Pythagorean Chaos III

First, let's write everything in terms of $a$ and $b$ :

$Re(z^2)=a^2-b^2$

$Im(z^2)=2ab$

$|z^2|=a^2+b^2$

If $a$ and $b$ share a common factor $k$ , then it is clear that $Re(z^2)$ , $Im(z^2)$ and $|z^2|$ will share a common factor $k^2$ . so $a$ and $b$ must be coprime.

The probability of this happening is $\frac{6}{{\pi}^2}$ . Why?

Now, from the second and third statements in Pythagorean Chaos II , we see that $Re(z^2)$ , $Im(z^2)$ and $|z^2|$ will be coprime iff one of $a$ and $b$ is odd.

The probability of this happening given two coprime integers is $\frac{2}{3}$ . An intuitive way to see why (more formal reasoning in the comments) is noticing that a coprime pair could be one of three cases: even,odd, odd,even, and odd,odd and for every odd,odd pair, you could multiply each number by $2$ separately to create an even,odd, or an odd,even pair.

So the probability we want is $P( \text{Two random integers are coprime}) \times P(\text{one of them is even})= \large \frac{6}{{\pi}^2} \cdot \frac{2}{3} = \frac{4}{{\pi}^2}$

And the answer is $4$ .