Some Pythagorean triples proofs I'm studying about number theory by myself. So I'm sorry if the question seems to be unclear.
I just want to know how to prove that one member of a Pythagorean triple is always divisible by 5 and that the area of an integer-sided right-angled triangle is always divisible by 6.
I searched for similar questions but didn't find an understandable proof for me.
 A: We are considering integer solutions to
$$
a^2+b^2=c^2
$$

Let us consider the equation $a^2+b^2=c^2$ modulo $2$:
$$
\begin{array}{c|cc}
+&0^2&1^2\\
\hline
0^2&0^2&1^2\\
1^2&1^2&0^2
\end{array}
$$
we see that in all four cases at least one of the three numbers is zero modulo $2$. So at least one in the triple is divisible by $2$.

Modulo $3$ we have $1^2=2^2$ so the same table applies. At least one is divisible by $3$.

Modulo $4$ we have $0^2=2^2$ and $1^2=3^2$ so again the same table applies. At least one is divisible by $4$.

Modulo $5$ we have $1^2=4^2=1$ and $2^2=3^2=4$ and so
$$
\begin{array}{c|ccc}
+&0^2&1^2&2^2\\
\hline
0^2&0^2&1^2&2^2\\
1^2&1^2&\times&0^2\\
2^2&2^2&0^2&\times
\end{array}
$$
where $\times$ indicates that result cannot be a square, since $1^2+1^2=2$ and $2^2+2^2=3$ are not values of squares modulo $5$. They are quadratic non-residues mod $5$.
All $7$ cases that are actually possible contain at least one zero. So at least one of the three numbers is divisible by $5$.

Finally, regarding the area being divisible by $6$, we see from the formulas for generating the triples
$$
(a,b,c)=(m^2-n^2,2mn,m^2+n^2)
$$
that the area must be
$$
T=\tfrac12 ab=(m^2-n^2)mn
$$
and if neither $m$ nor $n$ is divisible by $2$ we have $m=n=1$ modulo $2$ and therefore $m^2-n^2$ must be divisible by $2$.
If neither $m$ nor $n$ is divisible by $3$, we see that either $m,n\in\{1,2\}$ modulo $3$ so that $m^2-n^2=0$ modulo $3$. So indeed the area is divisible by both $2$ and $3$, and hence by $6$.
A: From the link:
"The case to consider is m = ±1 (mod 5) or m = ±2 (mod 5). And the same is true for n."
Here, we want to ultimately look at the squares, because we have already dealt with m or n being divisible by 5 on the previous condition.
"It then follows that both $m^2$ and $n^2$ (edited here) may only be 1 or 4 modulo 5."
Just do the multiplication and reduce modulo five for all of the cases.
"If they are equal modulo 5, then m2 - n2 = 0 (mod 5). Otherwise, m2 + n2 = 0 (mod 5)."
When the first statement is established and we have reduced to only the case where the squares of the elements can only be 1 or 4 mod 5, then either they are equal or the aren't.  If they are, then the subtraction case applies; if not, then the addition one does.
If there is still some confusion, it might be worth looking up modular arithmetic.
A: 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.
