One Diophantine equation 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.
 A: 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:
$$\begin{cases}
\lambda &= 1\\
\alpha  &= s - p,\\
\beta   &= 2(s+p+k),\\
\gamma  &= k+p,\\
\delta  &= k+s.
\end{cases}$$
A: 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 $(a,b,c,d)=(1,4,9,28)$
A: 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.
A: 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.
