Question about proving there are infinitely many primes of the form 4n+1

Question

Infinitely number of primes in the form $4n+1$ proof

I saw this answer which says Suppose n>1 is an integer. We define N=(n!)^2+1. Suppose p is the smallest prime divisor of N. Since N is odd, p cannot be equal to 2. It is clear that p is bigger than n (otherwise p∣1). "If we show that p is of the form 4k+1 then we can repeat the procedure replacing n with p and we produce an infinite sequence of primes of the form 4k+1."

I dont get how that last sentence makes sense. if p replaces n, then N=(p!)^2+1, then we can always find another smallest prime p' of N such that p'=4t+1? (so I thought the author meant N would be prime, but I guess not)

The proof does not state that $N$ is prime, just that the smallest prime divisor of $N$ is of the form $4k+1$. — Peter Foreman, Sep 26 '19 at 09:52
If $n>1$ then $every$ prime divisor of $N$ is of the form $4k+1$ and is greater than $n.$ — DanielWainfleet, Sep 27 '19 at 15:58

José Carlos Santos · Answer 1 · 2019-09-26T09:58:14.487

1

The author just said “we can repeat the procedure”, but did not say that $N$ would then be prime. If we repeat the procedure, we get a prime $p'\neq p$. And if we repeat it again, we get a prime $p''$ such that $p''\neq p'$ and $p''\neq p$. And so on…

edited Sep 26 '19 at 09:58

answered Sep 26 '19 at 09:52

José Carlos Santos

427,504

2

Each is shown to be larger than the previous, so there will not be any repeats and it produces an infinite sequence of these primes. – Jaap Scherphuis Sep 26 '19 at 09:57
so once we find smallest prime p of N, we can place back p in place of n. But how would that gurantee there is a smallest prime p' of N? N=((4k+1)!)^2+1 always has prime p' such that it has a form 4t+1? – Samuel Kim Sep 27 '19 at 07:02
ohh since p was arbitrary, and I found that p is in the form 4k+1, I can do so for another bigger number! thx! – Samuel Kim Sep 27 '19 at 07:10
@SamuelKim If my answer was useful, perhaps that you could mark it as the accepted one. – José Carlos Santos Sep 27 '19 at 07:40
how do I do that? there is no button – Samuel Kim Oct 05 '19 at 02:56
Yes, there is. It is the ✓ sign below the downvote arrow. – José Carlos Santos Oct 05 '19 at 07:52

score 1 · Answer 2 · answered Sep 26 '19 at 10:17

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We want to show that there are infinitely many such $p$. This can be expressed as:

For all $n\in\Bbb N$, there exists such a $p$ with $p>n$.

As explained, working with $N=(n!)^2+1$ in the proof guarantees $p>n$.

answered Sep 26 '19 at 10:17

Hagen von Eitzen

374,180

& we don't need to consider the $least$ prime divisor of $N.$ If $n>1$ then any prime divisor of $N$ is greater than $n$ and is $\equiv 1\mod 4.$ – DanielWainfleet Sep 27 '19 at 16:03
Corollary: There is no largest prime. ;) – DanielWainfleet Sep 27 '19 at 16:06

Question about proving there are infinitely many primes of the form 4n+1

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