Number theory: confused about the phrase an integer of the form

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Homework Help Overview

The discussion revolves around proving that any prime of the form 3k+1 is also of the form 6k+1, within the context of number theory.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants express confusion regarding the notation "an integer of the form," questioning whether the variable k must be the same across different expressions or if they represent different integers. There is an exploration of whether the values for 6k+1 are a subset of those for 3k+1.
  • Some participants suggest considering cases based on the parity of k, specifically examining the implications of k being odd or even on the forms of the integers.

Discussion Status

The discussion is ongoing, with participants exploring various interpretations and approaches to the problem. Some guidance has been offered regarding the relationship between the forms and the implications of k's parity, but no consensus has been reached.

Contextual Notes

Participants are navigating the constraints of the problem, particularly the definitions and implications of the forms given, as well as the nature of the integers involved.

doubleaxel195
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Number theory: confused about the phrase "an integer of the form"

Homework Statement


Prove that any prime of the form 3k+1 is of the form 6k+1


Homework Equations





The Attempt at a Solution



I'm not sure where to start at all. I tried rewriting 3k+1 as 6k+2=6k+(6-4)=6(k+1)-4. But that gets me no where really, and doesn't utilize the fact that it must be prime and now I made my prime number not prime.

But first and foremost, I am confused about the phrase/notation "an integer of the form." For instance, in this problem, does it mean that k must have the same value or are those 2 different k's where k is just any integer but not necessarily the same integer. Well, obviously for k=1, 3(1)+1=4 whereas, 6(1)+1=7 so they are not equal. So why is the same variable k used? Or is k used to refer to the set of values that 3k+1 and 6k+1 can take on? Like the set of values for 6k+1 is contained within 3k+1, like a subset. If this is true, then do I just use p=3k+1 and just show that an integer exists, not necessarily k, such that p=6q+1?
 
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doubleaxel195 said:

Homework Statement


Prove that any prime of the form 3k+1 is of the form 6k+1


Homework Equations





The Attempt at a Solution



I'm not sure where to start at all. I tried rewriting 3k+1 as 6k+2=6k+(6-4)=6(k+1)-4. But that gets me no where really, and doesn't utilize the fact that it must be prime and now I made my prime number not prime.
I don't believe they intended for the k on one side to be the same as the k on the other side. What I think they meant was that any prime of the form 3k + 1 is also of the form 6m + 1, for some integers k and m.
doubleaxel195 said:
But first and foremost, I am confused about the phrase/notation "an integer of the form." For instance, in this problem, does it mean that k must have the same value or are those 2 different k's where k is just any integer but not necessarily the same integer. Well, obviously for k=1, 3(1)+1=4 whereas, 6(1)+1=7 so they are not equal.
Bad example, since 4 isn't a prime.

A better choice is 3(6) + 1 = 19, which is a prime. 19 can also be written as 6(3) + 1.
doubleaxel195 said:
So why is the same variable k used? Or is k used to refer to the set of values that 3k+1 and 6k+1 can take on? Like the set of values for 6k+1 is contained within 3k+1, like a subset. If this is true, then do I just use p=3k+1 and just show that an integer exists, not necessarily k, such that p=6q+1?
 


If the prime of the form 3k+1 can't be written in the form 6m+1, what does that say about k?
 


Is k an integer? I can't really see a way out of this. However if you can prove that every 3k + 1 number can be written as 6m + 1 after a certain value of k, then it shouldn't be too difficult.

EDIT: Oops! I think you can show that if k is even, then your 3k + 1 is odd and can be written as 6m + 1, which means that all odd numbers of the form 3k + 1 can be written as 6m +1. Try that.
 


Split it up into two cases:

Case 1: k is odd. What happens if k is odd? Is this possible?

Case 2: k is even. How can we write k?
 

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