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Mar
3
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
I mean introduction of a new variable, whose role and meaning is unexplained.
Mar
3
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
I am extremely uncomfortable with how much you left implied, such as the instantiation of $K$, the references to GCD, and the very non-obvious invocation of Fermat's little theorem for $K$. Only after rereading your proof a few times on separate days did I start to believe that it plausibly leads to an answer at all. -1.
Mar
3
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
I didn't understand how your answer fits with my question. Please see the other answers as examples of proofs that I did understand.
Mar
2
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Other than that, I'm sorry but I don't understand your argument at all.
Mar
2
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Why are all your variables in uppercase?
Feb
29
accepted Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Feb
28
awarded  Editor
Feb
28
revised Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Added better answer
Feb
27
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Alternate explanation: $p$ divides $2^n-1$, so $2^n-1$ divided by $p$ leaves no remainder, thus $2^n-1 \equiv 0 \mod p$.
Feb
27
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Ohh right, I understand now. I guess I'm not comfortable enough with modular arithmetic that such an elementary fact slipped past me, heh.
Feb
27
comment Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
You said "$p$ divides $2^n-1$, so $2^n \equiv 1 \pmod p$". How does this step work?
Feb
27
answered Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Feb
27
asked Prove that if $2^n-1$ is prime, then $n$ divides $2^n-2$
Aug
11
comment Is there an everywhere discontinuous increasing function?
I admit in full honesty that I asked this question out of curiosity; it is not a homework problem for me as my calculus course did not go this far.
Aug
11
awarded  Student
Aug
11
suggested rejected edit on Is there an everywhere discontinuous increasing function?
Aug
11
awarded  Scholar
Aug
11
accepted Is there an everywhere discontinuous increasing function?
Aug
11
comment Is there an everywhere discontinuous increasing function?
I should note for myself that $\lim_{x \rightarrow a^-} f(x)$ exists because of the least upper bound axiom for real numbers.
Aug
11
awarded  Supporter