Direct proof of n^2 is even implies n is even

How indeed...

You have k = (n/2)n, but to conclude from that that k is a multiple of n is tantamount to saying that (n/2) is an integer, which is begging the question, since if you know n/2 is an integer, then you already know n is even.

Instead of your route, you could trying branching off in another direction after step 2.

You have: 2 | n*n.

Are you allowed to use the little theorem that says: if you have prime p, and integers a and b such that p | ab, then p | a or p | b ?

If n^2 is even, then n^2 - 1 is odd, therefore the product (n+1)(n-1) is odd.
Now n+1 and n-1 are consecutive integers on either side of n, which are either both even or both odd. So since their product is odd, they are both odd, which proves n is even,

Ok.
Yes, I can use it. So what would be the next step?

Thank you!

If you can use it, then you're basically done.

According to that theorem, if 2 | n*n, then (2 | n or 2 | n), since 2 is prime.
(You're letting 2 be p, and you're letting n be both a and b.)

2 | n => there exists integer m such that n = 2m.

right. ok, thank you.

But supposing I can't use this theorem? (for the fun of mathematics)

Well, I guess I'll defer to fresco in that case.

(His answer doesn't use that theorem.)