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Show any prime of the form $3k+1$ is of the form $6k+1$.

I came up with my own solution that made perfect sense to me, but when I read the text's solution, it argued that for the primes that are of the particular form are $6k+1 = 3(2k)+1$. But doesn't that really say the primes in the form of $3k+1$ are in the form of $6m+1$? It seems to me as though there's some misuse of notation here -- allowing $k = 2k$. So should the exercise be phrased as $6m+1$ instead?

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I think "of the form" is the key phrase here, not equal to. The letter is essentially a dummy variable. –  Daniel Rust May 3 '13 at 22:49
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3 Answers 3

up vote 4 down vote accepted

I think it would have made more sense to pose the problem as:

"Show that any prime of the form $3n+1$ is of the form $6k + 1$"

to distinguish the integers $n, k$, and allow for subsequently proving this is the case for $n = 2k$: primes of the form $3n + 1 = 3(2k) + 1$ are thus of the form $6k + 1$.

But just like indexing variables, I suspect that "$k$" as used in the actual problem statement was intended to be a "dummy" variable standing in for "some integer", much like $x$ in the expressions "$\forall x P(x)$, and $\forall x Q(x)$" each use $x$ independently of its use in the corresponding assertion. But this is not standard.

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Yeah, but when they used $k$ in both, it reads that the $k$'s are going to be the same. Is this sort of thing standard notation? –  AlanH May 3 '13 at 22:55
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The more standard notation would be that of congruence modulo an integer and so 6n+1=3(2k)+1 would more readily be read as $x=1 \mod 6\Rightarrow x=1 \mod 3$ –  Daniel Rust May 3 '13 at 23:00
    
I agree with you, AlanH. That's why I suggested the highlighted statement and find it preferable. The problem statement as you posted is ambiguous, perhaps even misleading. I agree with @Daniel Rust, wholeheartedly. –  amWhy May 3 '13 at 23:01
    
@amWhy: Looks like you clarified this one! +1 –  Amzoti May 4 '13 at 0:24
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Another way to phrase it is "If k is a positive integer such that 3k+1 is prime then k is even".

The proof, of course, is easy: If k is odd, then k=2h+1 for some integer h. But 3k+1 = 3(2h+1)+1 = 6h+4 is even and therefore not prime.

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The only $k$ for which $3k+1=6k+1$ is $k=0$, which doesn't even give a prime. The statement means "if a prime is a multiple of three plus one, then it is a multiple of six plus one". The $k$ is a dummy variable; although it is not proper to us $k$ twice, the statement is not much clear if you use two different variables.

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