As I said, I haven't learned about "modular arithmetic" yet. That's may be why I'm not getting your answers which seem to be obviously clear.
Despite this, I tried to prove it on my "Elementary" way and that was what I came up with.
"one member of a Pythagorean triple is always divisible by 5"
First, every integer (m, n) can be written on one of these forms (5k, 5k+1, 5k+2, 5k+3, 5k+4). We take the last four forms and square them. We will end only with two different forms(5k+1, 5k+4). Note that: (5k+9) can be rewritten to be (5k+4) with a different value of k.
If m^2 and n^2 can be written on the same form (5k+1, 5k+1) or (5k+4, 5k+4) then m^2 - n^2 should be divisible by five. If m^2 and n^2 are written on different forms (5k+1, 5k+4) then m^2 +n^2 should be divisible by 5 cause (4 + 1 = 5).
Now we have done for the last four forms and still have the first form which is (5k). Simply, if m can be written on the form (5k) and n is written on any other form then m * n should be divisible by five, Therefore, 2mn also should be divisible by 5.
Note: 1) My English isn't perfect so I'm sorry if there are some mistakes.
2) The proof is clear for me now and I can understand the main parts of your proofs. However, I won't close the question right now. So if you have any comments on what I have written, kindly, inform me.