<s>[INST] 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.
 [/INST] 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 $6$.
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