This is my question with my thoughts about how to approach it:
For all integers m and n, 4 divides (m2-n2) if and only if m and n are both even or m and n are both odd.
So obviously, I have to prove it from both directions.
--> For all integers m and n, 4 divides (m2-n2) if m and n are both even or both odd.
<-- If m and n are both even or both odd, then 4 divides (m2-n2) for all integers m and n.
Assuming these directions are correct, I would attempt to prove this the following way. Let m and n be even integers, say 2m and 2n. By substitution it is 4m2-4n2 and by algebra that is equal to 4(m2-n2). m2-n2 is an integer by definiton of integers while the 4 in front of the integer makes the number divisible by 4. That's for the evens. For the odds 2m+1 and 2n+1, by substitution and algebra I have 4(m2+m-n2-n). Again, that is an integer with a factor of 4 making it divisible by 4.
In terms of the second direction...in past biconditionals I remember taking the contrapositive of the 2nd. So...
There exists integers m and n that are both even or both odd such that 4 does not divide (m2-n2) for all integers m and n.
At this point, I am stuck. I might have done the contrapositive wrong because I think the "or" needs to change to "and". But I could be wrong on that...