Proving that a sum is even if and only if the stated conditions hold I am so frustrated about this question and I have been working on it the whole day before I decided to post this question. 
Let $w,x,y,z$ be nonnegative integers. 
$$w+x+y=z$$ 
Prove that $z$ is even iff $w$, $x$ and $y$ are even or exactly one of $w, x, y$ is even.  
What I have so far: 
$w, x, y$ can possibly be odds but if there are two odd numbers then the addition will be even. 
Also there is a hint saying $x$ is even iff $x^2$ is even. 
Thank you so much. 
 A: You want to start by writing down the definition of an even number and odd number.
Then you want to prove that only when one of w, x or y is even, or when all three are even, is z even. You can substitute in your definitions of even and odd numbers into w, x and y, and then test each combination of even/odd numbers (remember to use different variables for each w, x and y so that they are not the same number).
For example:
An even number x is defined as x = 2s, where s is an integer.
An odd number y is defined as y = 2k + 1, where k is an integer.
Case 1: One even number, two odd numbers
Let w = 2s, x = 2k + 1, y = 2j + 1 (Where s, k and j are integers)
w + x + y = z
2s + 2k + 1 + 2j + 1 = 2(s + k + j + 1) = z
Which is even, because z is divisible by 2.
The other cases should be easy to do if you follow the same method.
A: Just do it by cases.
There are either zero, one, two or three even numbers and the rest are odd.
Test them all.
1) They are all odd.
Then:
odd + odd + odd =
(odd + odd) + odd = 
even[1] + odd = odd[2].
$z$ can not be even if $x,y,w$ are all odd.
2) One is even the rest are odd.
You have
even + odd + odd = 
(even + odd) + odd = 
odd + odd = even.
So $z$ can be even if exactly one of $x,y,w$ is even.
3) two are even and the third is odd.
Then 
even + even + odd =
(even + even) + odd =
even[3] + odd = odd.
So $z$ can not be even if two are even.
4) All three are even.
even + even + even =
(even + even) + even=
even + even = even.
$z$ can be even if all three are even.
That's all possibilities.
.....
[1]  odd + odd = even
Proof.  Let $a = 2n + 1 $ and $b = 2m + 1$ be two odd numbers. Then $a + b = 2n + 2m + 2 = 2(n + m + 1)$
[2] odd + even = odd
Proof:  ... I don't really have to prove these do I?
....
