Can you prove n2 - n is even for any natural number n?

Answer 1

There are two cases here: one where n is even and one where n is odd.

Let's consider the case where n is even:

If n is even then n2 has to be even (since multiple of an even number must always be even.) In this case, we are subtracting an even number from and even number, the result must be even. This proves than n2 - n is even when n is even.

Now let's consider the case where n is odd:

If n is odd, then n2 must be odd. This is because an odd number times an odd number is always odd.

(You could think if this as an odd number times and even number and then adding an odd number. For example, say that n is odd. n-1 is then even, and n2 = n(n-1) + n. n(n-1) must be even, since it is a multiple of an even number. And even number plus and odd number then has to be odd.)

Now we know we have and odd number minus and odd number, which has to be even. So this proves that n2 - n is even when n is odd.

Since we have proved that n2 - n is even for both when n is even and when n is odd, and there are no other cases, n2 - n must be even for any natural number n.

or

Let n be a natural number. Then n can be even or odd.

We want to show that n2 - n = 2m where m is any positive nteger (by the def. of even).

Case 1:

Let n be even. Then n = 2k (def. of even), where k is any positive integer. Then,

n2 - n

= (2k)2 - (2k)

= 2(2k2 - k); let 2k2 - k = m

= 2m

Therefore, n2 - n is even.

Case 2:

Let n be odd. Then n = 2k +1 (def. of odd), where k is any positive integer. Then,

n2 - n

= (2k - 1)2 - (2k - 1)

= 4k2 - 4k + 1 - 2k + 1

= 4k2 - 6k + 2

= 2(2k2 - 3k + 1); let 2k2 -3k + 1 = m

= 2m

Therefore, n2 - n is even.

Therefore, for any natural number n, n2 - n is even.