GS. 4 Let $F$ be the number of integral solutions of $x^{2}+y^{2}+z^{2}+w^{2}=3(x+y+z+w)$. Find the value of $F$.
$(x, y, z, w)=(0,0,0,0)$ is a trivial solution.
$$
\begin{array}{l}
\left(x^{2}-3 x+\frac{9}{4}\right)+\left(y^{2}-3 y+\frac{9}{4}\right)+\left(z^{2}-3 z+\frac{9}{4}\right)+\left(w^{2}-3 w+\frac{9}{4}\right)=9 \\
\left(x-\frac{3}{2}\right)^{2}+\left(y-\frac{3}{2}\right)^{2}+\left(z-\frac{3}{2}\right)^{2}+\left(w-\frac{3}{2}\right)^{2}=9 \\
(2 x-3)^{2}+(2 y-3)^{2}+(2 z-3)^{2}+(2 w-3)^{2}=36
\end{array}
$$

Let $a=2 x-3, b=2 y-3, c=2 z-3, d=2 w-3$, the equation becomes $a^{2}+b^{2}+c^{2}+d^{2}=36$
For integral solutions of $(x, y, z, w),(a, d, c, d)$ must be odd integers.
In addition, the permutation of $(a, b, c, d)$ is also a solution. (e.g. $(b, d, c, a)$ is a solution)
$\because a, b, c, d$ are odd integers and $a^{2}+b^{2}+c^{2}+d^{2} \geq 0$
If one of the four unknowns, say, $a>6$, then L.H.S. $>36$, so L.H.S. $\neq$ R.H.S.
$$
\therefore a, b, c, d= \pm 1, \pm 3, \pm 5
$$

When $a= \pm 5$, then $25+b^{2}+c^{2}+d^{2}=36 \Rightarrow b^{2}+c^{2}+d^{2}=11$
The only integral solution to this equation is $b= \pm 3, c= \pm 1=d$ or its permutations.
When the largest (in magnitude) of the 4 unknowns, say, $a$ is $\pm 3$, then $9+b^{2}+c^{2}+d^{2}=36$
$\Rightarrow b^{2}+c^{2}+d^{2}=27$, the only solution is $b= \pm 3, c= \pm 3, d= \pm 3$ or its permutations.
$\therefore$ The integral solutions are $(a, b, c, d)=(5,3,1,1)$ and its permutations $\ldots(1) \times P_{2}^{4}=12$
$$
(3,3,3,3) \ldots(2) \times 1
$$

If $(a, b, c, d)$ is a solution, then $( \pm a, \pm b, \pm c, \pm d)$ are also solutions.
There are 16 solutions with different signs for $( \pm a, \pm b, \pm c, \pm d)$.
$$
\begin{aligned}
\therefore F & =(12+1) \times 16 \\
& =208
\end{aligned}
$$