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I am working on the following problem:

Prove that there are infinitely many primes in the sequence: $5, 11,17,23, 29, 35, 41...$ Hint: these numbers satisfy $x \equiv 5 \pmod 6$. Try a proof similar to Euclid

My approach:
We notice that the numbers in the sequence are either prime of the form $6x + 5$ or composite of the form $(6x + 5)(6x + 1)$ e.g.
$35 = 5 \cdot 7$
$65 = 5 \cdot 13$
$77 = 11 \cdot 7$
$95 = 5 \cdot 19$

We can see that $(6x + 5)(6x + 1) = (6x)^2 + 6x + 5\cdot6x + 5 = 6(6x^2 + x + 5) + 5 = 6k + 5$ where $k = 6x^2 + x + 5$
So indeed we have closure if we multiply numbers of this form.
We also notice that the product of primes of the form $(6x +5)$ belongs to a different equivalent class depending on if the total of numbers multiplied is odd vs even.
I.e.
Even number:
$(6x + 5)(6x + 5) = (6x)^2 + 6\cdot 5x+ 5\cdot 6x + 25 = 6k + 25 \equiv 1 \pmod 6$
Odd number:
$(6x + 5)(6x + 5)(6x + 5) = (6k + 25)(6x + 5) = (6kx)^2 + 6\cot 5k + 25 \cdot 6x + 5\cdot 25 = 6j + 5\cdot 25\equiv 5\cdot 1= 5 \pmod 6$ where $j = 6kx + 5k + 25x$

Assume that the set of primes of the form $6x + 5$ is finite i.e. $P = {p_1, p_2, p_3, ...p_n}$

Let $N$ be the product of the set of primes: $N = p_1\cdot p_2 \cdot p_3...p_n$

Note we have $2$ cases: Either $N\equiv 5 \pmod 6$ or $N \equiv 1 \pmod 6$ (depending on the size of the set.

Case 1:
Assume that $|P| \equiv 1 \pmod 2 \implies N \equiv 5 \pmod 6$
This means that $N$ is of the form $6x + 5$.
To create a composite number we multiply by $(6m + 1)$ i.e.
$S = N \cdot (6m + 1) = (6x + 5)(6m + 1)$
But this means that $S$ is of the form $6k + 5$

Now a prime $p$ from the $P$ is a factor of $N$ and must also be a factor of $S$. I.e.
$p \mid N \And p \mid S \implies p \mid 6x + 5 - (6k + 5) \Leftrightarrow p \mid 6x - 6k = 6(x - k) \Leftrightarrow p \mid 6q$ where $q = x - k$

But none of the primes is a factor of $6$ since they are of the form $6x + 5$ so we end up with an impossible case.
Similar logic for case 2 also leads to impossibility.

Hence the primes in the sequence must be infinite.

Note:
I am interested to understand the problems in the thought process of my proof.
I found this post that seems to be similar, but so far, the book I am reading has not discussed about "quadratic reciprocity" at this point (also not familiar with it), and I'd like to understand if my proof is valid or what are the problems with it

