Find three numbers such that the sum of all three is a square and the sum of any two is a square It seems like it should be a well known problem.  Ive read that Diophantine himself first posed it.  I couldnt find a solution when i researched for one.
It asks to find ordered triples $(x,y,z) $ such that 
$$x+y+z=a^2 $$
$$x+y=b^2$$
$$x+z=c^2$$
$$y+z=d^2$$
As an example  (41, 80, 320).  Any guidance is appreciated.  The ideal would be some kind of parametric solution for x, y, and z
 A: Assume that $$2a^2 = u^2+v^2+w^2$$
holds for some $(a,u,v,w)\in\mathbb{N}^4$ and the system
$$ \left\{\begin{array}{rcl}y+z&=& u^2 \\ x+z&=&v^2 \\ x+y&=&w^2\end{array}\right. $$
has integer solutions. Then we are done, since $x+y+z=a^2$.
If $u,v,w$ are even numbers the system clearly has integer solutions ($x=\frac{v^2+w^2-u^2}{2}$ and so on), so every triple $(\alpha,\beta,\gamma)\in\mathbb{N}^3$ such that $2(\alpha^2+\beta^2+\gamma^2)$ is a square leads to a solution of the original problem. But, wait. If $2(\alpha^2+\beta^2+\gamma^2)$ is a square it is an even square, i.e. a number of the form $4n^2$. In particular we get a solution of the original problem for every solution of the Diophantine equation $\alpha^2+\beta^2+\gamma^2 = 2n^2$.
For instance, $(\alpha,\beta,\gamma,n)=(0,1,7,5)$ leads to $2(10)^2 = 0^2+2^2+14^2$ and to the solution
$$ (x,y,z) = (-96,96,100). $$
The solution found by the OP, $(x,y,z)=(41,80,320)$, is associated with $11^2+19^2+20^2=2\cdot 21^2$. Another solution is $(x,y,z)=(-111,120,280)$, which is associated with $3^2+13^2+20^2=2\cdot 17^2$.
A: A family of solutions can be obtained from
\begin{eqnarray*}
x&=&48m^2+8m+1 \\
y&=&96m^2+16m \\
z&=&16m(6m+1)(6m^2+m-1)
\end{eqnarray*}
One can easily verify that 
\begin{eqnarray*}
x+y&=&(12m+1)^2 \\
x+z&=&(24m^2+4m-1)^2 \\
y+z&=&(4m(6m+1))^2 \\
x+y+z&=&(24m^2+4m+1)^2.
\end{eqnarray*}
A: For the General case of formula there.  https://artofproblemsolving.com/community/c3046h1172008_combinations_of_numbers_in_squares
The system of equations:            
$$\left\{\begin{aligned}&a+b=x^2\\&a+c=y^2\\&b+c=z^2\\&a+b+c=q^2\end{aligned}\right.$$
Solutions have the form:            
$$a=4t((2t-p)k^2+2(2t-p)^2k-2p^3+9tp^2-14pt^2+8t^3)$$
$$b=4(p^2-3pt+2t^2)k^2+8(4t^3-8pt^2+5tp^2-p^3)k+$$      
$$+4(p^4-6tp^3+15p^2t^2-18pt^3+8t^4)$$            
$$c=k^4+4(2t-p)k^3+4(p^2-3pt+3t^2)k^2-8(p^2-3pt+2t^2)tk+$$      
$$+4t(2p^3-9tp^2+14pt^2-7t^3)$$            
$$x=2(2t-p)(k+2t-p)$$            
$$y=k^2+2(2t-p)k+2t^2$$            
$$z=k^2+2(2t-p)k+2(t-p)^2$$            
$$q=k^2+2(2t-p)k+6t^2-6tp+2p^2$$            
$k,t,p$ - integers asked us.
