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I am reading the following problem:

Given the sequence: $T=3, 7, 11, 15, 19, 23, 27 ...$ prove that there are infinitely many prime numbers that are in $T$ (hint: multiply by $4$ and subtract by $1$)

My approach:
Assume that the prime numbers that are in $T$ are finite and are represented by the set $L$.
We have $N=\prod{i}\space L$ i.e. $N$ is the product of all the primes in $L$.
If we take: $4N - 1$ that is also a positive number $\gt 1$ (for $N \gt 0$). If we take a factor $p \in L$of $4N - 1$ we know that also $p \mid N$.
Hence $p | (4N - 1) - N \rightarrow p | (3N - 1) $

Now I can see that $3N - 1 = 2, 5, 8 ,11, 14...$ and none of which can have any factor of the form $4N +3$ which are the factors in $L$ but I am stuck on how to make that formal connection between $3N-1$ and $4N + 3$

Bill Dubuque
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Jim
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  • argue that $4N-1$ can not be even and it can not be be the product of odd primes of the form $4k+1$. – lulu Jun 30 '21 at 14:40
  • 11 is a prime of the form $4k + 3$. – Mees de Vries Jun 30 '21 at 14:44
  • @lulu: how does $4N -1$ not being even helps? – Jim Jun 30 '21 at 14:48
  • @MeesdeVries: ah I didn't notice that! Now I am really stuck, what the hint is about – Jim Jun 30 '21 at 14:48
  • Any $p\in L$ also divides $4N$ so you can say $p\mid 4N-1-4N= -1$ to get your contradiction. Like @lulu said, to find $p\in L$ dividing $4N-1$ in the first place, you should notice that $4N-1= 3 \pmod 4$. Breaking down $4N-1$ into a product of odd primes— must it contain a prime factor of the form $4k-1$? – klein4 Jun 30 '21 at 14:48
  • If $x_1, x_2, \cdots, x_n$ are each congruent to $1 \pmod{4}$, then so is $\prod_{i=1}^n x_i.$ This assertion is routinely proven by induction. Now, assume that there are only a finite # of primes in $T$, and use that assumption to arrive at a contradiction. – user2661923 Jun 30 '21 at 14:51
  • @MeesdeVries, Jim, $4n+3=4(n+1)-1$ so the hint is probably saying the sequence is $4n-1$ for $n=1,2,..$ which is equivalent to $4n+3$ for $n=0,1,..$. – John Douma Jun 30 '21 at 15:38
  • @klein4: How does $p \in L$ also divide $4N$? $p$ is of the form $4N + 3$ – Jim Jun 30 '21 at 15:42
  • You subtracted $N$ once to get $p \mid 3N-1$. What happens when you keep subtracting $N$? – arbashn Jun 30 '21 at 19:42
  • @arbashn: Do you mean keep subtracting $N$? I would get $2N - 1$ and then $N$. Which would mean that $p$ divides $N$ but we already know that. So what do you mean? I didn't get it – Jim Jul 01 '21 at 07:44
  • You don't get $N$ when you subtract $N$ from $2N-1$. This is meant to illustrate that eventually you get $ p \mid -1$ which is impossible, which means that $N$ and $4N-1$ do not share prime factors. This is what the first sentence of @klein4 's comment was saying. Then you may notice that multiplying two numbers of the form $4t+1$ gives you another number of the firm $4k+1$. By induction, any product of numbers of such a form gives you $4k+1$. (If you know modular arithmetic, this is trivial). But $4N-1$ is of the firm $4k+3$, and therefore, it has a prime factor of the form $4k+3$. – arbashn Jul 01 '21 at 09:09
  • @arbashn: So $p | N \space \And \space p | 4N\space$ because $N$ is of the form $4k + 3$? But then why does $p$ divide $4N - 1$? – Jim Jul 01 '21 at 15:23
  • @arbashn: I guess because $4N - 1 = 4N -4 + 3 = 4(N - 1) + 3 = 4K + 3$ and it must be divisible by a prime of $T$? – Jim Jul 01 '21 at 16:41

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