J. W. Tanner
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Jim
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    It is hard to separate the "long story" from the "proof". Try to rewrite the proof in a compact way. Inventing a lot of letters is not very helpful. Then check explicitly why should your "proof" work to show that the set $P$ with the elements $5$, $11$, $17$, $23$, $29$, $41$, $47$, $53$, $59$, $71$, $83$, $89$ is not the set of all primes congruent to $5$ modulo six. (Take even $P={5,11,17}$.) Note that there may be a lot of other primes (congruent to $1$ modulo six). You should avoid the oddest prime $2$, and even its best friend $3$. – dan_fulea Jul 11 '21 at 13:36
  • The "Primes in arithmetic sequence" section (p. 16-19) of this book might be relevant. – ryang Jul 11 '21 at 13:44
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    Which is the similarity to Euclid's argument exactly? Euclid assumes there are only finitely many primes building a finite set $P$ and considers $$M=\prod_{p\in P}p\color{red}{+1}\ .$$ He adds one to insure no prime in $P$ divides $M$. In your construction i see only a multiple of the product, no analogue of "adding one"... If the question is if the proof is correct, the answer is simple, it is not correct. – dan_fulea Jul 11 '21 at 13:45
  • Does this answer your question? Prove that there are infinitely many primes of the form $6n + 5$ –  Jul 11 '21 at 13:48
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    @JitendraSingh Could you please indicate which is the question in the OP answered by the above link? The OP is trying to show that there are infinitely primes congruent to $5$ modulo six, he has some construction, however there is no question. – dan_fulea Jul 11 '21 at 13:56
  • A side note: It's counterproductive to multiply out $(6x+5)(6y+5)$ when you know modular arithmetic. Just do $5^2 \equiv 25 \equiv 1 \pmod 6$ or the even easier $5^2 \equiv (-1)^2 \equiv 1 \pmod 6$. – arbashn Jul 11 '21 at 14:39
  • @dan_fulea: Euclid constructs a new number by adding $1$. I tried to construct a new number by multiplying something that creates a composite number, but it turned out that the premise is all wrong – Jim Jul 11 '21 at 21:46
  • @arbashn: But how do I ignore the other parts and only focus on the constants? I mean if I don't do the multiplication, how would I be sure that the first part is a multiple of $6$? Wasn't obvious to me – Jim Jul 11 '21 at 21:48
  • @Jim One you prove that if $a \equiv b \pmod n$, then $a+c \equiv b+c \pmod n$ and $ac \equiv bc \pmod n$, you'll realize that addition and multiplication works the same in modular arithmetic. So you only have to work with the residues. (In this case since $6x+5 \equiv 5 \pmod 6$, you can replace it with $5$ when adding or multiplying). – arbashn Jul 12 '21 at 07:21
  • @arbashn: For $a \equiv b \pmod n \implies ac \equiv bc \pmod n)$ to be able to hold shouldn't $gcd(c, n) = 1$? – Jim Jul 12 '21 at 10:44
  • No, you're thinking of $\frac ac \equiv \frac bc \pmod n$ which is generally true only if $\text{gcd}(c,n)=1$. As for my assertion, $a \equiv b \pmod n \implies n \mid (a-b) \implies n \mid (a-b)c \implies ac \equiv bc \pmod n$. – arbashn Jul 12 '21 at 11:38
  • I don't understand why my question was marked as duplicate. My question was about my specific proof and writing not how to solve the question in general – Jim Jul 14 '21 at 14:33

4 Answers4

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I admit to struggling to follow your proof OP, and there is a simpler more consise way.

HINT: Let $p_1, p_2,\ldots, p_n$ be the first $n$ primes satisfying $p_i \equiv_6 5$ for each $i=1,\ldots n$.

If $n$ is even, then $$N \doteq \left(\prod_{i=1}^n p_i\right)+4$$ is odd and satisfies $N \equiv_6 5$ and so there is a prime $p$ satisfying both $p \equiv_6 5$ and $p|N$ [in fact there is precisely an odd number of such $p$]. Can $p$ be any of $p_1,p_2,\ldots,p_n$?

