Pythagoras theorem $$a^2+b^2=c^2$$
Let $(p,q,r)$ and $(x,y,z)$ are any two pair of Pythagoras triples.
We have $px+qy+rz=X^2$.
For examples:
$(3,4,5)$ and $(20,21,29)$ then $\color{blue}{3\times20+4\times21+5\times29=17^2}$
$(8,15,17)$ and $(9,40,41)$ then $\color{red}{8\times9+15\times41+17\times41=37^2}$
$(48,55,73)$ and $(65,72,97)$ then $\color{blue}{48\times65+55\times72+73\times97=119^2}$
$(19,180,181)$ and $(28,195,197)$ then $\color{red}{19\times28+180\times195+181\times197=267^2}$
$(21,220,221)$ and $(120,209,241)$ then $\color{blue}{21\times120+220\times209+221\times241=319^2}$
$(69,260,269)$ and $(68,285,293)$ then $\color{red}{69\times68+260\times285+269\times293=397^2}$
Noticing that $X$ seems to be of ONLY of prime number.
Here I found an eample proof to be false, $X$ is not always a prime number!
$(20,21,29)$ and $(11,60,61)$ then $\color{green}{20\times11+21\times60+29\times61=57^2=(3\times17)^2}$
My question is: Are there infinitely many of this equation, $px+qy+rz=X^2$?
There are infinitely many primitive sextuples as requested. All that is necessary is to pair the odd legs times the even legs. Here are integers $a,b,c,d,$ $a > b \geq 1,$ $\gcd(a,b) = 1,$ $a+b$ is odd, $c > d \geq 1,$ $\gcd(c,d) = 1,$ $c+d$ is odd. $$ a^2 - b^2, \; \; 2ab, \; \; a^2 + b^2 $$ $$ 2cd, \; \; c^2 - d^2, \; \; c^2 + d^2 $$ $$ \color{red}{ 2cd \left(a^2 - b^2\right) + 2 a b \left( c^2 - d^2\right) + \left(a^2 + b^2\right) \left( c^2 + d^2\right) = \left( a (c + d) + b (c - d) \right)^2} $$
The main ordering in the computer run below is that $r < z.$ I think I finally got it to list everything with the bounds I put on $a,b,c,d.$ Given any two distinct primitive triples, your construction can be made to work, just match $p \equiv y \pmod 2,$ $q \equiv x \pmod 2$ in order to write the multiplications as you wanted.