Problem 1790

Let call the numbers satisfying the given condition as "good" number. Consider some good number $n$. One can see that if $n$ is even, then $d = 2$ is a divisor of $2$, however, then $1^2$ and $2026^2$ both divide $2027$ with remainder $1$, which does not satisfy. Therefore, $n$ is odd. We see that $n = 1$ satisfies the given condition.

Now consider $n > 1$ and set $n = p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$ with $p_1, p_2, \dots, p_k$ being distinct odd primes and $a_1, a_2, \dots, a_k$ positive integers. Since $\tau(n)^2 \mid n$, $\tau(n)$ is also odd.

We have $\tau(n) = (a_1+1)(a_2+1)\cdots(a_k+1)$ which is odd, so $a_1, a_2, \dots, a_k$ are all even. Therefore, $n$ must be a perfect square. Note that if $k \ge 3$ then $n \ge (3 \cdot 5 \cdot 7)^2 > 2027$, is not satisfied. Hence $k \in \{1; 2\}$. We consider the following cases.

1. If $k = 1$ then $n = p^{2m}$ with $p$ being an odd prime and $m \in \mathbb{Z}^+$. Note that if $2m \ge 6$ then $\tau(n) \ge 7$ so $p \ge 7$ and $n \ge 7^6 > 2027$, does not satisfy. Therefore, we reduce to considering if $2m = 2$ then $p = 3$ so $n = 9$; and if $2m = 4$ then $\tau(n) = 5$ so $p = 5$, which implies $n = 5^4 = 625$. We can check that the numbers $n = 9, n = 625$ all satisfy condition (ii).

2. If $k = 2$ then $n = p^{2a}q^{2b}$ with $p, q$ being distinct odd primes and $a, b \in \mathbb{Z}^+$. Note that if both $2a$ and $2b \ge 4$ then $n \ge (3 \cdot 5)^4 > 2027$, is not satisfied. Therefore, we can assume $2b = 2$, which implies that $3|\tau(n)$ so $p = 3$ or $q = 3$.

a. If $q = 3$ then $n = 9p^{2a}$, then if $2a = 2$ then $\tau(n) = 9$ so $9^2|n$, is not satisfied; therefore $2a \ge 4$ so $n \ge 9 \cdot 5^4 > 2027$, is also not satisfied.

b. If $p = 3$ then $n = 3^{2a}q^2$; if $2a = 2$ then the same absurdity exists, so $2a > 2$. But if $2a \ge 6$ then $n \ge 3^6 \cdot 5^2 > 2027$, is not satisfied. Therefore $2a = 4$ and $5|\tau(n)$ so $q = 5$, implies $n = 2025$. We see that this number satisfies condition (ii).

Therefore $n \in \{9; 625; 2025\}$. We will prove that these numbers are good.

Consider an arbitrary positive divisor $d$ of $n$ then $d$ is odd. Suppose there exist $u, v \in \{1; 2; \dots; 2026\}$ and $u \neq v$ such that $u^d \equiv v^d \pmod{2027}$. Let $s$ be the inverse of $v$ modulo $2027$ then let $t = us$, we have $t^d \equiv 1 \pmod{2027}$. Next, let $h = \text{ord}_{2027}(t)$ then $h \nmid d$ and by Fermat's little theorem $h \nmid 2026$ so $h \nmid \text{gcd}(d, 2026)$. Note that $2026 = 2 \cdot 1013$ with $1013$ being a prime number, it is easy to see that $\text{gcd}(d, 2026) = 1$ and $h = 1$. Then, we will have
$$t \equiv 1 \pmod{2027} \iff us \equiv 1 \pmod{2027}.$$
Multiplying both sides of this congruence by $v$ gives $u \equiv v \pmod{2027}$, which is absurd. Thus, all positive divisors $d$ of these $n$ satisfy condition (i).

So all the good numbers are $1, 9, 625, 2025$. $\square$