Is this a valid proof? Discrete Mathematics I feel like there are many solutions for some proofs, and I want to make sure I'm still getting everything correct. This is the problem:
$$m^2 = n^2 (P)\quad\text{if and only if}\quad m = n (Q)\quad\text{or}\quad m = -n (Z)$$
So I need to prove $P \Rightarrow (Q \lor Z)$ and $(Q \lor Z) \Rightarrow P$ because this is a bi-conditional statement, yes? This is my proof.
Proving $P \Rightarrow (Q \lor Z)$:
Let's assume $m$ doesn't equal $n$ (proving by contra-position). It is said that $m^2 = n^2$. Take the square root of both sides and you are left with $m = n$. However, we are assuming $m$ doesn't equal $n$ so there is a contradiction. 
And I pretty much did the reverse for $(Q \lor Z) \Rightarrow P$ by squaring both sides to reach the contradiction.
Is this a valid proof or is something going over my head?
 A: 
So I need to prove $P \to (Q \vee Z)$ and $(Q \vee Z) \to P$ because this is a bi-conditional statement, yes?

Yes.

Let's assume $m$ doesn't equal $n$ (proving by contra-position). It is said that $m^2 = n^2$. Take the square root of both sides and you are left with $m = n$. However, we are assuming $m$ doesn't equal $n$ so there is a contradiction.

There are a couple of flaws here.  First, the outline of what you did is "We want to prove $R$.  So assume $\neg R$.  Then, blah blah blah, $R$.  But we assumed $\neg R$.  This is a contradiction."  But you only contradicted your original assumption $\neg R$.  In the middle, you proved $R$.  So in proofs like that you don't need the proof-by-contradiction framework.  You just did a direct proof.
Or did you?  You said $m^2 = n^2 \implies m = n$ “by taking the square root of both sides.”  But this isn't valid.  For $2^2 = (-2)^2$ while $2 \neq -2$. So something went wrong, and it was that you lost a solution by assuming $m$ and $n$ were both positive. 
With nonlinear algebraic equations, it's often safer to set the equation equal to zero and factor.  In this case:
$$
    m^2 = n^2 \implies m^2 - n^2 = 0 \implies (m-n)(m+n) = 0
$$ 
Two numbers have a product of zero if and only if one of them is zero.
So either $m-n = 0$ (in which case $m=n$) or $m+n=0$ (in which case $m=-n$).
A: Others have pointed out the flaws in your proof.
Hint for a correct proof: Show that $$(n-m)(n+m)=n^2-m^2$$
So if $n^2=m^2$ then what?
Bigger hint: You need to know (or prove) the rule that $ab=0$ if and only if $a=0$ or $b=0$.
