Does $\displaystyle \frac{2^x-1}{3}=n$, n natural, have odd solutions in the naturals for x? How do you know (does Euler's theorem tell me this)?

Printable View

- Jun 7th 2008, 03:25 PMsleepingcatSimple congruence
Does $\displaystyle \frac{2^x-1}{3}=n$, n natural, have odd solutions in the naturals for x? How do you know (does Euler's theorem tell me this)?

- Jun 7th 2008, 04:29 PMTheEmptySet
All odd natural numbers (n) can be written $\displaystyle n=2k+1, k \in \mathbb{N} \cup 0$

For the above to have solutions in the odd naturals this must be true for some k

$\displaystyle (2^{n}-1 \equiv 0 \mod \\\ {3}) \iff (2^{2k+1}-1 \equiv 0 \mod \\\ {3}) \iff (2\cdot 4^{k}-1 \equiv 0 \mod \\\ {3}) $

but $\displaystyle [4^k]=[4]^k=[1]^k$ mod 3

$\displaystyle (2\cdot 4^{k}-1 \equiv 0 \mod \\\ {3})\implies (2\cdot 1^{k}-1 \equiv 0 \mod \\\ {3}) \implies (2-1 \equiv 0 \mod \\\ {3})$

This false so there is not a solution. - Jun 7th 2008, 06:17 PMThePerfectHacker
Note that $\displaystyle 2^x - 1 = 1+2+2^2+...+2^{x-1}$ (this identity appears in

*The Elements*).

Next, $\displaystyle 2 \equiv -1 (\bmod 3)$.

Thus, $\displaystyle 1+2+2^2+...+2^{x-1} = \pm 1(\bmod 3)$ (depending whether $\displaystyle x$ is even or odd).

Thus, it is impossible to have this congruent to $\displaystyle 0$ mod 3.