A New, Possible Proof of the Infinitude of the Primes?

Question

$$1=1$$

$$2=2$$

$$3=3$$

$4=2\cdot2$ At $4$, the first prime number, $2$, is there as a factor. So I say that at the square of $2$, $2$ comes into play as a prime factor. At this point, $2$ is the only prime factor in play, and therefore, until $3$ comes into play at $9$, there will be a prime at every other integer.

\begin{align} 5 & = 5 \\ 6 & = 2\cdot3 \\ 7 & = 7 \\ 8 & =2^3 \end{align}

$9$ is the square of $3$, where $3$ comes into play at. With 2 and 3 being the only prime factors in play between 9 and 25, there have to be primes in this interval because 2 and 3 can only hit at most three out of four consecutive integers.

When 5 comes into play at 25, 2,3, and 5 are the only prime factors in play until the square of the next prime, 7, which comes into play at 49. Out of six consecutive integers, 2,3, and 5 can hit at most five of them. So the biggest prime gap between 25 and 49 is 6.

Each prime comes into play as a factor at its own square. So each time we arrive at the square of a prime, one more prime factor is in play, however, the gaps between the squares of the primes grow such that there will always be another prime, at least one in each gap.

Is this sufficient to prove the infinitude of the primes? Are the mathematical concepts presented here correct? Can the way this is presented be improved? Can the wording of the mathematical facts be improved?

An explanation is in order. Let us examine some of the gaps between consecutive squares of primes. Quite simply, the number of prime factors in play before the square of the first prime is 0. Therefore each integer between 1 and 4 is prime. After 4 and before 9, 2 cannot hit every point in the gap. The size of this gap equals 2·2, so there are 4 integers of which 2 can at most hit 2. Likewise, between 9 and 25, the number of integers in the gap is equal to 25-9-1, which is 15. In any span of 4 integers, with 2 and 3 being the only prime factors in play, they will miss at least one space. There is a simple proof for this. Once 5 comes into play, the prime factors begin to follow a great pattern.

The great pattern of how the prime factors are arranged is such that they will miss infinitely many spaces. Factoring all composite numbers from 4 to 30 will reveal this great pattern. Note that 30 is the product of the first three primes.

Answer 1

OK, here's a guess as to what is meant: among the numbers $1,2,3,4,\ldots,(p_1\cdots p_n)$, where $p_1<\cdots<p_n$ are the $n$ smallest primes, only $p_1\cdots p_n - (p_1-1)\cdots(p_n-1)$ of them can be divisible by at least one of the primes $p_1,\ldots,p_n$; therefore the others must be divisible by some other primes; hence there are more primes than $p_1,\ldots,p_n$.

But the argument is not clearly expressed here.

Answer 2

The approach seems to be:

Consider the $n$-th prime $p$. Between $p+1$ and $p^2$ every number is either prime or divisible by one of the first $n$. TAMO. Therefore, there is at least one prime between $p+1$ and $p^2$, hence there are infinitely many primes.

I don't know what the miraculous step in the middle is supposed to be. I mean, you can't use the Chinese Remainder Theorem since you'd get $$ \prod_{\text{primes }q\le p} \frac{q-1}{q} $$ primes but they'd be spread out among the numbers up to $$ p\#=\prod_{\text{primes }q\le p}q $$ which is much larger than $p^2$ unless $p$ is tiny.

Let me put it another way. Suppose all you knew about prime divisors was that in a range of $n$ numbers the prime $p$ divides either $\lfloor n/p\rfloor$ or $\lceil n/p\rceil$ of the numbers. So if you look at the range 6 to 30 (25 numbers) you know that 12 or 13 are even, 8 or 9 are divisible by 3, and exactly 5 are divisible by 5. But that's not enough to conclude that there are any primes in that interval, since $13+9+5=27\ge25.$ Now it happens that the numbers don't conspire that way and there are primes (like 7) in that interval. But how could you know that a conspiracy doesn't happen with larger intervals? If you look at the interval $(97,\ 97^2]$ you'll find that there are 16777 to 16799 primes 2-97 dividing the 9312 numbers in that range so it's entirely possible (a priori) that every number would be covered and there wouldn't be a single prime.