Pythagorean Theorem Special Case I was hoping you could help me prove the following:
for the case of all $3, 4, 5$ special right triangles, like $6, 8, 10$ and $3\sqrt2, 4\sqrt2, 5\sqrt2$, etc., the relationship 
$a/3+b=c$
is always true.
Or can you find a counterexample? I believe it to be true always but am looking for a formal proof.
It is much easier to work with than
$a^2+b^2=c^2.$
Please prove me right :)

Consider the two cases:
$$1) \ \begin{cases}a=m^2-n^2\\ b=2mn \\ c=m^2+n^2\end{cases} \qquad 2) \ \begin{cases}a=2mn\\ b= m^2-n^2\\ c=m^2+n^2\end{cases}$$
For case 1:
$$\frac{a}{3}+b=c \iff \frac{m^2-n^2}{3}+2mn=m^2+n^2 \iff m^2+2n^2=3mn \iff \\
(m-n)^2=m(m-n) \iff m=2n\\
\begin{cases}a=3n^2\\ b=4n^2\\ c=5n^2\end{cases}$$
For case 2:
$$\frac{a}{3}+b=c \iff \frac{2mn}{3}+m^2-n^2=m^2+n^2 \iff mn=3n^2 \iff m=3n\\
\begin{cases}a=6n^2\\ b=8n^2\\ c=10n^2\end{cases}$$
Note: $n\in \mathbb R, n\ne 0$, in particular, for $n=\sqrt[4]{2}$, case 1 is $3\sqrt{2},4\sqrt{2},5\sqrt{2}$.
