Let $d = n + 5$. Then $d \mid n^4 + 1395$.

We have $n = d - 5$, so
$$n^4 + 1395 = (d - 5)^4 + 1395.$$
Expand $(d - 5)^4$:
$$(d - 5)^4 = d^4 - 4d^3 \cdot 5 + 6d^2 \cdot 25 - 4d \cdot 125 + 625 = d^4 - 20d^3 + 150d^2 - 500d + 625.$$
So
$$n^4 + 1395 = d^4 - 20d^3 + 150d^2 - 500d + 625 + 1395 = d^4 - 20d^3 + 150d^2 - 500d + 2020.$$
We want $d \mid d^4 - 20d^3 + 150d^2 - 500d + 2020$.

But $d \mid d^4 - 20d^3 + 150d^2 - 500d + 2020$ iff $d \mid -20d^3 + 150d^2 - 500d + 2020$ (since $d \mid d^4$).

So $d \mid -20d^3 + 150d^2 - 500d + 2020$.

Let us write $-20d^3 + 150d^2 - 500d + 2020 = d(-20d^2 + 150d - 500) + 2020$.

So $d \mid 2020$.

Therefore, $n + 5 \mid 2020$.

We want the largest positive integer $n$ such that $n + 5 \mid 2020$.

The largest positive divisor of $2020$ is $2020$ itself, so $n + 5 = 2020 \implies n = 2015$.

Check: Is $2015 + 5 = 2020$ a divisor of $2015^4 + 1395$?

But our argument shows that for $n = 2015$, $n + 5 = 2020$ divides $n^4 + 1395$.

Thus, the answer is $\boxed{2015}$.

$$4(2^{m-1} + 1)(2^m + 1)(2^{2m-1} + 2^m + 1) = (p - 1)(p + 1).$$
Since $p$ is odd, the right-hand side is the product of two consecutive even numbers, so it is divisible by 8. The left-hand side is not divisible by 8, unless $m = 1$.
It follows that the only solution is $(p, m, n) = (11, 1, 3)$.
3.3. Let $G$ be the set of all cities in the country. We call a pair $(A, Z)$, where $A$ and $Z$ are disjoint subsets of $G$ good if all cities in the set $A$ can be visited using only bus such that no city is visited twice and all cities in the set $Z$ can be visited using only train such that no city is visited twice.
Let $(A, Z)$ be a good pair such that the set $A \cup Z$ has the maximum number of elements. If we prove $A \cup Z = G$, the statement of the problem holds.
Let us assume the opposite, i.e. there is a city $g$ which isn't from $A$ nor $Z$. Without loss of generality we can assume that $A$ and $Z$ are non-empty, because otherwise we can transfer any city from a non-empty set to an empty one.
Let $n$ be the number of cities in the set $A$, and $m$ the number of cities in the set $Z$. Let us arrange the cities from $A$ in the series $a_1, \dots, a_n$ such that every two consecutive cities in that series are connected by a direct bus line. Also, let us arrange the cities from $Z$ in the series $z_1, \dots, z_m$ such that every two consecutive cities in that series are connected by a direct train line.
Since we assumed that the pair $(A, Z)$ is maximum, the cities $g$ and $a_1$ have to be connected by train (otherwise the pair $(A \cup \{g\}, Z)$ would be a good pair whose union would have more elements than $A \cup Z$), and $g$ and $z_1$ have to be connected by bus (otherwise the pair $(A, Z \cup \{g\})$ would be a good pair whose union would have more elements than $A \cup Z$).
The cities $a_1$ and $z_1$ have to be connected by bus or by train.

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## Croatia2016_booklet — Page 19