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
On
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 $3^2+13^2+20^2=2\cdot 17^2$.
On
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*}
On
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.
I did the complete solution for multiplier 3 here: http://math.stackexchange.com/questions/1964607/when-will-a-parametric-solution-generate-all-possible-solutions/1965805#1965805
Back to 2:
With Jack's variables, let $p+q+r+s$ be odd and $\gcd(p,q,r,s) = 1,$ then define $$ a = p^2 + q^2 + r^2 + s^2, $$ $$ u = 2(-pr + qr +ps+qs), $$ $$ v = p^2 - q^2 + r^2 - s^2 + 2 pq + 2rs, $$ $$ w = p^2 - q^2 - r^2 + s^2 - 2 pq + 2rs. $$ This gives $$ u^2 + v^2 + w^2 = 2 a^2 $$ and should give all primitive solutions. Checking, and then proving, that these are all, takes longer than finding the formula.
Notice that $u \equiv 0 \pmod 4,$ because $$ -pr + qr +ps+qs \equiv pr + qr +ps+qs \equiv (p+q)(r+s) \pmod 2. $$ As we demanded that $p+q+r+s$ be odd, it is not possible to have both $p+q$ and $r+s$ odd. One of $p+q$ and $r+s$ is odd, while the other is even, meaning the product is even.
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Raw search 2 a^2 = u^2 + v^2 + w^2, with odd a,v,w, even u, and v >= w.
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