Simple congruence

**sleepingcat** · June 7th 2008, 03:25 PM

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)?

**TheEmptySet** · June 7th 2008, 04:29 PM

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.

**ThePerfectHacker** · June 7th 2008, 06:17 PM

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.

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