If $n$ is odd then instead let $$N \doteq \left(\prod_{i=1}^n p_i \right)+6.$$

Mike
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  • Better: Let $N=6\Pi-1$. – Oscar Lanzi Jul 12 '21 at 00:26
  • @Mike: Can $p$ be any of $p_1,p_2,\ldots,p_n$?. Shouldn't that be the starting assumption? I.e. that a $p \in p_1,p_2,\ldots,p_n$ must be a divisor of $N$ and then somehow end up in a contradiction? I think there is a step in your hints that I am not aware and makes me think in loops back to my original approach – Jim Jul 12 '21 at 10:40
  • Keep in mind that none of $p_1,\ldots, p_n$ divide $N$ though. Indeed, lets consider the case where $N=(\prod_{i=1}^n p_i)+4$. Then each $p_i$ divides $(\prod_{i=1}^n p_i) = N-4$. So the only way that $p_i$ would be able to divide $N$ as well as $N-4$, is if $p_i$ divides 4, and this is impossible for any $p_i$ that is $5$ mod $6$. Can you see the case where $N=(\prod_{i=1}^n p_i) +6$ instead? – Mike Jul 12 '21 at 11:28
  • I can understand that $p_i \mid \prod - 4$. But I am confused because my thinking process is the following. Following Euclid, we start the proof assuming a finite set of primes. These primes are obviously a factor of the product of that finite set. Then following Euclid, we create a new number, which in this case would be $\prod + 4$. But the premise of the proof is that we have a finite set of primes. So if $\prod + 4$ is a prime, we have a contradiction. We have a new prime created. If it is composite it must have a prime in the finite set. Then we should disprove it. – Jim Jul 12 '21 at 13:05
  • But here you highlight that we have a composite but you consider as obvious that it can't be a factor of $\prod$ which is what we need to prove. So I am not clear what is the problem with my original approach which tries to mimic, Euclid's logic but reaches a dead-end – Jim Jul 12 '21 at 13:06
  • @Jim, forget about $\pi$ for a moment. Let $M$ be any integer and $p$ a prime such that $p|M$. Now let $a$ be any integer such that $p$ does not divide $a$. Then $p$ does not divide $M+a$; indeed, $(M+a) \mod p$ is $a \mod p$ which is nonzero because $p$ does not divide $a$. Now getting back to this, each $p_i$ as above divides $\prod_{i=1}^N p_i$. However, no such $p_i$ divides $4$, and no such $p_i$ divides $6$. Thus, no such $p_ i$ divides $(\prod_{i=1}^N p_i)+4$, and likewise, no such $p_ i$ divides $(\prod_{i=1}^N p_i)+6$. – Mike Jul 12 '21 at 17:59
  • @Jim, forget about $\Pi$ for a moment. Let $M$ be any integer and $p$ a prime such that $p|M$. Now let $a$ be any integer such that $p$ does not divide $a$. Then $p$ does not divide $M+a$; indeed, $(M+a) \mod p$ is $a \mod p$ which is nonzero because $p$ does not divide $a$. Now getting back to the above post, each $p_i$ as above divides $\prod_{i=1}^N p_i$. However, no such $p_i$ divides $4$, and no such $p_i$ divides $6$. Thus, from the above in this comment, no such $p_ i$ divides $(\prod_{i=1}^n p_i)+4$, and likewise, no such $p_ i$ divides $(\prod_{i=1}^n p_i)+6$. – Mike Jul 12 '21 at 18:00
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You should imitate Euclid more closely. In your construction of the number $S$, you set $$S:=N\times(6m+1),$$ for some unspecified $m$. Of course every prime factor of $N$ will also be a prime factor of $S$. In particular, you are not guaranteed to get new primes of the form $6x+5$. You say you arrive at an impossible case, but you simply arrive at $p\mid(x-k)$, which is not impossible at all.

Instead, pick an integer $m$ wisely and set $$S:=N+m,$$ so that $S$ is guaranteed to have a prime factor of the form $6x+5$ that does not divide $N$. As you have noticed, what is a good choice for $m$ may depend on whether the number of primes in the product $N$ is even or odd.

Servaes
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  • I see, so basically I have not dis proven that p is not a divisor of $x - k$. I thought that since it can't divide $6$ it wouldn't divide a multiple of $6$ would suffice. – Jim Jul 11 '21 at 13:30
  • @Jim Unfortunately that's not the case; for example $5$ does not divide $6$, but it does divide $30=6\times5$. It seems you've got some part backwards, because the idea fails spectacularly. – Servaes Jul 11 '21 at 13:32
  • you are not guaranteed to get new primes of the form 6+5. but all the numbers in the sequence has to be of the form $6x + 5$ so how is it possible to get a prime of a different form? – Jim Jul 11 '21 at 21:39
  • pick an integer wisely and set :=+ what I thought from your sentence is that for the case of $N = 6x + 5$ for $S$ the $m$ should be $6$. So it would be: $$S = N + 6 = 6x + 5 + 6 = 6(x + 1) + 5$$. But the primes in the finite list, they are of the form $6i + 5$ and the largest one $p_n=6n + 5$ it should be $n < x + 1$ and as a result the prime factor of $S$ is larger than all primes in the finite set. Hence it must be a new prime which means the set of primes is not finite. Have I got it right? Or did I mess up? – Jim Jul 11 '21 at 21:43
  • $S$ has to be composite because if it is prime the proof is finished. If it is composite it must have a prime $p$ that is part of the finite set and divides $N$. Now I am stuck – Jim Jul 11 '21 at 22:35
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Your case distinction makes a lot of sense. I find that it helps me personally to think about $\mathbb Z_6$ as $\{-2,-1,0,1,2,3\}$ or so. At least to me, this suggests more strongly that getting a factor of $2$ or $3$ into any product will propagate to the end: if any of the prime factors of a composite number is $\equiv\pm2$ or $\equiv\pm3$, then the whole composite number must be congruent to one of $\{-2,0,2,3\}$. If on the other hand a composite number is congruent to $1$ or $-1\equiv 5$, then you know all of its factors (both prime and composite) will be $\equiv 1$ or $\equiv -1$. So looking at only these two kinds of factors you can tell from the remainder of the composite number whether you have an even or an odd number of $-1$ factors, thus the case distinction.

