Pythagorean Chaos II

If $(a,b,c)$ is a pythagorean tripple, then there is some gaussian integer $z$ and a real number $k$ such that $z^2=k(a+bi)$

Correct: Sonsider the number $z=|a+bi|+a+bi$ . We know it's a gaussian integer because |a+bi| is an integer, and since $arg(z+|z|)=\frac{arg(z)}{2}$ and $arg(z^2)=2arg(z)$ , we see that $arg((|a+bi|+a+bi)^2)=arg(a+bi)$ . They have the same argument, so their ratio is real.

If $Im(z)$ and $Re(z)$ are coprime integers, and $(|Re(z^2)|,|Im(z^2)|,|z^2|)$ isn't a primitive pythagorean triple, then $(\frac{|Re(z^2)|}{2},\frac{|Im(z^2)|}{2},\frac{|z^2|}{2})$ is.

Correct: say $z=a+bi$ , then $(Re(z^2),Im(z^2),|z^2|)=(a^2-b^2,2ab,a^2+b^2)$ . Suppose some prime $p>2$ divides $2ab$ then, $p$ divides either $a$ or $b$ . but this means $p$ doesnt divide $a^2-b^2$ . So the highest possible value of $p$ is $2$ .

If $Im(z)$ and $Re(z)$ are coprime integers and only one of them is odd, $(|Re(z^2)|,|Im(z^2)|,|z^2|)$ is a primitive pythagorean triple.

Correct: In the previous statement, the only way for $2$ to be a comon factor of $Re(z^2)$ and $Im(z^2)$ is for both $Re(z)$ and $Im(z)$ to be odd.

If $n\equiv3\pmod4$ then $n$ can never be the hypotenuse of a primitive pythagorean triangle.

Correct: if $z$ is a primitive pythagorean number with $n=|z|$ (a pythagorean number with coprime real and imaginary parts), the previous statements show that it is either the square of a gaussian integer $g$ with coprime $Re(g)$ and $Im(g)$ , or exactly half the square of such a gaussian integer. Let's consider the two cases:

Case 1: $z=g^2$ for some gaussian integer $g=a+bi$ with $\gcd(a,b)=1$ In this case, for $|z|$ to be odd (it's odd because $|z|\equiv3\pmod4$ , it has to be that only one of $a$ and $b$ is odd. Without losing generality, say $a=2m+1$ and $b=2n$ for integers $m$ and $n$ . Then, $|z|=(2m+1)^2+(2n)^2=4m^2+4m+1+4n^2$ looking at it $\mod4$ we get $3\equiv1$ .

Case 2: $z=\frac{g^2}{2}$ for some gaussian integer $g=a+bi$ with $\gcd(a,b)=1$ As we saw earlier, for this to happen, it must be that $a$ and $b$ are odd. w.l.o.g. $a=2m+1$ and $b=2n+1$ . $|z|=\frac{4m^2+4m+1+4n^2+4n+1}{2}=2m^2+2m+2n^2+2n+1=2(m^2+m)+2(n^2+n)+1$ since $x$ and $x^2$ have the same parity, both $m^2+m$ and $n^2+n$ are even. so we can say w.l.o.g that $|z|=2(m^2+m)+2(n^2+n)+1=4c+4d+1$ . Finally, looking at it $\mod4$ again gives $3\equiv1$ .

I wonder if there are simpler, more intuitive proofs for this, or at least a more direct one, If someone knows one, writing it as a solution or as a comment to this one would be highly apreciated

If $z$ is a pyhtagorean number, then $z^3$ is too.

Correct: $|z|$ is an integer, then so is $|z^3|$ . The real and imaginary parts are, of course, also integers.

If $z^3$ is a pythagorean number, then $z$ is too.

Wrong: $4+3i$ doesn't have gaussian integer cube root, for example.