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

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.
