Integer Pythagoras

Suppose our triple is $a,b,c$ ; we know that $a^2 + b^2 = c^2$ .

Suppose none of $a,b,c$ is divisible by $3$ . Notice that all squares are congruent to $0,1 \pmod 3$ ; moreover, if a number $n$ is not divisible by $3$ , then $n^2$ is not divisible by $3$ , so it must be congruent to $1 \pmod 3$ . But adding together, $a^2 + b^2 \equiv 1 + 1 \equiv 2 \not\equiv 1 \equiv c \pmod 3$ , contradiction.

We can do a similar method for divisibility by $5$ . We'll get $0,1,4 \pmod 5$ , and throwing away $0$ , we have $1,4 \pmod 5$ . Thus by checking all four cases, $a^2 + b^2 \equiv 2, 5, 8 \equiv 0, 2, 3 \pmod 5$ , which will never be congruent to $c^2$ .

For divisibility by $4$ , this is a little tougher. We cannot just take quadratic residues modulo $4$ , because $2 \not\equiv 0 \pmod 4$ but $2^2 \equiv 0 \pmod 4$ . Instead, we need to take modulo $8$ ; if $n$ is not divisible by $4$ , then $n^2$ is not divisible by $8$ . Also, the residues of squares modulo $8$ are $0,1,4$ ; throwing $0$ , we have $1,4 \pmod 8$ . Thus we use the same method as with modulo $5$ , getting $c^2 \equiv 2,5,8 \equiv 0,2,5 \pmod 8$ , which is also impossible.

Thus we have derived a contradiction from assuming that all side lengths are not divisible by any of $3,4,5$ . Thus all three statements are correct .

We know that Pythogorean triples can only be of the form $k(m^{2} - n^{2}), 2kmn$ and $k(m^{2}+n^{2})$ .

Checking the divisibility with respect to different numbers given can lead us to the conclusion :

$\textbf{Case 1.}$ First we check the divisibility by $3$ . If $m \equiv n \pmod{3}$ ,then $3$ divides $m^{2} - n^{2}$ .

Also if any of $m$ or $n$ is divisible by $3$ we are done since $3 | 2kmn$ .

So the only case left to prove is $m\equiv 2 \pmod{3}$ and $n\equiv 1 \pmod{3}$ (Without the loss of generality).

In this case clearly $m^{2} - n^{2} \equiv 4-1 \equiv 3 \equiv 0 \pmod{3}$ .

hence the sides of the triangle are always divisible by 3.

The cases for $4$ and $5$ can also be nailed down using similar arguments.

This triangle is a right triangle.so First let us assign common Pythagorean Triples $3,4,5$ And we know that the numbers $3,4,5$ meet the criteria of been divisible by 3,4 and 5 thus the answer is $\boxed{1,2.3}$

Just filter the equation $x^2+y^2=z^2$ with the quadratic residues modulo $3,4,5$ and see that all of the statements are correct.

we could have a triangle with 3²+4²=5² then one side is divisible by 3 the other by 4 and the last one by 5 so all the statements are correct

I'm not sure how to prove it, but looking through the first 50 or so Pythagorean Triples, we can see that all of them have terms divisible by three, four, and five.