I wonder now that the following Diophantine equation:
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
have only this formula describing his decision?
$a=-(k^2+2(p+s)k+p^2+ps+s^2)$
$b=2k^2+4(p+s)k+3p^2+3ps+2s^2$
$c=3k^2+4(p+s)k+2p^2+ps+2s^2$
$d=2k^2+4(p+s)k+2p^2+3ps+3s^2$
$k,p,s$ - what some integers.
By your question, I mean what that formula looks like this. Of course I know about the procedure of finding a solution, but I think that the formula would be better.
I don't think your formula exhaust all solutions of the Diophantine equation
$$2(a^2+b^2+c^2+d^2) = (a+b+c+d)^2\tag{*1}$$
Your equation can be rewritten as
$$(a+b+c+d)^2 = (-a+b-c+d)^2 + (-a+b+c-d)^2 + (-a-b+c+d)^2$$
This is the equation for a Pythagorean quadruple. The set of all Pythagorean quadruples can be parametrized by 5 integers $\alpha,\beta,\gamma,\delta$ and $\lambda$:
$$\begin{cases} \hphantom{-}a + b + c + d &= \lambda (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)\\ -a + b - c + d &= \lambda(\alpha^2+\beta^2 - \gamma^2 - \delta^2)\\ -a + b + c - d &= 2\lambda(\alpha\delta + \beta\gamma)\\ -a - b + c + d &= 2\lambda(\beta\delta - \alpha\gamma) \end{cases} $$ Using this, we see all solutions of $(*1)$ must have the form $$ \begin{cases} a &= \frac{\lambda}{2} \left(\gamma^2 + \delta^2 + (\alpha-\beta)\gamma - (\alpha+\beta)\delta\right)\\ b &= \frac{\lambda}{2} \left(\alpha^2 + \beta^2 + (\alpha+\beta)\gamma + (\alpha-\beta)\delta\right)\\ c &= \frac{\lambda}{2} \left(\gamma^2 +\delta^2 - (\alpha-\beta)\gamma + (\alpha+\beta)\delta\right)\\ d &= \frac{\lambda}{2} \left(\alpha^2+\beta^2 - (\alpha+\beta)\gamma - (\alpha-\beta)\delta\right) \end{cases}\tag{*2} $$ Furthermore, if one substitute any integers $\lambda, \alpha,\beta,\gamma,\delta$ into $(*2)$ and if the resulting $a, b, c, d$ are integers, then it will be a solution of $(*1)$. It is not hard to check this happens when and only when
$$\lambda(\alpha+\beta+\gamma+\delta)\quad\text{ is an even number }\tag{*3}$$
Conclusion - all solutions of $(*1)$ can be parametrized by $(*2)$ subject to the constraint $(*3)$.
The formula you have is a special case of above parametrization. It can be reproduced by following substitutions:
$$\begin{cases} \lambda &= 1\\ \alpha &= s - p,\\ \beta &= 2(s+p+k),\\ \gamma &= k+p,\\ \delta &= k+s. \end{cases}$$
EDITTTT: two research level articles on this, one by Kontorovich and one by Fuchs, can be downloaded for free at AMS BULLETIN APRIL 2013. There is also a short survey article by Peter Sarnak, in the April 2011 MAA Monthly. I have a pdf of that, if anyone is interested.
Alright, I see part of what is going on. The standard recipe, stereographic projection around a given integral solution, is guaranteed to give all rational solutions; this method parametrizes solutions with four components by three parameters, or generally $k$ components by $k-1$ parameters. And, some choices of central point for projection give more efficient recipes than others; the one individ chose was very good.
However, it appears that such a recipe is not going to give all integral solutions, nor will a finite number of such recipes. So, here is my version, with rational solutions. That is, I took coprime $(x,y,z)$ and calculated
$$ a = x^2 + y^2 + 2 z^2 - y z - z x - 2 x y, $$
$$ b = x^2 - y z + z x,$$
$$ c = y^2 + y z - z x,$$
$$ d = 2 z^2 + y z + z x.$$
Then I took $$ g = \gcd(a,b,c,d), $$ and divided all four of $(a,b,c,d)$ by that. As a result, I found all the "root" solutions given at TABLE. I am encouraged that $g$ was always the sum of two squares. Hmmm; actually, it is easy to show that $\gcd(x,y,z) = 1$ implies that $ g = \gcd(a,b,c,d) $ is not divisible by $4$ or by any prime $q \equiv 3 \pmod 4,$ so we can always write $g = u^2 + v^2$ with integers and $\gcd(u,v)=1.$
