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## Homework Statement

Let "a" be an element of any integer. Let "n" be an element of any natural number that is

**greater than or equal to 2**. The integer "a^n" is an even number

**if and only if**"a" is an even number. Prove this bi-conditional statement.

## Homework Equations

Definition of an even number: Let x be an even number. Then there exists an integer y such that x = 2y.

Definition of an odd number: Let x be an odd number. Then there exists an integer y such that x = 2y + 1.

## The Attempt at a Solution

Note that it is a bi-conditional statement, this means that we have to prove:

1. If a^n is an even number, then a is an even number.

2. If a is an even number, then a^n is an even number.

The most straightforward approach I had was proving it directly (I only know direct proof, contrapositive, and contradiction for now):

1. Since a^n is even, then there is an integer y such that a^n = 2y. This brings forth the following relationships:

a^n = 2y

a^n = 2 * (2^(n-1) * k^n) - where k is an integer.

a^n = 2^n * k^n

a = 2k

Thus, a is even.

2. Since a is even, then there is an integer y such that a = 2y. This brings forth the following relationships:

a^n = (2y)^n

(2^n)*(y^n) = 2 * (2^(n-1)) * y^n

a^n = 2 * (2^(n-1) * y^n)

a^n = 2m - where m is an integer

Thus, a^n is even.

The reason I substitute y = (2^(n-1) * k^n) or m = (2^(n-1)) * y^n) was under the closure property of integers.

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I am not even sure if I proved this statement correctly. Furthermore, I am also having doubts about how I applied the closure property of integers. Can anyone give me feedback?

On a side note, I already know that I can easily prove (2.) by using contradiction.

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