Given that $x$ is some prime number, show that, for values of $x$ other than $2$ or $3$, at least one of $2x-1$ and $4x-1$ cannot be prime. I'm having difficulty proving that this is the case for all $x$ other than $2$ or $3$, so any help related to this question will be appreciated.
Show that either $2x-1$ or $4x-1$ is not prime for prime values of $x$ other than $2$ or $3$
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number-theory
prime-numbers
proof-writing
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3Hint: assume $x>3$ is prime. Then it must be $3k+1$ or $3k-1$ for some $k$. Try both. – 2017-02-13
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0@lulu Thank you so much, this made the answer so much clearer. – 2017-02-13
1 Answers
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Following from lulu's excellent comment: If we have $x>3$ prime, then $x$ must be of the form $3k \pm 1$ for some integer $k$.
In the first case $x = 3k+1$ we have $4x - 1 = 6k +4 - 1 = 6k + 3 = 3(2k + 1)$ which is the product of two integers, so not prime.
In the second case $x = 3k-1$ we have $2x -1 = 6k-2-1 =6k-3 = 3(2k-1)$ which is the product of two integers again, so not prime.
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0Note that $2$ is of the form $3k-1$, yet both $2\cdot 2-1$ and $4\cdot 2-1$ are prime. Perhaps you should state somewhere where you actually require $x>3$. – 2017-02-13