Note: if a number $a$ is a divisor of $N$ , then the number $b$ is also a divisor of $N$ , where $a \cdot b = N$ .
The second important observation: $\min(a, b) \le \sqrt{N}$ , that is, at least one of the divisors in the pair is always not greater than $\sqrt{N}$ .

This means that we can iterate only through numbers from $1$ to $\sqrt{N}$ . If we find a divisor $x$ , then we also add $\frac{N}{x}$ to the set of divisors.
Thus, we can find all divisors of $N$ in $O(\sqrt{N})$ .

Code example:

Primality test

To test a number for primality, it is enough to count the number of its divisors.
If the number of divisors is two, then the number is prime; otherwise, it is composite.

Factorization

Factorization is the decomposition of a number into prime factors.

The factorization algorithm is based on the same ideas as the primality test or finding divisors described above.
If a number has a prime divisor different from itself, then it does not exceed $\sqrt{N}$ .

To factorize a number, you need to iterate through all numbers in the range from $2$ to $\sqrt{N}$ and successively divide $N$ by the found divisors.

for (int i = 1; i <= N; i++) {
    if (N % i == 0) {
        ...
    }
}

vector<int> divisors;
for (int a = 1; a * a <= N; a++) {
    if (N % a == 0) {
        int b = N / a;
        divisors.push_back(a);
        if (a != b) {
            divisors.push_back(b);
        }
    }
}

vector<int> factors;

for (int i = 2; i * i <= N; i++) {
    while (N % i == 0) {
        factors.push_back(i);
        N /= i;
    }
}

if (N > 1) {
    factors.push_back(N);
}