Oh, root solutions have $$ a \leq 0 \leq b \leq c \leq d, $$ $$ a+b+c+d > 0, $$ $$ a + b + c \geq d. $$ These were defined in Graham, Lagarias, Mallows, Wilks, Yan, Apollonian circle packings: Number Theory, Journal of Number Theory, volume 100 (2003) pages 1-45. It was shown that every integral solution can be connected to such a root solution by Vieta jumps, thus dividing the solutions into a forest of countably many rooted trees. An unusual feature is that these Vieta jumps are (they must be) elements in a group of $4$ by $4$ invertible integral matrices, called the Apollonian Group, which is just the (orthogonal or rotation or automorphism) group for the quadratic form/indefinite lattice given by $$ a^2 + b^2 + c^2 + d^2 -2ab-2ac-2ad-2bc-2bd-2cd, $$ with Gram matrix $$ H \; = \; \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ -1 & 1 & -1 & -1 \\ -1 & -1 & 1 & -1 \\ -1 & -1 & -1 & 1 \end{array} \right). $$ As a matrix, eigenvalues are $-2,2,2,2$ with orthogonal (but not orthonormal) eigenvectors as columns of $$ W \; = \; \left( \begin{array}{rrrr} 1 & -1 & -1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & 0 & 2 & -1 \\ 1 & 0 & 0 & 3 \end{array} \right). $$
a b c d g x y z
-1 2 2 3 gcd was 1 1 0 1
-2 3 6 7 gcd was 5 -3 -6 2
-3 4 12 13 gcd was 10 4 12 -3
-3 5 8 8 gcd was 1 2 3 -1
-4 5 20 21 gcd was 5 -2 1 -7
-4 8 9 9 gcd was 29 -1 12 9
-5 6 30 31 gcd was 26 -6 -30 5
-5 7 18 18 gcd was 10 -4 2 -9
-6 7 42 43 gcd was 37 -7 -42 6
-6 10 15 19 gcd was 13 -10 -15 6
-6 11 14 15 gcd was 2 0 -4 -3
-7 8 56 57 gcd was 13 -3 2 -19
-7 9 32 32 gcd was 17 5 -3 16
-7 12 17 20 gcd was 5 -8 -9 -4
-8 9 72 73 gcd was 65 9 72 -8
-8 12 25 25 gcd was 25 -14 -27 8
-8 13 21 24 gcd was 5 7 11 -4
-9 10 90 91 gcd was 82 -10 -90 9
-9 11 50 50 gcd was 5 -4 -17 3
-9 14 26 27 gcd was 26 -16 -28 9
-9 18 19 22 gcd was 17 -1 13 11
-10 11 110 111 gcd was 25 -4 3 -37
-10 14 35 39 gcd was 10 -6 2 -13
-10 18 23 27 gcd was 1 -4 -5 2
-11 12 132 133 gcd was 122 -12 -132 11
-11 13 72 72 gcd was 37 -7 5 -36
-11 16 36 37 gcd was 34 -18 -38 11
-11 21 24 28 gcd was 5 -7 1 -7
-12 13 156 157 gcd was 145 -13 -156 12
-12 16 49 49 gcd was 5 -6 -17 4
-12 17 41 44 gcd was 1 -3 -7 2
-12 21 28 37 gcd was 25 21 28 -12
-12 21 29 32 gcd was 2 -6 -8 3
-12 25 25 28 gcd was 1 -5 -5 -2
-13 14 182 183 gcd was 41 -5 4 -61
-13 15 98 98 gcd was 50 -8 6 -49
-13 18 47 50 gcd was 2 -2 -8 -5
-13 23 30 38 gcd was 26 -16 2 -19
-14 15 210 211 gcd was 197 -15 -210 14
-14 18 63 67 gcd was 53 -18 -63 14
-14 19 54 55 gcd was 1 -2 1 -5
-14 22 39 43 gcd was 37 -24 -41 14
-14 27 31 34 gcd was 10 -16 -18 7
-15 16 240 241 gcd was 226 -16 -240 15
-15 17 128 128 gcd was 13 -6 -43 5
-15 24 40 49 gcd was 34 -24 -40 15
-15 24 41 44 gcd was 10 2 16 11
-15 28 33 40 gcd was 2 0 -6 -5
-15 32 32 33 gcd was 2 8 8 -3
a b c d g x y z
The above equation which is mentioned below has another parametrisation,
$2(a^2+b^2+c^2+d^2)=(a+b+c+d)^2$
$a=(5k+2)^2$
$b=(3k-4)^2$
$c=(2k+6)^2$
$d=4(19k^2+10k+28)$
For $k=0$ we get $(a,b,c,d)=(1,4,9,28)$
You can consider another equation:
$2(a^2+y^2+c^2+d^2+u^2)=(a+y+c+d+u)^2$
And write the formula to solve this equation.
$a=-(k^2+2(q+t+b)k+b^2+q^2+t^2+bq+bt+qt)$
$y=k^2+2(q+t+b)k+2b^2+q^2+t^2+2bq+2bt+qt$
$c=k^2+2(q+t+b)k+b^2+2q^2+t^2+2bq+bt+2qt$
$d=k^2+2(q+t+b)k+b^2+q^2+2t^2+bq+2bt+2qt$
$u=2k^2+2(q+t+b)k+b^2+q^2+t^2$
$k,q,t,b$ - what some integers.
My more general method of calculation. Enables us to solve and other factors. Find out whether or when given coefficients solutions and immediately write the formula.
For example, consider the equation:
$4(a^2+b^2+c^2+d^2)=3(a+b+c+d)^2$
Then the solutions are of the form:
$a=-(p^2+4(k+s)p+2k^2+2ks+2s^2)$
$b=p^2+4(k+s)p+6k^2+6ks+2s^2$
$c=p^2+4(k+s)p+2k^2+6ks+6s^2$
$d=3p^2+4(k+s)p+2k^2-2ks+2s^2$
I think the best result by a direct solution of these equations. Brute force search of the law or does not make sense. Spend a lot of effort, but the result does not always work.