What doesn't make sense is your approach of multiplying to get a composite number. Yes, sure, you get a composite that way, but not one with any new prime factors. Euclid added one to the product of all primes. So you too should focus on adding. Adding one, however, will not align well with your modulo group.

If you have a product that is $N\equiv-1\pmod 6$ then adding $2$ will lead to $N+2\equiv1\pmod 6$ which you know contains an even number of prime factors of the desired kind. Perhaps none. So that's actually the harder case.

If on the other hand you have $N\equiv1\pmod6$ then you might go for example for $N-2$ or $N+4$. For either of these numbers you would need to show that it must contain at least one factor that is $\equiv-1$ and coprime to all the factors in $N$.

Can you also think of a way to use the simpler case to solve the hard case?

MvG
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  • this suggests more strongly that getting a factor of 2 or 3 into any product will propagate to the end what do you mean by this? I didn't understand this sentence – Jim Jul 12 '21 at 09:32
  • which you know contains an even number of prime factors of the desired kind. Perhaps none why none? The only primes are of the form $6k + 5$ and it has to have a prime factor right? – Jim Jul 12 '21 at 09:33
  • @Jim The only primes in $N$ are of the form $6k+5$ but that doesn't hold for $N+2$. So $N+2$ might contain only factors of the form $6k+1$. There is nothing anywhere saying that all existing primes will be of the form $6k+5$ and clearly that is not the case. I'll update my answer around the factors $2$ and $3$, – MvG Jul 13 '21 at 12:36
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I can still not find a question in the OP, so here is an answer - assuming the question is "How to prove there are infinitely many primes $\equiv -1$ modulo six along the lines of Euclids proof of the infinity of the set of primes?"

Assume the set $\mathbb P$ of primes $p$ with $p\equiv -1$ modulo six is finite. Its elements are $5,11,17,\dots$, at any rate $\mathbb P$ is not empty. Let $P$ be

  • the product of the elements of $\mathbb P$, if $\mathbb P$ has odd cardinality,
  • $5$ times the product of the elements of $\mathbb P$, if $\mathbb P$ has even cardinality.

So $P$ is divisible only by prime numbers which are in $\mathbb P$. The primes $2,3$ do not divide $P$. By design, $P$ is - taken modulo six - an odd product of minus ones, so it is minus one (modulo six). Let $q$ be the number: $$ q=P+6\ . $$ Then $q$ is bigger than all the numbers in $\mathbb P$, and is minus one when taken modulo six. By assumption it is not a prime. By the unique factorization theorem, we can and do write it as a product of primes. In such a product there are no $2$ and/or $3$ factors. (Since $2,3$ divide the term $6$ in $q=P+6$, but do not divide the term $P$.) In such a product there is also no $p\in\mathbb P$ factor. (Since $p$ divides the term $P$ in $q=P+6$, but does not divide the term $6$.) But any prime number $\ne 2,3$ is of the shape $6k\pm 1$, so each of the factors in the factorization of $q=P+6$ is congruent to one modulo six. So their product $q=P+6$ is one modulo six. By construction it is minus one modulo six. Contradiction.

Our assumption is false, leading to the wanted conclusion.

dan_fulea
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  • The way you generalized the use cases to $\equiv -1$ was really helpful. I didn't understand this part: But any prime number $\ne 2,3$ is of the shape $6k\pm 1$, so **each** of the factors in the factorization of $q=P+6$ is congruent to one modulo six. Why is it congruent to $1 \pmod 6$? It could also be $-1$ right?I mean the prime factors could be $(6k + 1)(6x - 1) \equiv -1$ or $(6k + 1)(6x + 1) \equiv 1$ or – Jim Jul 14 '21 at 14:33
  • Assume $()$ there is (at least) one factor $p\equiv -1$ modulo six dividing $q=P+6$. Then $p$ is in our list $\mathbb P$ of all (assumed to be finitely many) primes which are $\equiv -1$ modulo six. So $p$ divides the product $P$ of such primes (where the factor $5$ may be "doubled"). This $p$ also divides $q$, as assumed $()$. So it divides the difference $6$, contradiction. – dan_fulea Jul 14 '21 at 17:01