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.
On
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:
$$\begin{cases} \lambda &= 1\\ \alpha &= s - p,\\ \beta &= 2(s+p+k),\\ \gamma &= k+p,\\ \delta &= k+s. \end{cases}$$
On
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.
On
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
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.