Three pythagorean triples Are there any solutions for $a, b, c$ such that:
$$a, b, c \in \Bbb N_1$$
$$\sqrt{a^2+(b+c)^2} \in \Bbb N_1$$
$$\sqrt{b^2+(a+c)^2} \in \Bbb N_1$$
$$\sqrt{c^2+(a+b)^2} \in \Bbb N_1$$
 A: Here is a complete parametrization of all rational solutions to $\sqrt{a^2+(b+c)^2}\in\mathbb{Q}$, $\sqrt{b^2+(c+a)^2}\in\mathbb{Q}$, and $\sqrt{c^2+(a+b)^2}\in\mathbb{Q}$, where $a,b,c\in\mathbb{Q}$.  If $p,q,r\in\mathbb{Q}_{\geq 0}$ be such that $$\frac{2p}{1+2p-p^2}+\frac{2q}{1+2q-q^2}+\frac{2r}{1+2r-r^2}=1\,,\tag{*}$$
then $(a,b,c)=\left(\frac{2p}{1+2p-p^2}x,\frac{2q}{1+2q-q^2}x,\frac{2r}{1+2r-r^2}x\right)$ for some $x\in\mathbb{Q}$ (namely, $x=a+b+c$).  All rational solutions $(a,b,c)$ are of this form.  There exists a positive integer solution $(a,b,c)$ associated to $(p,q,r)$ iff $0< p,q,r<1+\sqrt{2}$.
Frankly, I don't know if solving (*) is any easier than using the method mentioned by Tito Piezas III, but at least, there is one equation to be solved now, and with only $3$ rational variables.  (However, if you try to write $p=\frac{m_1}{n_1}$, $q=\frac{m_2}{n_2}$, $r=\frac{m_3}{n_3}$, where $m_i,n_i\in\mathbb{Z}$ for $i=1,2,3$, then you will end up with $6$ variables, but the method mentioned by Tito Piezas III can reduce the number of variables to $5$.)  There may be an algebraic-geometry/algebraic-number-theory method to solve (*), but I'm not so knowledgeable in these fields.  Here is an example: $(a,b,c)=(108,357,368)$ is given by $(p,q,r,x)=\left(\frac{2}{27},\frac{1}{3},\frac{8}{23},833\right)$, where $\frac{2p}{1+2p-p^2}=\frac{108}{833}$, $\frac{2q}{1+2q-q^2}=\frac{3}{7}=\frac{357}{833}$, and $\frac{2r}{1+2r-r^2}=\frac{368}{833}$.
A: $(108,357,368),(216,714,736)$ and its permutations are the unique solutions of your equation in positive integers below $1000$ (found using a computer search). Of course the latter triple is just a constant multiple of the former one. By multiplying by a constant we can find infinite number of solutions, but so far I have found only one primitive solution (i.e. with coprime numbers).
Edit: I have extended the range and I found one more primitive solution: $(564,748,1425)$. Right now I have looked at all triples with $a\leq b\leq c,a\leq 1000,b\leq 1500,c\leq 2000$.
Edit 2: Following Batominovski's suggestion on how to speed up calculation, I have ran a search with range $a\leq 3000$ odd, $b\leq c,b\leq 3000,c\leq 4000$ divisible by $4$ and here is the list of solutions in that range, in order of apprearance:
$$(357,108,368)\\
(975,348,2380)\\
(1071,324,1104)\\
(1425,564,748)\\
(1785,540,1840)\\
(2499,756,2576)$$
Here is the code I have used (quite ugly, I know):
#include<Windows.h>
#include<conio.h>
#include<iostream>
#include<Windows.h>
#include<math.h>

using namespace std;

long int main(){
    for(double a=1;a<3000;a+=2){
    for(double b=4;b<3000;b+=4){
    for(double c=b;c<4000;c+=4){
        if(floor(sqrt(a*a+(b+c)*(b+c)))==sqrt(a*a+(b+c)*(b+c))&&floor(sqrt(b*b+(a+c)*(a+c)))==sqrt(b*b+(a+c)*(a+c))&&floor(sqrt(c*c+(b+a)*(b+a)))==sqrt(c*c+(b+a)*(b+a))){
            cout<<a<<"|"<<b<<"|"<<c<<endl;
        }
    }
    }
    }
    cout<<"done";
    return getch();
}

