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doubleaxel195
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Number theory: confused about the phrase "an integer of the form"
Prove that any prime of the form 3k+1 is of the form 6k+1
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?
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?