Geometric proof that a Pythagorean triple has a number divisible by $3$?

Question

Can I infer from

“If $a,b \in \mathbb{N}$ such that $(a,3)=(b,3)=1$ then $a^2+b^2\neq c^2$ for any $c\in\mathbb{N}$”

that any pythagorean triple has a number which is divisible by $3$?

If so, how can this relationship be proved geometrically?

Answer 1

Simple graphical solution to the problem.

Red region: Has unit squares divisible by 3.

Green region: All unit squares together are a multiple of 3.

Yellow squares: Left out square(s).

If you have two squares of type 1, you will have 2 yellow squares, which don't fit into any type of square.

Answer 2

Primitive solution of Pythagorean triple is given below.

$(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$.

Assuming $gcd(a,b)=1$, then $ab = 2mn(m^2-n^2) \equiv 0 \pmod{3}$ (Fermat's little theorem)

Hence $a$ or $b$ is divisible by $3$.

Thus, if $a^2+b^2=c^2$ then $a$ or $b$ is divisible by $3$.

Answer 3

A slightly detailed(irrelevant) approach: If any of, m, n are divisible by 3, then we are done as $\ (a,b,c)=({ m }^{ 2 }-{ n }^{ 2 },\quad 2mn,\quad { m }^{ 2 }+{ n }^{ 2 }) $ and $\ 3|b $. Hence, assume both m and n aren't multiples of 3. Therefore $$\ m\equiv 1,2\quad (mod\quad3)\\ n\equiv 1,2\quad (mod\quad3)$$ This implies that $$\ { m }^{ 2 }\equiv { n }^{ 2 }\equiv 1(mod\quad 3) $$ Hence $$\ { m }^{ 2 }-{ n }^{ 2 }=a\equiv 0(mod\quad 3)$$

Answer 4

If in the Pythagorean triangle $(a,b,c),$ where $c$ is the hypotenuse, we have $c=3m\pm1$ and $a=3n\pm1,$ then $$(3\,|\, (c^2-1)) \text{ and } (3\,|\, (a^2-1))\Rightarrow 3\,|\,(c^2-a^2)\Rightarrow 3\,|\,b^2\Rightarrow 3\,|\,b.$$

Geometrically.

If in the Pythagorean triangle the hypotenuse and leg do not divide to $3$, then the squares difference divides to $3$.