Need help with proof of sets of integers

1.From the problem statement
Let Z⁺ be the set of all positive integers; that is,
Z⁺ = {1,2,3,...}
Define Z⁺ x Z⁺ = {(a₁, a₂) : a₁ is an element of Z⁺ and a₂ is an element of Z⁺ }



2. If S is contained in Z⁺ and |S| >=3, prove that there exist distinct x,y that are elements of S such that x+y is even.



3. I need some help on where to start as I'm quite lost.
I can see how this would be true. for example the set S = {1,2,3,4}
The pairs (1,3) and (2,4) both satisfy the condition of x+y being even.


the are extensions of this homework problem as well, but I'm having trouble with the first bit...

 

how does this sound?
there is a set S with three elements that are integers. Assume that any two of these elements are added together and always produce an odd value. (ie x₁+x₂ or x₁+x₃ or x₂+x₃ is always odd).

assuming that all three elements of S have even values, take the sum of any two of these elements.
k is even which makes k+2 even.
k+k+2 = 2k+2. This value is even
k+2 is even which makes k+2+2 even also.

k+2+(k+2+2) = 2k+6. This value is even.

assuming that all three elements of S have odd values, take the sum of any two of these elements.
k is odd which makes k+2 odd.
k+k+2 = 2k+2. This value is even
k+2 is even which makes k+2+2 even also.

k+2+(k+2+2) = 2k+6. This value is even.

next assume that the set of S has elements some of which are even and some of which are odd.
if two of the values are even, sum those two values. the result will be an even number.
if two of the values are odd, sum those two values. the result will be an even number.

Therefore, the original assumption that any two elements of S that are summed will produce an odd value is false.

how is that?
the third criteria seems weak.

 

ok
so it's not really a proof by contradiction then.
is it still sufficient though?
I still feel like the wording could be better.