}:=4294967297, etc

we have the recurrence relation F_{k+1}:=2+\prod^{k}_{i=0}F_{i}

From the recurrence relation we have the following

Lemma: Any two F_{i} and F_{j} for distinct i, j are co-prime,
i.e. have no factors in common


Now let F:={F_{0},F_{1},F_{2},...} be the set of all Fermat numbers and this set is countable,
i.e. has the cardinality of N:={1,2,3,4,...}, the set of the natural numbers

Therefore no finite P_{r}:={2,3,5,...,p_{r}} can meet the requirements
of our Lemma, there would simply be not enough primes.

 

[RamaWolf]

Let P:={p_{i}} = {2,3,5,7,...} and
E_{k}:=1+\prod^{k}_{i=1}p_{i} be the Euclid numbers with E_{1}:=3, E_{2}:=7, E_{3}:=31. etc

We have the following

Lemma All E_{i} are odd, i.e. 2 is not a factor of E_{i}.

and it is not difficult to prove the following

Theorem Any two E_{i}, E_{j} for different i,j are co-prime, i.e. have no common factors.

Now supose, P:={2,3,5,7,...,p_{r}}, then the corresponding {E_{i}} must be the numbers {2,3,5,7,...,p_{r}}
in some order (our Theorem), but that is impossible (our Lemma)

This proof was inspired by the Polya proof

 

[RamaWolf]

From the Euler \zeta-function \zeta(s):=\sum^{\infty}_{i=1}(\frac{1}{i})^{s}

we know, that \zeta(2) is \frac{\pi^{2}}{6}

This quantity is irrational, since \pi itself is irrational

From the Euler product formula, we have

\zeta(2):=\prod(1 - \frac{1}{p^{2}})^{-1}, where the product is taken over all primes

If there were only a finite number of primes, this product would
be a rational number, in contradiction to the irrational nature of \frac{\pi^{2}}{6},
which concludes the proof.

 

[disregardthat]

I think not.

Recall the original version of Euclids proof:

Let Q_{r}:={q_{1}, q_{2},...,q_{r}} be the set a r distinct
prime numbers, and set q:=1+q_{1}*...*q_{r}

(Note: the r primes must NOT be the first r primes 2,3,..,p_{r} !)

Then q is NOT divisible by any of the q_{i} and must therefore be an new
prime or at least a composite number formed with additional primes.

Therefore: from any Q_{r} you can construct a Q_{r+1}, and that proves the infinitude of the primes

Now compare this to the Kummer proof I cited.

Yes, it is not equivalent, but "essentially" equivalent. It doesn't contain any new ideas.