Sure, let's break this down:

1. Assume n is an even integer. This means that n can be written in the form 2k, where k is an integer.

2. If we substitute n = 2k into the expression n^3 + 5, we get (2k)^3 + 5 = 8k^3 + 5.

3. The term 8k^3 is always even because it's a product of two even numbers (8 and k^3, since the cube of any integer is an integer).

4. Therefore, 8k^3 + 5 is the sum of an even number and an odd number, which is always odd. So, if n is even, then n^3 + 5 is odd.

Now, let's prove the other direction:

1. Assume n^3 + 5 is odd. This means that n^3 must be even (since the sum of an even and odd number is odd).

2. The cube of an odd number is always odd, and the cube of an even number is always even. Therefore, if n^3 is even, then n must be even.

So, we have shown that n is even if and only if n^3 + 5 is odd.

Question

Sure, let's break this down:

1. Assume n is an even integer. This means that n can be written in the form 2k, where k is an integer.

2. If we substitute n = 2k into the expression n^3 + 5, we get (2k)^3 + 5 = 8k^3 + 5.

3. The term 8k^3 is always even because it's a product of two even numbers (8 and k^3, since the cube of any integer is an integer).

4. Therefore, 8k^3 + 5 is the sum of an even number and an odd number, which is always odd. So, if n is even, then n^3 + 5 is odd.

Now, let's prove the other direction:

1. Assume n^3 + 5 is odd. This means that n^3 must be even (since the sum of an even and odd number is odd).

2. The cube of an odd number is always odd, and the cube of an even number is always even. Therefore, if n^3 is even, then n must be even.

So, we have shown that n is even if and only if n^3 + 5 is odd.

Knowee AI · Accepted Answer