TITLE: Three pythagorean triples
QUESTION [10 upvotes]: Are there any solutions for $a, b, c$ such that:
$$a, b, c \in \Bbb N_1$$
$$\sqrt{a^2+(b+c)^2} \in \Bbb N_1$$
$$\sqrt{b^2+(a+c)^2} \in \Bbb N_1$$
$$\sqrt{c^2+(a+b)^2} \in \Bbb N_1$$

REPLY [2 votes]: 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 $\frac{2r}{1+2r-r^2}=\frac{368}{833}$.