# Simple congruence

• Jun 7th 2008, 03:25 PM
sleepingcat
Simple congruence
Does $\frac{2^x-1}{3}=n$, 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 PM
TheEmptySet
Quote:

Originally Posted by sleepingcat
Does $\frac{2^n-1}{3}$ have odd solutions in the naturals? How do you know?

All odd natural numbers (n) can be written $n=2k+1, k \in \mathbb{N} \cup 0$

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

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

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

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

This false so there is not a solution.
• Jun 7th 2008, 06:17 PM
ThePerfectHacker
Quote:

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

Note that $2^x - 1 = 1+2+2^2+...+2^{x-1}$ (this identity appears in The Elements).

Next, $2 \equiv -1 (\bmod 3)$.
Thus, $1+2+2^2+...+2^{x-1} = \pm 1(\bmod 3)$ (depending whether $x$ is even or odd).
Thus, it is impossible to have this congruent to $0$ mod 3.