Are any prime gaps end larger than the square of the prime?

Question

I am not trained in mathematics at all and I was thinking about a certain problem. I'm not sure if there's a name for this problem or how to search for work done on it before.

 

Assume the following:

N and X are both positive integers
A is the set of integers {1, 2, 3 ... N}
B is the set of integers {1, 2, 3 ... X}
P is the set of all unique products that can be made from two integers in A

1: Find the lowest possible N such that P contains all integers in B and N < X
2: Additionally, determine if there is any X and N such that P does not contain any integers greater than X

After manually testing N = {1, 2 & 3} I was able to come to the following conclusions:

The largest integer in P will be N²
Any N where N² > X will fail 2 and any N where N² < X will fail 1
If N = 1 it fails N < X, assuming X must also be 1
If N ≠ 1, N² = X and B contains a prime that is missing from A (Is greater than N): that prime will also be missing from P

Question: Is there any gap of primes that is larger than the square of the prime that started the gap?

Said another way, assume Y and Z are primes and Z is the prime following Y. Is there any Z > Y²?

Just using these conclusions and the fact that solving 1&2 hinges on answering a question about large prime gaps leads me to think that if there is an N that satisfies 1&2, it is very large. After looking at known prime gaps, which are all very small compared to the primes that start them I'm guessing there is no N that satisfies 1&2.

It seems like there should be a proof somewhere that I can use to show whether this is possible or not. I attempted to read about the limits of prime gaps but I was not able to comprehend enough of it to come to any useful conclusions.

Answer 1

I'm going to try B={1,2,3...28,29,30} so that x=30.

 

30 has a prime factorization of 2*3*5.

 

I was going to conclude that all composite number less than 30 can be made from these numbers, but that's not true since I can't make 14, 21, 28. So I need the 7 in there as well. I also need 11 so that I can make 22, I also need 13 so that I can make 26...and now I can make all composite numbers less than equal to 30. So to just get all the composites less than equal to 30, Set A must go up to 13.

 

The problem is I cant make the prime numbers in this set unless my first set contains that prime number. So for my example, I cannot multiply anything to make 29 unless 29 is in set A. So if set B is all the consecutive integers less than equal to 30, Set A MUST go up to 29.

 

So the lowest possible N to insure that set P contain all elements of set B is the largest prime that is less than x.

In other words, P is a set of composite numbers plus all the primes less than N.

 

2. The answer is no.Let's takes B = to the numbers 1 through 6. In order for P to contain all the elements of B, A must contain the element 5. 5 times any of the elements other than 1 will be greater than 6.

 

If B = {1,2,3}, then A must be the same set and the largest element of P would be 6.

If b={1,2,3,...,99,100}, then A must go up to 97 and 97 * 2 is greater than 100.

 

So larger examples are definitely out.

 

If B = {1,2} than A = {1,2} and P={2}.

 

This brings up a problem with your question. N<X!! In fact, if x is a prime number, then A must contain that prime number so N=X, which contradicts one of your requisites.

 

About the prime gaps between 2 consecutive primes, my general intuition is that the square of the lower one will certainly be larger than the higher one,

 

I just looked up prime gaps list. They use the definition of a prime gap to be the number of composites between the two primes rather than the difference of the 2 primes.

For example 7 and 11. There are 3 composites between 7 and 11 so they call this prime gap of 3 where if you subtract the numbers, you get 4. in any case this table should convince you that the square of the smaller one is always larger than the larger one.

 

For example, according to this table, 31397 and 31397+71 +1=31469 are consecutive primes. The square of the lower one is definitely greater than the larger one since the gap is only 71.

 

Hope this helped.