Let $\omega(n)$ be the number of prime factor of $n$. Out of the first $9.2 \times 10^9$ Pythagorean triplets $a^2 + b^2 = c^2$ (in ascending order of the hypotenuse) I found about $1.03 \times 10^7$ triplets which satisfied
$$ \omega(a)^2 + \omega(b)^2 = \omega(c)^2 $$
Each of these $1.03 \times 10^7$ solutions had one thing in common which is they are all the smallest Pythagorean triplet $(3,4,5)$. For example
$$ 4870421759^2 + 12656832000^2 = 13561578241^2 \\ \omega(4870421759)^2 + \omega(12656832000)^2 = \omega(13561578241)^2 = 3^2 + 4^2 = 5^2 $$
Note: Theoretically, for $c > 6.5 \times 10^9$ we can have some $\omega(c) = 10$ so we expect to find the triplet $(6,8,10)$. The computation so far has reached $9.2 \times 10^9$ and I found only $(3,4,5)$ but no $(6,8,10)$. So even the second multiple of $(3,4,5)$ is must be rare.
Question: Is it true that if $a^2 + b^2 = c^2$ and $\omega(a)^2 + \omega(b)^2 = \omega(c)^2$ then the triplet $(\omega(a),\omega(b),\omega(c))$ is always some multiple of $(3,4,5)$?
The triple \begin{align*} a&=16736821055633718703\\ b&=54526056082320564460271199635604\\ c&=54526056082320564460271202204295 \end{align*} does the trick, with \begin{align*} a&=31\cdot 41\cdot 43\cdot 47\cdot 6515700430933\\ b&=2^2\cdot 3^2\cdot 7\cdot 11^3\cdot 19\cdot 23\cdot 31\cdot 41\cdot 43\cdot 47\cdot 44453\cdot 3257850215467\\ c&=5\cdot 13\cdot 17\cdot 29\cdot 31\cdot 37\cdot 41\cdot 43\cdot 47\cdot 109^2\cdot 677\cdot 1433\cdot 1553253593. \end{align*}
Here's the sketch as to how I found this:
To find a triple with $\omega$-values $(5,12,13)$, it suffices to find a triple with $\omega$-values $(1,8,9)$ and then multiply each element by four new primes. Since Pythagorean triples with one prime element are heavily restricted, this means we are looking for a triple of the form $$\left(p,\frac{p^2-1}2,\frac{p^2+1}2\right)$$ where $p^2\pm1$ both have a lot of factors. We are somewhat constrained in the prime factors of $p^2+1$, since we want there to be $9$ of them, all of which must be $1\bmod 4$ (besides $2$). We thus consider the following sets $$S_1=\{3,7,11,19,23\},\ S_2=\{5,13,17,29,37\}$$ of a few small primes (all elements of $S_2$ are $1\bmod 4$) and seek $p$ so $$p^2\equiv (-1)^{i+1}\bmod q,\ \ \forall q\in S_i.$$ We can find a few acceptable values of $p$ modulo the product of all primes in $S_1$ or $S_2$; from here I just checked a bunch until I found one that worked.