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This question already has an answer here:

Since $n\geq 4$ is even, we can let $n=2k$. Then $2k\geq 4$ or $k\geq 2$ which can be substituted where we have $2^{2k}-1 = (2^k)^2 - 1^2 = (2^k+1)(2^k-1)$. Since $k \geq 2$, we have that $2^k\geq 4$.

This is where things get derailed. Where do I go from here?

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marked as duplicate by Bill Dubuque elementary-number-theory Jun 26 '17 at 13:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ you are (almost) done already $\endgroup$ – supinf Jun 26 '17 at 11:40
  • $\begingroup$ $\gcd(2^k+1,2^k-1)=\gcd(2^k+1,2)=1$ $\endgroup$ – Jack D'Aurizio Jun 26 '17 at 14:02
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If $k>1$ then $$ 4^k-1=\underbrace{(2^k+1)}_{>1}\underbrace{(2^k-1)}_{>1}. $$ If $k=1$ then $4^k-1$ is prime.

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You are able to write a number as product of two numbers none of them equals to $\pm 1$.

This says it is not a prime.

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Just to give a different approach, if $n=2m\ge4$, then $2^n-1\gt3$, but $2^{2m}-1\equiv(-1)^{2m}-1\equiv0$ mod $3$, so $3\mid2^n-1$.

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