1. The picture shows several circles 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 $a$ and $b$ are 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-11.jpg?height=137&width=234&top_left_y=611&top_left_x=928)
Answer: $n=65$.

Solution. First, let's make three observations.

1) $n$ is odd. Indeed, suppose $n$ is even. Among six 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, therefore, form a three-link cycle.
3) If $q$ is a prime divisor of $n$, and in the chain $(a, b, c, d)$ of sequentially connected numbers, the sum of the squares of any pair of adjacent numbers is divisible by $q$, then $(a, b, c, d)$ forms a cycle. Indeed,

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

that is, the numbers $a$ and $d$ are also connected.

From 1) and 3), it follows that the number $n$ has at least two distinct 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 hence $a$ and $b$ are divisible by $p$. By 2), this condition can be satisfied by only one pair. From the remaining six pairs of connected numbers, a non-cyclic three-link chain can be formed (for example, passing along the left and top or right and bottom sides of the large rectangle). In this chain, the sum of the squares of each pair of adjacent numbers is divisible by $q$, which contradicts 3).

Thus, if $n$ has exactly two distinct 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-11.jpg?height=200&width=311&top_left_y=1922&top_left_x=884)