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KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3Well, suppose towards a contradiction that \(n=a\cdot b\) is composite with \(a,b>1\). Then, \[2^n1=2^{ab}1=(2^a)^b1\]Now substitute \(2^a=x\) to get \(x^b1\), and use the formula I helped you prove by induction earlier. Make sense?

johnny0929
 2 years ago
Best ResponseYou've already chosen the best response.0I understand all the way up to the former formula. I know that it would look like\[(x1)(x^{b1}+x^{b2}+...+x^{2}+x+1)\] but why does that make \(2^{n}1\) not prime?

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3\[2^n1=(2^a)^b1=(2^a1)(2^{a(b1)}+2^{a(b2)}+...+2^a+1)\]Since \(a>1\), \(2^a1>1\), and so we have that \((2^a1)(2^n1)\).

johnny0929
 2 years ago
Best ResponseYou've already chosen the best response.0thank you so much i understand it now. Proof writing is a very difficult task for me :(

KingGeorge
 2 years ago
Best ResponseYou've already chosen the best response.3Proof writing is definitely an acquired skill :P
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