Hi all,

This is my question with my thoughts about how to approach it:

For all integers m and n, 4 divides (m^{2}-n^{2}) 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 (m^{2}-n^{2}) if m and n are both even or both odd.

<-- If m and n are both even or both odd, then 4 divides (m^{2}-n^{2}) 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 4m^{2}-4n^{2}and by algebra that is equal to 4(m^{2}-n^{2}). m^{2}-n^{2}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(m^{2}+m-n^{2}-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 (m^{2}-n^{2}) 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...