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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

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  • $\begingroup$ Does a Euler brick exist?? $\endgroup$ – Bumblebee Aug 5 '17 at 21:23
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    $\begingroup$ @Nil: that is not accurate: here we do not need that $x,y,z$ are squares. $\endgroup$ – Jack D'Aurizio Aug 5 '17 at 21:26
  • $\begingroup$ @JackD'Aurizio: I just wanted to mention something related. Some times it is helpful to solve problems by looking at similar problems. $\endgroup$ – Bumblebee Aug 5 '17 at 21:28
  • $\begingroup$ @Nil. That's a cool reference. Nonetheless I'm having trouble seeing how this helps. $\endgroup$ – AmateurMathPirate Aug 5 '17 at 21:31
  • $\begingroup$ @AmateurMathGuy it's a subset for the last three equations involving squares is one way. based on the equations alone you get,x,y and z must all be a difference of squares. $\endgroup$ – user451844 Aug 5 '17 at 21:35
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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.

=================================

? a = p^2 + q^2 + r^2 + s^2
%1 = p^2 + (q^2 + (r^2 + s^2))
? 
? 
? u = 2 * ( -p * r + q * r + p * s + q * s    )
%2 = (-2*r + 2*s)*p + (2*r + 2*s)*q
? 
? v = p^2 - q^2 + r^2 - s^2 + 2 * p * q + 2 * r * s
%3 = p^2 + 2*q*p + (-q^2 + (r^2 + 2*s*r - s^2))
? 
? w = p^2 - q^2 - r^2 + s^2 - 2 * p * q + 2 * r * s
%4 = p^2 - 2*q*p + (-q^2 + (-r^2 + 2*s*r + s^2))
? 
? 
? 
? 
? u^2 + v^2 + w^2 - 2 * a^2
%5 = 0
? 
? 

===========================

Raw search 2 a^2 = u^2 + v^2 + w^2, with odd a,v,w, even u, and v >= w.

  1       0   1   1
  3       4   1   1
  5       0   7   1
  5       4   5   3
  7       4   9   1
  7       8   5   3
  9       4  11   5
  9       8   7   7
 11       4  15   1
 11       8  13   3
 11      12   7   7
 13       0  17   7
 13       8  15   7
 13      12  13   5
 13      16   9   1
 15       8  19   5
 15      16  13   5
 15      20   7   1
 17       0  23   7
 17       4  21  11
 17       8  17  15
 17      20  13   3
 17      24   1   1
 19       4  25   9
 19      12  17  17
 19      12  23   7
 19      16  21   5
 19      24  11   5
 21       4  29   5
 21       8  23  17
 21      16  25   1
 21      20  19  11
 23       4  31   9
 23      12  25  17
 23      16  21  19
 23      24  19  11
 23      28  15   7
 23      32   5   3
 25       0  31  17
 25       4  35   3
 25       8  31  15
 25      20  27  11
 25      20  29   3
 25      24  25   7
 25      28  21   5
 25      32  15   1
 27       8  35  13
 27       8  37   5
 27      16  29  19
 27      20  23  23
 27      28  25   7
 29       0  41   1
 29       4  35  21
 29       8  33  23
 29      12  37  13
 29      20  29  21
 29      28  27  13
 29      36  19   5
 29      40   9   1

