Prove that if $n\geq 4$ is even then $2^n -1$ is not prime. [duplicate]

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?

marked as duplicate by Bill Dubuque elementary-number-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jun 26 '17 at 13:41

• $\gcd(2^k+1,2^k-1)=\gcd(2^k+1,2)=1$ – Jack D'Aurizio Jun 26 '17 at 14:02
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.
You are able to write a number as product of two numbers none of them equals to $\pm 1$.
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 $3\mid2^n-1$.