0

I am going through this youtube tutorial on euclid's technique for proving that there are an infinite number of primes - https://www.youtube.com/watch?v=pNhbW1Hrjcs

This question is also there on the site (although closed as off-topic)

I have tried to work it out as follows -

  1. Assume all numbers beyond 16 as "infinity"
  2. So the set of prime numbers would be - 2,3,5,7,11,13
  3. Their products (up to that element in the set) are as follows - 2, 6, 30, 210, 2310, 30031
  4. For the above results (products), for elements until 13, all entries (n+1) yields a prime - 3,7,31,211,2311 but 30031 is composite ( 59 * 509 ).
  5. As can be seen the smallest number 59 is in the "infinite" section as per our assumption (1)
  6. So the question is - what is the smallest number P(i) in the set of primes (2,3,...i,...n) such that the primorial (the prime equivalent of factorial) plus 1 (going as per euclid's proof) has a factor that is less than or equal to P(i) ?

PS - I've tried using a spread sheet but a primorial grows quite quickly (similar to a factorial). So thought I would ask if there is a known sample.

Ravindra HV
  • 135

2 Answers2

1

The first one you omitted is composite. After that, it's all composites for a while.

Here are the first $20$ cases. As you see, you got all the primes in that range.

lulu
  • 70,402
  • Sorry! I accepted this answer but then realized none of the factors are within the set at that point ! For example primorial(17) is 1997277 but the smallest factor (19) is still greater than 17. The case is more obvious for others. This helps me generate the factors but will keep the answer open for an exact answer ! – Ravindra HV Jul 25 '21 at 20:57
  • 1
    I don't understand. Obviously there can't be any prime factors of primorial$(n)$+1 which are less than or equal to the $n^{th}$ prime. – lulu Jul 25 '21 at 20:59
  • Great ! That helps too ! How is that (initially it seemed so to me as well - but could not think of 'how' ) ? Basically then the step in the proof where in we divide the factor P(i) = x (x being the factor) with the primorial would not hold ! The proof itself would be invalid ! (I am referring to 07:31 timestamp in the video). The linked SO post only has words. The video illustrates. – Ravindra HV Jul 25 '21 at 21:22
  • 2
    The first $n$ primes all divide primorial$(n)$. Hence none of them can divide primorial$(n)+1$. – lulu Jul 25 '21 at 21:33
  • That sort of occurred to me too initially but then I could not get back to it. So then please update that as the answer. Will accept that instead! Thanks a lot! – Ravindra HV Jul 25 '21 at 21:56
  • Also is my understanding correct that it invalidates the approach we use to prove primality in the given video? If not why not? Can post it as a separate question though. – Ravindra HV Jul 25 '21 at 21:57
0

Note - The answer I am looking for is in the comments. To quote lulu -

The first n primes all divide primorial(n). Hence none of them can divide primorial(n)+1.

Ravindra HV
  • 135