Discrete Root

The problem of finding discrete root is defined as follows. Given a prime $n$ and two integers $a$ and $k$, find all $x$ for which:

$x^k \equiv a \pmod n$

The algorithm

We will solve this problem by reducing it to the discrete logarithm problem.

Let's apply the concept of a primitive root modulo $n$. Let $g$ be a primitive root modulo $n$. Note that since $n$ is prime, it must exist, and it can be found in $O(Ans \cdot \log \phi (n) \cdot \log n) = O(Ans \cdot \log^2 n)$ plus time of factoring $\phi (n)$.

We can easily discard the case where $a = 0$. In this case, obviously there is only one answer: $x = 0$.

Since we jnow that $n$ is a prime, any number between 1 and $n-1$ can be represented as a power of the primitive root, and we can represent the discrete root problem as follows:

$(g^y)^k \equiv a \pmod n$


$x \equiv g^y \pmod n$

This, in turn, can be rewritten as

$(g^k)^y \equiv a \pmod n$

Now we have one unknown $y$, which is a discrete logarithm problem. The solution can be found using Shanks' baby-step-giant-step algorithm in $O(\sqrt {n} \log n)$ (or we can verify that there are no solutions).

Having found one solution $y_0$, one of solutions of discrete root problem will be $x_0 = g^{y_0} \pmod n$.

Finding all solutions from one known solution

To solve the given problem in full, we need to find all solutions knowing one of them: $x_0 = g^{y_0} \pmod n$.

Let's recall the fact that a primitive root always has order of $\phi (n)$, i.e. the smallest power of $g$ which gives 1 is $\phi (n)$. Therefore, if we add the term $\phi (n)$ to the exponential, we still get the same value:

$x^k \equiv g^{ y_0 \cdot k + l \cdot \phi (n)} \equiv a \pmod n \forall l \in Z$

Hence, all the solutions are of the form:

$x = g^{y_0 + \frac {l \cdot \phi (n)}{k}} \pmod n \forall l \in Z$.

where $l$ is chosen such that the fraction must be an integer. For this to be true, the numerator has to be divisible by the least common multiple of $\phi (n)$ and $k$. Remember that least common multiple of two numbers $lcm(a, b) = \frac{a \cdot b}{gcd(a, b)}$; we'll get

$x = g^{y_0 + i \frac {\phi (n)}{gcd(k, \phi (n))}} \pmod n \forall i \in Z$.

This is the final formula for all solutions of discrete root problem.


Here is a full implementation, including routines for finding the primitive root, discrete log and finding and printing all solutions.

int gcd (int a, int b) {
    return a ? gcd (b%a, a) : b;

int powmod (int a, int b, int p) {
    int res = 1;
    while (b)
        if (b & 1)
            res = int (res * 1ll * a % p),  --b;
            a = int (a * 1ll * a % p),  b >>= 1;
    return res;

int generator (int p) {
    vector<int> fact;
    int phi = p-1,  n = phi;
    for (int i=2; i*i<=n; ++i)
        if (n % i == 0) {
            fact.push_back (i);
            while (n % i == 0)
                n /= i;
    if (n > 1)
        fact.push_back (n);

    for (int res=2; res<=p; ++res) {
        bool ok = true;
        for (size_t i=0; i<fact.size() && ok; ++i)
            ok &= powmod (res, phi / fact[i], p) != 1;
        if (ok)  return res;
    return -1;

int main() {

    int n, k, a;
    cin >> n >> k >> a;
    if (a == 0) {
        puts ("1\n0");
        return 0;

    int g = generator (n);

    int sq = (int) sqrt (n + .0) + 1;
    vector < pair<int,int> > dec (sq);
    for (int i=1; i<=sq; ++i)
        dec[i-1] = make_pair (powmod (g, int (i * sq * 1ll * k % (n - 1)), n), i);
    sort (dec.begin(), dec.end());
    int any_ans = -1;
    for (int i=0; i<sq; ++i) {
        int my = int (powmod (g, int (i * 1ll * k % (n - 1)), n) * 1ll * a % n);
        vector < pair<int,int> >::iterator it =
            lower_bound (dec.begin(), dec.end(), make_pair (my, 0));
        if (it != dec.end() && it->first == my) {
            any_ans = it->second * sq - i;
    if (any_ans == -1) {
        puts ("0");
        return 0;

    int delta = (n-1) / gcd (k, n-1);
    vector<int> ans;
    for (int cur=any_ans%delta; cur<n-1; cur+=delta)
        ans.push_back (powmod (g, cur, n));
    sort (ans.begin(), ans.end());
    printf ("%d\n", ans.size());
    for (size_t i=0; i<ans.size(); ++i)
        printf ("%d ", ans[i]);