Query regarding a theorem in Number Theory

Question

Please refer to Theorem $8$ in the attached picture.

I do not understand the significance of the condition $(n+1)!+k>k$. Is it not always true since the inequation implies $(n+1)!>0$ which is always true. Also why are the numbers taken as $(n+1)!+k$ and not $n!+k$. Why does $k$ start from $2$ and not from $1$? Also what is the meaning of the line: "Thus, there are arbitrarily large gaps in the sequence of primes."? How are the gaps "arbitrarily large"?

Answer 1

Stating the obvious

The only reason the proof mentions that $(N+1)!+k>k$ is that it is not enough to know that:

1. $k\geq 2$
2. $k$ divides $D$.

to conclude that $D$ is composite. Specifically, if $k$ is prime, and $k=D$, then (1) and (2) are true, but $D$ is not composite.

If we also know also that $D>k,$ then we can conclude that $D$ is composite.

So, even though it is trivially true that $(N+1)!+k>k,$ we need to mention it in proving that $(N+1)!+k$ is composite.

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Prime gaps

If $2=p_1<p_2<p_3<\cdots<p_n<\cdots$ are all the primes, then a prime gap is the difference between two consecutive primes $p_{n+1}-p_n.$ If $p_{n+1}-p_n=N+1$ then $p_{n}+1,p_{n}+2,\dots,p_{n}+N$ are $N$ consecutive composite numbers.

On the other hand, assume $k,k+1,\dots,k+N-1$ are $N$ consecutive composite numbers.

We can let $p=p_n$ be the largest prime such that $p\leq k-1.$

Now, it is not possible that $p_{n+1}\leq k-1,$ or we'd contradict that $p_n$ is the largest prime $\leq k-1.$ So $p_{n+1}\geq k.$ But $k,k+1,\dots,k+N-1$ are not prime, so $p_{n+1}\geq k+N.$

Then

$$p_{n+1}-p_n \geq (k+N)-(k-1)=N+1.$$

So there exists $N$ consecutive composite numbers if and only if there is a prime gap of size at least $N+1.$

The quoted proof proves that we can find $N$ consecutive composite numbers for any $N,$ namely with $k=(N+1)!+2.$ so we can make prime gaps of size at least $N+1$ for any $N,$ which is what we mean by saying that the prime gaps can be arbitrarily large.

Probably an easier way to phrase:

There are arbitrarily large prime gaps.

is to say

The prime gap function $f(n)=p_{n+1}-p_n$ is not bounded above.

Answer 2

a few answers:

• $k$ starts at 2, because $y!+1$ can be prime examples are $y=2,3$, and $k=1$ being proved to cause divisibility by 1 shows nothing about it potentially being composite.
• $n+1$ is used to save cases like $n=4$ from the results of $k>1$
• The gaps get as large as you want, based on $n$, because $n+1$ grows with $n$. Otherwise we wouldn't need to choose or be given, any n in the positive integers.