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Let $$X(n)=\frac{2^n-1}n;n\in\Bbb N; n>1$$ Prove that $X(n)\notin\Bbb N$.

My try: check this for $n=2$. Then $$X(n+1)=\frac{2^n-1}{n+1}+\frac{2^n}{n+1}$$

So if $n+1$ is odd then $\frac{2^n}{n+1}$ can't be integer, if $n+1$ is even $\frac{2^n-1}{n+1}$ can't be integer.

And then I thought proof is over, however, I realized – this doesn't cover cases where sum of remainders of both fractions is equal to 1.

Parcly Taxel
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Joe Half Face
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1 Answers1

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The statement to prove is equivalent to$$2^n\not\equiv1\bmod n,n\in\Bbb N,n>1$$ We split the proof into two cases depending on the parity of $n$.

Case 1: $n$ is even. Then the proof is obvious, since $2^n\bmod n$ must be even as well.

Case 2: $n$ is odd, so $\gcd(2,n)=1$. By Carmichael's theorem $2^{\lambda(n)}\equiv1\bmod n$, where $\lambda$ denotes the Carmichael function. But $2\mid\lambda(n)$ for $n\ge3$, so $\lambda(n)\nmid n$ since $n$ is odd. This implies $2^n\not\equiv1\bmod n$ since $\lambda(n)$ is the least exponent that will yield a residue of 1 and any higher such exponents are multiples of it.

Therefore the statement is true for all $n>1$.

Parcly Taxel
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  • Nice proof! Is there a proof that doesn't involve Carmichael's theorem? – Carl Schildkraut Oct 02 '16 at 04:04
  • @CarlSchildkraut Thought of using Euler's theorem, but then realised the exponent given by that isn't minimal. Interestingly enough I just finished a number theory module in my high school, and that module taught Euler's theorem but not Carmichael's. It's a pretty easy strengthening though. – Parcly Taxel Oct 02 '16 at 04:10
  • Yeah, Euler's Theorem isn't strong enough. Pretty much all you need is that the minimal exponent is even. – Carl Schildkraut Oct 02 '16 at 04:12
  • @CarlSchildkraut Yes, Carmichael is a bit overkill here, see e.g. this answer for a proof avoiding it. – Bill Dubuque Nov 02 '16 at 22:53