Find the problem with this proof. The following attempts to prove that if $n^2$ is even, then $n$ is even.
Suppose $n^2$ is even. Then $n^2$ = 2$k$ for some integer $k$. Let $n$ = 2$m$ for some integer $m$. This shows that $n$ is even.
I believe the problem is the believing the let $n$ = 2$m$ step proves the statement. This seems to put a lot of weight on a mere assumption. For example, say I was trying to prove "If 3$n$ + 2 is odd, then $n$ is odd" by saying " let n = 2$k$ + 1 for some integer $k$. This shows $n$ is odd." I guess circular reasoning would be the best way to describe it, the argument beings what it wants to end with. It's along the same logical lines of saying "Marcus is tall, therefore Marcus is tall". Am I on point here guys?
 A: I'm not sure exactly what you're trying to prove, but it looks like you're trying to show $$ n^2 \text{ is even} \implies n \text{ is even}$$
I think what is missing is since you're doing cases, you probably want to do a proof by contradiction for when n is odd. A better way (in my opinion) is a proof by contrapositive. Assume $ \exists q \in \mathbb{Z}. \; n = 2q+1,$ and show why $n^2$ cannot be even.
EDIT: Also you just assumed that $n$ is even, but there is no relation to $n^2$, so you can't prove anything.
A: The problem is that you make an assumption. If you say let $n=2m$ then of course $n$ is even. To put another way: you're assuming that what you want to prove is true. This way everyting can be true. The argument in the OP goes similar to this:

Suppose all dogs are the same colour. Let all dogs be  green. This shows all dogs are the same colour.

Wich is obviously wrong.
You can only assume something if it is without loss of generality, or if it is the opposite of what you want to prove (in wich case you are proving by contradiction). Neither applies to your case, since you assumption loses all generality and is exactly what you want to prove.
