1. In the picture, several circles are drawn, connected by segments. Tanya chooses a natural number n and places different natural numbers in the circles so that for all these numbers the following property holds:

if numbers a and b are not connected by a segment, then the sum $a^{2}+b^{2}$ must be coprime with n; if connected, then the numbers $a^{2}+b^{2}$ and n must have a common natural divisor greater than 1. For what smallest n does such an arrangement exist?

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-07.jpg?height=217&width=217&top_left_y=600&top_left_x=931)
Answer: $n=65$.

Solution. First, let's make three observations.

1) $n$ is odd. Indeed, suppose $n$ is even. Among five numbers, there are always three numbers of the same parity, and by the condition, they must be pairwise connected. But there are no cycles of length 3 in the picture.
2) If $d$ is a divisor of $n$, then no more than two numbers are divisible by $d$. Suppose there are three such numbers. Then they must be pairwise connected, and thus form a three-link cycle.
3) If $q$ is a prime divisor of $n$, then among four sequentially connected numbers, there exists a pair of adjacent numbers whose sum of squares is not divisible by $q$. Suppose there is a chain $(a, b, c, d)$ of sequentially connected numbers such that the sum of squares of any pair of adjacent numbers is divisible by $q$. Then

$$
a^{2}+d^{2}=\left(a^{2}+b^{2}\right)-\left(b^{2}+c^{2}\right)+\left(c^{2}+d^{2}\right) \vdots q
$$

This means that the numbers $a$ and $d$ are connected, so there is a cycle of length 4 in the picture, which does not exist.

From 1) and 3), it follows that the number $n$ has at least two different odd prime divisors. Let there be exactly two (say, $p$ and $q$). We will show that they are different from $3, 7, 11$. Suppose, for example, that $p \in \{3, 7, 11\}$. Note that for any natural number $a$

$$
a^{2} \bmod 3 \in \{0,1\}, \quad a^{2} \bmod 7 \in \{0,1,2,4\}, \quad a^{2} \bmod 11 \in \{0,1,3,4,5,9\}
$$

Therefore, $a^{2}+b^{2}$ is divisible by $p$ if and only if $a^{2}$ and $b^{2}$ are divisible by $p$, and thus $a$ and $b$ are divisible by $p$. By 2), this condition can be satisfied by only one pair. From the remaining four pairs of connected numbers, a three-link chain can be formed in which the sum of squares of each pair of adjacent numbers is divisible by $q$. But this contradicts 3).

Thus, if $n$ has exactly two different odd prime divisors, then $n \geqslant 5 \cdot 13=65$. If there are more than two such divisors, then $n \geqslant 3 \cdot 5 \cdot 7=105>65$. The arrangement for $n=65$ is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_6bce8335c6b594326f89g-07.jpg?height=339&width=354&top_left_y=1915&top_left_x=862)