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

    a       u    v    w      p  q  r  s
    1       0    1    1      1  0  0  0
    3       4    1    1      0  1  1  1
    5       0    7    1      0  0  2  1
    5       0    7    1      0  0 -2 -1
    5       4    5    3      2  0 -1  0
    7       4    9    1      2  1  1  1
    7       8    5    3      1 -1 -2 -1
    9       4   11    5      2  0 -2 -1
    9       8    7    7      0  1  2  2
   11      12    7    7      3  0 -1  1
   11       4   15    1      3  1 -1  0
   11       8   13    3      3  1  0  1
   13       0   17    7      0  0  3  2
   13       0   17    7      0  0 -3 -2
   13      12   13    5      3  0 -2  0
   13      16    9    1      1  2  2  2
   13       8   15    7      2  1  2  2
   15      16   13    5      2 -1 -3 -1
   15      20    7    1      3  1 -1  2
   15       8   19    5      1  1  3  2
   17       0   23    7      4  1  0  0
   17      20   13    3      0  2  3  2
   17      24    1    1      3  0 -2  2
   17       4   21   11      2  0 -3 -2
   17       8   17   15      4  0 -1  0
   19      12   17   17      0  1  3  3
   19      12   23    7      3  0 -3 -1
   19      16   21    5      4  1 -1  1
   19      24   11    5      3 -1 -3  0
   19       4   25    9      4  1  1  1
   21      16   25    1      3  2  2  2
   21      20   19   11      4  1  0  2
   21       4   29    5      1  0 -4 -2
   21       8   23   17      4  0 -2 -1
   23      12   25   17      2  1  3  3
   23      16   21   19      3  1  2  3
   23      24   19   11      1  2  3  3
   23      28   15    7      3  2  1  3
   23      32    5    3      1  3  2  3
   23       4   31    9      3  1  3  2
   25       0   31   17      0  0  4  3
   25       0   31   17      0  0 -4 -3
   25      20   27   11      2 -1 -4 -2
   25      20   29    3      4  2  1  2
   25      24   25    7      4  0 -3  0
   25      28   21    5      1 -2 -4 -2
   25      32   15    1      2 -2 -4 -1
   25       4   35    3      2  1  4  2
   25       8   31   15      4  1  2  2
   27      16   29   19      1 -1 -4 -3
   27      20   23   23      5  0 -1  1
   27      28   25    7      3 -1 -4 -1
   27       8   35   13      5  1 -1  0
   27       8   37    5      4  1 -3 -1
   29       0   41    1      5  2  0  0
   29      12   37   13      3  0 -4 -2
   29      20   29   21      5  0 -2  0
   29      28   27   13      0  2  4  3
   29      36   19    5      4  2  0  3
   29      40    9    1      3 -2 -4  0
   29       4   35   21      2  0 -4 -3
   29       8   33   23      4  0 -3 -2
    a       u    v    w      p  q  r  s

=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=

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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 $3^2+13^2+20^2=2\cdot 17^2$.

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  • $\begingroup$ Also I generally do not know how to solve, diophantically, the equation $$2a^2=b^2+c^2+d^2$$. Any references or guidance on that? $\endgroup$ – AmateurMathPirate Aug 5 '17 at 21:55
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    $\begingroup$ @AmateurMathGuy: such solutions are associated with decompositions $2n^2=a^2+b^2+c^2$ where $a,b,c$ have roughly the same magnitude. For instance, $$(x,y,z)=(57,112,672)$$ is associated with $13^2+27^2+28^2=2\cdot 29^2$. $\endgroup$ – Jack D'Aurizio Aug 5 '17 at 21:57
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    $\begingroup$ And by Legendre's theorem (en.wikipedia.org/wiki/Legendre%27s_three-square_theorem), any number that is not of the form $4^k(8m+7)$ can be represented as the sum of three squares, so there are plenty of solutions. $\endgroup$ – Jack D'Aurizio Aug 5 '17 at 21:58
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    $\begingroup$ Outstanding. Your answer has shown me where I need to study. Thank you. $\endgroup$ – AmateurMathPirate Aug 5 '17 at 22:02
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    $\begingroup$ Jack, all primitive solutions of $a^2 + b^2 + c^2 = 2 d^2$ can be parametrized by $\bar{q} v q,$ where $v$ is a quaternion with real part zero and norm 2, such as $v = i+j;$ meanwhile, $d$ is the norm of $q.$ Here is multiplier 3: math.stackexchange.com/questions/1964607/… $\endgroup$ – Will Jagy Aug 6 '17 at 0:26
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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*}

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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.

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