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Sorry if this is trivial!

Consider the set 3, 6, 9, 15, 21, 30, 36, 51, 54, 69,

These are all such that $2x-1$ and $2x+1$ are both prime.

Why are they all divisible by 3?

And if say $yx-1$ and $yx+1$ are both prime (i.e. generated by the twin-prime-averages that are divisible by $y$), then do all these $x$'s have a common divisor?

Why?

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2 is not divisible by 3, even though $2\times 2 -1$ and $2\times 2 +1$ are prime –  Henry Sep 24 '12 at 16:04
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4 Answers

All primes are of the form $6k \pm 1$ (except $2$ and $3$) i.e. $2(3k) \pm 1$. No prime can be of the form $6k+3$ (except $3$) since $3 \mid 6k+3$. So twin primes will always be of the form $(6k-1, 6k+1)$.

Hence, if we have twin primes $(2x-1,2x+1)$, then they should also be of the form $(6k-1,6k+1)$ and hence $3$ divides $x$

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Since one of every three consecutive natural numbers is divisible by $3$ and $2x-1$ and $2x+1$ being prime(not divisible by $3$ except trivial case when one of them is $3( x=1)$) $\implies 2x$ is divisible by $3$ and since $3\not|$ $ 2\implies 3$ divides $x$

For part 2, Number between twin primes is always a multiple of $6$, so if you take twin primes of the form $3x-1$ and $3x+1$, then, $2$ surely divides $x$; but in general for $yx-1$ and $yx+1$ as prime, i think there is very little chance of having a common factor of all the $x'$s

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I really like the informal answer in the first paragraph. The answer does not require a lot of fancy math notation. –  phkahler Sep 24 '12 at 19:10
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Let $p$ be an integer greater than 3, and consider the remainder when you divide $p$ by 6. If the remainder is 0, 2 or 4, then $p$ is divisible by 2, and so isn't prime; if the remainder is 3, then $p$ is divisible by 3, and so isn't prime.

Therefore, if $p$ is prime, it must have a remainder of 1 or 5 when divided by 6.

If $p$ and $q$ are primes with $q=p+2$, the only way to arrange the remainders is for $p$ to have a remainder of 5 and $q$ to have a remainder of 1 when divided by 6; then $p$ must have the form $6k-1$ (remainder of 5) and $q$ the form $6k+1$ (remainder of 1).

You are considering $x$ such that $2x-1 = p$ and $2x+1 = q$, so therefore $z=3k$, and that is your answer.

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If $2x=3z \pm 1, (2x-1)(2x+1)=(2x)^2-1=(3z\pm 1)^2-1=3(3z^2\pm 2z)$ which is clearly divisible by $3$.

If $2x-1>3,$ so is $2x+1$, then at least one of them is divisible by $3$, hence is not prime.

Regarding the 2nd question, the average of $yx_1-1,yx_1+1$ is $yx_1$

and the average of $yx_2-1,yx_2+1$ is $yx_2$ where $x_1,x_2$ are integers.

So, $(yx_1,yx_2)=y(x_1,x_2)$ is always divisible by $y $

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