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

  • #1
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?
 

Answers and Replies

  • #2
33,496
5,188


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.
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.
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?
 
  • #3
360
0


If the prime of the form 3k+1 can't be written in the form 6m+1, what does that say about k?
 
  • #4
256
2


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.
 
  • #5
gb7nash
Homework Helper
805
1


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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