- For all n > 1, n does not divide
2n−1 .
Proof: Suppose not. Then n is odd as2n−1 is odd. Letp be the smallest prime dividing n. We have(p−1,n)=1 . By division algorithm,n=q(p−1)+k for some integers q and k with0<k<p . By Fermat’s Little Theorem,2p−1≡1 (mod p), so2k≡(2p−1)q2k≡2n≡1 (mod p). This results in a contradiction by infinite descent.
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