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I have two expressions (both of which have a term raised to the power of $n$) and I am trying to prove that they can't be prime numbers at the same time for $n>2$.

I can't post the expressions, but I was wondering if there was someway to prove it by saying: "Let the first and the second be prime numbers then because their product is equal to something they can't be prime numbers at the same time.

Is their a statement or law that I can use?

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    $\begingroup$ Well, if the product has at least three factors ... $\endgroup$ – Hagen von Eitzen Jul 30 '15 at 6:44
  • $\begingroup$ it has three factors. The product is 63 for n=3. 63=3*3*7. How can I generalize that for n>2? $\endgroup$ – DoubleOseven Jul 30 '15 at 6:45
  • $\begingroup$ If they are distinct and both prime, one would have to be $3$ and the other $7$, and then the product would not be $63$. But there is a simpler solution to the general problem I think you are starting from. $\endgroup$ – André Nicolas Jul 30 '15 at 6:54
  • $\begingroup$ If this is the real question, I urge you to edit the post. $\endgroup$ – Yves Daoust Jul 30 '15 at 7:10
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Of any three consecutive integers, one is divisible by $3$. In particular, one of $2^n-1$, $2^n$, and $2^n+1$ is divisible by $3$. But $2^n$ is not divisible by $3$, so one of $2^n-1$ and $2^n+1$ is divisible by $3$.

If $n\gt 2$, then $2^n-1$ and $2^n+1$ are both bigger than $3$. One of them is divisible by $3$ and greater than $3$, so is not prime.

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Hint: Primes other than 2,3 always have the form $6k + 1$ or $6k + 5$ for $k \in \mathbb Z$.

To prove this, notice that numbers that have the form $6k, 6k+2$ or $6k+4$ are divisible by 2, and numbers that have the form $6k+3$ are divisible by 3.

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  • $\begingroup$ I don't think this will work here. Look above at my comment where you can see the two expressions $\endgroup$ – DoubleOseven Jul 30 '15 at 7:09
  • $\begingroup$ Can you, using this information, prove that the sum of the numbers in a twin prime (other than $(3,5)$), is always divisible by 6? $\endgroup$ – wythagoras Jul 30 '15 at 7:11
  • $\begingroup$ Nash, $6k+1,6k+5$ is not a twin prime. Rather you should have $6k-1, 6k+1$ as twin prime for a total sum of $12k$. But, if $(2^k-1,2^k+1)$ was a twin prime, what would divide their sum? $\endgroup$ – wythagoras Jul 30 '15 at 7:18
  • $\begingroup$ would that be 6? $\endgroup$ – DoubleOseven Jul 30 '15 at 7:19
  • $\begingroup$ Yes. Can't you get a contradiction from this? $\endgroup$ – wythagoras Jul 30 '15 at 7:20

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