Pythagorean Chaos I

If $n\equiv3 \pmod4$ , it can never be the hypotenuse in a pythagorean triangle.

This statement is false (not this one, that one, let's avoid paradoxes :) ). An easy counter example is $15^2=9^2+12^2$ . It is true, however, that you can't write a number that's one less than a multiple of $4$ as a sum of two squares.

If $z$ is a complex number with distinct positive integer real and imaginary parts, then there is a pythagorean triangle with side lengths $|Im(z^2)|$ , $|Re(z^2)|$ , and $|z^2|$ .

Correct: When we square a complex number, we square it's magnitude, and since the magnitude of a complex number is just $\sqrt{(Re(z))^2+(Im(z))^2}$ , when we square it, it becomes an integer. it is also equal to $\sqrt{(Re(z^2))^2+(Im(z^2))^2}$ , of course. So we have $|z^2|=\sqrt{(Re(z^2))^2+(Im(z^2))^2}$ Which gives $|z^2|^2=(Re(z^2))^2+(Im(z^2))^2$

If $(a,b,c)$ is a pythagorean triple, it can be rewritten as $(Re(z^2),Im(z^2),|z^2|)$ , for some complex number $z$ with integer real and imaginary parts.

Incorrect: as a counter-example, you can't write $6+8i$ that way, in other words, $\sqrt{6+8i}$ isn't a gaussian integer

If for a $z$ with distinct positive integer real and imaginary parts $(|Re(z^2)|,|Im(z^2)|,|z^2|)$ isn't a primitive pythagorean triple, $(\frac{|Re(z^2)|}{2},\frac{|Im(z^2)|}{2},\frac{|z^2|}{2})$ is.

Incorrect: take $z=4+2i$ , when we square it, we get $12+16i$ , which is not primitive, but $6+8i$ isn't either.

If $(a,b,c)$ is a primitive pythtagorian triple, then there is exactly $1$ even number, and it is either $a$ or $b$ .

Correct: If it were possible to make a primitive pythagorean triple by having a triangle with two odd legs, we'd have: $(2a)^2=(2b+1)^2+(2c+1)^2)$ . Expanding gives: $4a^2=4b^2+4b+4c^2+4c+2$ . Looking at it $\mod4$ gives $4\equiv2$ , which is a contradiction.