Task 1. Determine all values of the real numbers $a, b, c$ and $d$ such that

$$
a^{2}+b^{2}+c^{2}+d^{2}=a(b+c+d)
$$
Solution. The equation can be written in the form

$$
\begin{aligned}
& \frac{a^{2}}{4}-a b+b^{2}+\frac{a^{2}}{4}-a c+c^{2}+\frac{a^{2}}{4}-a d+d^{2}+\frac{a^{2}}{4}=0 \\
& \left(\frac{a}{2}-b\right)^{2}+\left(\frac{a}{2}-c\right)^{2}+\left(\frac{a}{2}-d\right)^{2}+\frac{a^{2}}{4}=0
\end{aligned}
$$

The sum of squares of real numbers is zero if and only if each of them is zero. Therefore, the last equality is true if and only if

$$
\frac{a}{2}-b=0, \frac{a}{2}-c=0, \frac{a}{2}-d=0, \frac{a}{2}=0
$$

which means,

$$
b=c=d=\frac{a}{2}=0
$$

Thus, the only real numbers for which the original equation is true are $a=b=c=d=0$.