## Task 1.

Let $a, b, c,$ and $d$ be real numbers such that

$$
a^{2}+b^{2}+c^{2}+d^{2}=a b+b c+c d+d a+8
$$

Determine the smallest possible value of the largest number in the set $\{|a-b|,|b-c|,|c-d|,|d-a|\}$.
## Solution.

If we multiply the original equation by 2, we get

$$
2 a^{2}+2 b^{2}+2 c^{2}+2 d^{2}=2 a b+2 b c+2 c d+2 d a+16
$$

which is successively equivalent to

$$
a^{2}-2 a b+b^{2}+b^{2}-2 b c+c^{2}+c^{2}-2 c d+d^{2}+d^{2}-2 d a+d^{2}=16
$$

or

$$
(a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}=16
$$

If we assume that the maximum value among the numbers $|a-b|,|b-c|,|c-d|$, and $|d-a|$ is strictly less than 2, we have $(a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}<16$, which is a contradiction.

Therefore, the maximum value among the numbers $|a-b|,|b-c|,|c-d|$, and $|d-a|$ is greater than or equal to 2.

It can indeed be as low as 2, for example, for $(a, b, c, d)=(2,0,2,0)$ we have

$$
|a-b|=|b-c|=|c-d|=|d-a|=2
$$