Inspired by Rubal Singh Dhaliwal

Number Theory Level 2

Can we find a positive integer $N$ , such that $2^{N}+1$ is even?

No Yes Cannot be determined

1 solution

Sabhrant Sachan
Oct 11, 2017

If possible Let's assume $2^{N}+1$ can be even for a natural number . Let $N_{1}$ be the first Natural number such that $2^{N}+1$ is even , then

$2^{N_{1}} +1 = 2k_{1} \hspace{10mm} k_{1} , k_{2} \in \mathbb{N} \\ 2^{N_{1}-1}+1=2k_{2}-1$

Combining both the equations , we get

$2(2k_{2}-k_{1}) = 3$

Which is impossible since any integer multiplied by $2$ is an even number . Hence $2^{N}+1$ is odd $\forall N \in \mathbb{N}$

What about 0.

2^0 + 1 = 1 + 1 = 2 => even

Dem Illuminati - 3 years, 8 months ago

$N$ is a natural number , $0$ is not a natural number .

Sabhrant Sachan - 3 years, 8 months ago

Some definitions, including the standard ISO 80000-2 begin the natural numbers with 0

Dem Illuminati - 3 years, 8 months ago

@Dem Illuminati – Yes, there is no agreement on whether or not 0 is in the natural numbers. Therefore when stating problems like the poster above did, it would be helpful if this is clarified in the statement of the problem.

The term "natural number" refers either to a member of the set of positive integers 1, 2, 3, ... (OEIS A000027) or to the set of nonnegative integers 0, 1, 2, 3, ... (OEIS A001477; e.g., Bourbaki 1968, Halmos 1974). Regrettably, there seems to be no general agreement about whether to include 0 in the set of natural numbers. In fact, Ribenboim (1996) states "Let P be a set of natural numbers; whenever convenient, it may be assumed that 0 in P."

source: http://mathworld.wolfram.com/NaturalNumber.html

Warren Cowley - 3 years, 8 months ago

Good proof and nice writeup. Maybe you can add an alternative approach in your solution that $2^N$ is obviously even nad the number which succeeds an even number is obviously odd. :)

Ashish Menon - 3 years, 7 months ago

Nice solution.

Hana Wehbi - 3 years, 8 months ago

Inspired by Rubal Singh Dhaliwal

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