0

Show that $p\gt 3, 2p+1, 4p+1$ cannot be both prime

I know that for testing if an integer is odd, need check if its square is of the form $\exists k\in \mathbb{Z}, 8k+1$.

If $p\gt 3, 2\nmid p$, so $p$ is odd.
So, one way can be to show that the congruence class $8k+1$ cannot be split into $(2p+1)\cdot(4p+1) = 8p^2+6p+1 \ne \exists k'\in \mathbb{Z},\, 8k'+1$.

I am asking this for addtl. reason that the solution provided by a book states that :
A prime can only take the form $3k-1$, or $3k+1$.
If $p=3k+1$, then
$$2p+1=2(3k+1)+1=3(2k+1),$$ so that $2p+1$ is composite. If $p=3k-1,$ then$$4p+1=4(3k-1)+1=3(4k-1),$$ so that $4p+1$ is composite.

I believe the solution approach is faulty, for the reason that it is not necessary for the integers of the form $3k-1, 3k+1$ to be prime.

jiten
  • 4,524
  • 1
    is $p$ assumed to be a prime number? – Dr. Sonnhard Graubner Mar 12 '18 at 18:06
  • 3
    Is $p$ also supposed to be prime? If so, you can easily check that one of $2p+1,4p+1$ is divisible by three. – Jyrki Lahtonen Mar 12 '18 at 18:06
  • Yes, $p$ is prime. – jiten Mar 12 '18 at 18:09
  • 1
    In that case you should take my comment as a hint, and work it out1 Post is as an answer to get feedback on the details of your argument! – Jyrki Lahtonen Mar 12 '18 at 18:10
  • 1
    @JyrkiLahtonen Pleas check the edited OP. It is for reviewing the book's approach, my approach based on your earlier comment will follow soon. – jiten Mar 12 '18 at 18:12
  • 1
    Great, now we get to the heart of the matter (or not)! You are right in that not all the numbers of the forms $3k\pm1$ are primes. But, that is of no concern here. For the argument in the book to be on a solid footing it suffices that all the primes $>3$ are of the form $3k\pm1$. Do you see why that is true? The claim is true as long as $p$ is not divisible by three. It doesn't have to be a prime! – Jyrki Lahtonen Mar 12 '18 at 18:17
  • @JyrkiLahtonen Then your first comment is already implemented in the book's answer. However, book states $p$ to be a prime. I also want to extend the book's solution to : If $p=3k+1$, then $4p+1 = 4(3k+1)+1=12k+5$, and if $p=3k-1$, then $2p+1=2(3k-1)+1= 6k-1$. So, does the book mean that the integers $\gt 3$ with forms $1k+5. 6k-1$ are primes. – jiten Mar 12 '18 at 18:26
  • 1
    Yes, yes. No need for you to rewrite that. But, the logic of the solution only needs that the implication "$p>3$ is a prime" $\implies$ "$p$ is of the form $3k\pm1$". It never uses the false converse implication "$p$ is of the form $3k\pm1$" $\implies$ "$p$ is a prime". So the argument is solid. – Jyrki Lahtonen Mar 12 '18 at 18:31
  • @JyrkiLahtonen There is a big typo in my earlier comment's last line:"...$12k+5, 6k-1$ are primes". Regarding, your last comment, I have gathered that the form $12k+5, 6k-1$ can be either prime or not just like $3k\pm 1$ can be either prime or not. It is only the definite composite forms that need to check for as : $3(2k+1), 3(4k-1)$. Also my argument, as in the OP, that it is impossible to split the original in the stated forms' product, should have similar applicability here, as it applies for odd numbers' square's form $8k+1$, which can be composite/prime . – jiten Mar 12 '18 at 18:49
  • @JyrkiLahtonen I will be waiting eagerly for your comment/answer. – jiten Mar 12 '18 at 18:49

2 Answers2

2

For your initial attempt:

  • if $x$ is an odd number, then $x^2 \equiv 1 \pmod{8}$.

  • However, if $x$ is odd and $y$ is odd, it is possible that $xy \not\equiv 1 \pmod{8} $. For example $$9 \cdot 7 = 63 \equiv 7 \pmod{8}$$

For the book attempt:

  • If $p$ is a prime, $p>3$, then $p=3k-1$ or $p=3k+1$, otherwise, $p=3k$ and since $p>3$, it means $p$ would be a positive number that is divisible by $3$ and $p \neq 3$, hence $p$ can't be a prime.

Edit (for book's approach)

Case $1$: $p=3k$, since $p>3$, $k>1$, but this mean $p$ is not a prime, hence it is a contradiction. Case $1$ doesn't happen.

Case $2$: $p=3k-1$, then $4p+1=3(4k-1)$, hence $4p+1$ is not a prime.

Case $3$: $p=3k+1$, then $2p+1=3(2k+1)$ is not a prime.

