Question 847697
Assume (n^3) + 2 is odd.

Subtracting 2 does not change the parity of our integer.

So n^3 is odd.

If n^3 is odd, then let's look at the cube root of odd perfect cubes.

cube root of 1 = 1 <--- odd    (implies that n^3 that ends in 1, will have an n that ends in 1)

cube root of 343 = 7 <--- odd  (implies that n^3 that ends in 3 will have an n that ends in 7)

cube root of 125 = 5 <---- odd (implies that n^3 that ends in 5 will have an n that ends in 5)

cube root of 27  = 3 <---- odd (implies that n^3 that ends in 7 will have an n that ends in 3)

cube root of 729 = 9 <---- odd (implies that n^3 that ends in 9 will have an n that ends in 9)

Notice that no matter what odd n^3, we'll get an odd n.

What's left to show is that if n is odd, then n^2 is odd.

1^2 = 1 <-- odd  (if n ends in 1, n^2 will end in 1)

3^2 = 9 <-- odd  (if n ends in 3, n^2 will end in 9)

5^2 = 25 <--- odd (if n ends in 5,  n^2 will end in 5)

7^2 = 49 <--- odd (if n ends in 7, n^2 will end in 9)

9^2 = 81 <--- odd (if n ends in 9, n^2 will end in 1)

Given an odd n, we know that we will have an odd n^2.

This completes the proof.