Pythagorean Theorem Special Case

Question

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 :)

Answer 1

It's true! If $a=3s, b=4s, $ and $c=5s$, then $a/3+b=s+4s=5s=c$.

To avoid misunderstanding, it should be noted that there are Pythagorean triples not of this form;

one such example is $5, 12, 13$.

Answer 2

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 $3\sqrt{2},4\sqrt{2},5\sqrt{2}$.

Answer 3

Here is a function that may help you find the triples you want. I use the negative from the radical because the positive would yield $n=m$ and that would result in trivial triples such as $0,8,8$.

$$\frac{m^2-n^2}{3}+2mn=(m^2+n^2)$$ $$m^2-n^2+6mn=3m^2+3n^2)$$ $$4n^2-6mn+2m^2=0=2n^2-3mn+m^2$$

$$\text{For the }2^{nd}\text{ degree equation above }\quad a=2\quad b=-3m\quad c=m^2$$

$$\text{Solving for n we get: }\quad n=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{3m\pm\sqrt{9m^2-4*2*m^2}}{2*2}=\frac{3m\pm\sqrt{m^2}}{4}=\frac{m}{2}$$

You can generate a triple whenever $n=\frac{m}{2}$ is an integer so $m$ can be any even number. Here are samples with $m$ ranging from $2$ to $26$:

$$3,4,5\quad 12,16,20\quad 27,36,45\quad 48,64,80\quad 75,100,125\quad 108,144,180\quad 147,196,245\quad 192,256,320\quad 243,324,405\quad 300,400,500\quad 363,484,605\quad 432,576,720\quad 507,676,845\quad$$

Note, using the $+$ side of the equation generates $m=n$ resulting in a trivial triple where $A=0$.