Siong Thye Goh
  • 149,520
  • 20
  • 88
  • 149
  • I request elaboration for the book's attempt, as book showed that for a given 'possible' prime form, it is not possible for it to have both $2p+1, 4p+1$ forms. Your answer is taking the product of two numbers of given forms as constituting one number, which obviously cannot be a prime then, hence confuses me by its approach. 2. Regarding my attempt to try the $\forall x\in \mathbb{Z}, 2\nmid x, \exists k \in \mathbb{Z},, x^2\in 8k+1$ approach for a product of two different forms, your answer states that there are two different numbers, and not one number presented in two different forms.
    • – jiten Mar 13 '18 at 05:03
  • 1
    Note that here $p$ is fixed, $2p+1 \ne 4p+1$. – Siong Thye Goh Mar 13 '18 at 05:16
  • I have an unusual request, that would humbly hope to get attention. I have come across web-pages that contain a book's pages, which is very informative, and is different from all other conventional books on the subject dealt with it, in terms of opening a beginner to the basics of topics. This aspect can be easily checked by reading the given contents. The link is: https://www.math.utah.edu/~bertram/courses/4030/. Kindly help me in finding the name of the book as any amount of googling, with the content of the book, has failed. Thanking you in anticipation. – jiten Mar 14 '18 at 21:25
  • 1
    That is Professor Aaron Bertram's own note. It is a class taught at UTAH known as Foundation of Algebra – Siong Thye Goh Mar 14 '18 at 21:42
  • Sorry for late response. But, is there any way to access it for me. I feel it is too good, but found no way to access. – jiten Mar 15 '18 at 02:04
  • 1
    the link that you send me are the notes isn't it? – Siong Thye Goh Mar 15 '18 at 02:07
  • No, pages are too few, and have gaps. Also, it states on pg. 79 (https://www.math.utah.edu/~bertram/courses/4030/Fund.pdf) that : "(We will not even attempt the theorems of Abel and Galois until later.)". So, definitely the book is too big. The concerned professor's course is never so big to cover these topics ever, in any course on his web-site. – jiten Mar 15 '18 at 02:15
  • 1
    Page 44 is at the very next link. Page 61 should appear under the topic Euclidean Domain, to be sure, you might like to contact the prof. I believe it is a written by him to teach the class. – Siong Thye Goh Mar 15 '18 at 02:21
  • I have today posted one post, for which the second answer at : https://math.stackexchange.com/a/2694730/424260, is assuming me to be able to follow the hint given. But, the algorithmic approach used makes the solution's any example run clumsy. If you could please help me in that, then would be highly thankful. Also, the other answer leads to algebraic curves, for which need more hint/details. – jiten Mar 17 '18 at 15:12
  • Triangular numbers have 'sequence' s.t. the first row is having $1$ member, the 2nd row has $2$ elements (with the sum = 3, as given by: $\frac{n(n+1)}{2}$, with $n=2$). Then find sum of 'sequence' of elements of the form $\frac{n(n+1)}{2}$ with incrementally elements increasing at each row, as given at the post: https://math.stackexchange.com/a/2282274/424260, but is unclear as to why a form is chosen as: $A.\frac{n(n+1)}{2} + 1 = (Bn+C)^2$. – jiten Mar 23 '18 at 16:19
  • I request a formal proof for proving that $r_n =\gcd(a,b)$ is the largest common divisor, where $r_n$ is the last non-zero remainder found using the Euclidean algorithm. I request to use the fact that if a non-common divisor (let, $x$) is taken, then there are both possiblities that $\gcd(r_n,x)=1, \ne 1$. Something formal for me can be elaborated by an incomplete approach of mine is: There are two cases based on $\gcd(r_n,x)$ values- (i) if $\gcd(r_n,x)=1$, then..., (ii) if $\gcd(r_n, x)\ne 1$, then... I do not know hoe to pursue, but want such type of approach, or even more formal. – jiten Mar 29 '18 at 11:40
  • In continuation to the earlier comment, there is a post of mine at : https://math.stackexchange.com/q/2712980/424260, that concerns with the same issue, although not stated there the form of solution preferred as in the earlier comment. – jiten Mar 29 '18 at 21:40
  • Thanks, but your response was incomplete in providing pointer to the definition aspect of $\gcd$, and have placed a comment 'there' that requests the needed info. for detail. – jiten Mar 30 '18 at 05:11
  • 1
    beyond my ability.... – Siong Thye Goh Mar 30 '18 at 05:18
  • Please see my post of today at https://math.stackexchange.com/q/2720436/424260. Please post an answer to that. – jiten Apr 03 '18 at 15:30
  • Please help with my post as specified in the earlier comment. I am unclear about the mechanism used to derive the steps for both cases. – jiten Apr 03 '18 at 21:46