Just to be clear, I call an integer $c$ a 'primitive Pythagorean hypotenuse' if there exist coprime integers $a$ and $b$ satisfying $a^2+b^2 = c^2$. I noticed that the set of such primitive Pythagorean hypotenuses seems to form a closed set under multiplication (and indeed this is backed up by this Mathworld page -- see equations (25)-(26)). However, I can't find a proof of it (the link above provides a book reference but I don't have access to it), or come up with one myself.
Some thoughts: Euler showed that $c$ is a primitive Pythagorean triple iff it is odd and can be written as the sum of two squares $c = m^2+n^2$ for some coprime integers $m$ and $n$. It is straightforward to show that the product of a sum of two squares is also a sum of two squares (and hence the set of all Pythagorean hypotenuses is closed under multiplication), but I'm having trouble with the primitive/coprime part. In other words I can show that if $c_1 = m_1^2 + n_1^2$ and $c_2 = m_2^2 + n_2^2$ then there exist $m_3$ and $n_3$ such that $c_1c_2 = m_3^2 + n_3^2$, but I can't show that if $gcd(m_1,n_1)=gcd(m_2,n_2)=1$ then it is possible to choose $m_3$ and $n_3$ such that $gcd(m_3,n_3)=1$. Any hints or references would be appreciated!
Edit (in response to individ's comments): Note that the formulae $$ m_3 = m_1 n_2 - n_1 m_2 \ , \ n_3 = m_1 m_2 + n_1 n_2 $$ and $$ m_3 = m_1 n_2 + n_1 m_2 \ , \ n_3 = m_1 m_2 - n_1 n_2 $$ do not always produce coprime $m_3$ and $n_3$. The earliest example of this I could find is $$ m_1 = 8, n_1 = 1, m_2 = 31, n_2 =92 $$ for which the first formula above gives $m_3 = 705, n_3 = 340$ (both divisible by 5), and the second formula above gives $m_3=767, n_3=156$ (both divisible by 13). However $m_3^2 + n_3^2 = 612,625$ is still the hypotenuse of a primitive triple because $$ 199176^2 + 579343^2 = 612625^2 $$ and $gcd(199176,579343)=1$. (Sorry for the rather ugly example, but it was the simplest I could find (maybe because I don't know how to search efficiently!).)