# Homework Help: Prove that k is even

1. Oct 4, 2009

### zeion

1. The problem statement, all variables and given/known data

a) Let k be any integer. Prove that if k3 is even, then k is even.

2. Relevant equations

3. The attempt at a solution

If the hypothesis is true, then k3 cannot be even if k is odd.

Assume k is odd:
k = 2n + 1, such that n is any integer.
k3 = (2n + 1)3
k3 = (2n +1)(2n +1)(2n +1)
k3 = 2(k3 + 5k2 + 3k) + 1
Now, (k3 + 5k2 + 3k) is any integer.
Therefore k3 is odd if k is odd, hence if k3 is even if k is even.

2. Oct 4, 2009

### ravioli

Looks okay to me, unless you were trying to do a proof by contradiction. Your current proof is one of a contrapositive.

3. Oct 4, 2009

### ravioli

My bad, last statement is iffy. I think it should read if k cubed is even, then k is even

4. Oct 5, 2009

Ok thanks

5. Oct 5, 2009

### Staff: Mentor

The line above doesn't make sense to me. If you multiply out the right side, you get 8n3 + 12n2 + 6n + 1 = 2(4n3 + 3n + 3) + 1. This is clearly an odd integer, which means that k3 is an odd integer.
Another approach is to prove the statement directly.
Assume that k3 is even.
Then k3 = 2m for some integer m.
Because k occurs to the third power, there must be three factors of 2 on the right.
So k3 = 2*2*2*n for some integer n.
Again, because k occurs to the third power, n must have three equal factors, say n = p3.
Hence, k3 = 2*2*2*p*p*p, from which you can easily show that k = 2p, an even integer.