Computing square roots modulo a prime with the Tonelli-Shanks algorithm

Computing square roots modulo a prime with the Tonelli-Shanks algorithm

This post is about the problem of computing square roots modulo a prime number, a well-known problem in algebra and number theory. Nowadays multiple highly-efficient algorithms have been developed to solve this problem, e.g. Tonelli-Shanks, Cipolla’s algorithms. In this post we will focus on one of the most prominent algorithms, namely the Tonelli-Shanks algorithm, which is often described in a rather complicated way in the literature, although its derivation and the idea behind it is actually simple and elegant. The goal of this post is to showcase this beauty of the theory and to then show how the Tonelli-Shanks algorithm can be implemented in Python. We also discuss some interesting connections of this problem to cryptography.

Preliminaries

Formally, the problem of computing square roots modulo a prime is to solve the congruence b2amodpb^2 \equiv a \mod p

for bb, given an odd prime number pp and a(Z/pZ)a \in (\mathbb{Z}/p\mathbb{Z})^*, which is also called a quadratic residue if the congruence above holds. We first discuss some theory we will rely on in the next section about the Tonelli-Shanks algorithm itself.

Let qpq_p be the squaring endomorphism defined as follows qp:(Z/pZ)(Z/pZ)aa2\begin{aligned} q_p : (\mathbb{Z}/p\mathbb{Z})^* &\rightarrow (\mathbb{Z}/p\mathbb{Z})^* \\ a &\mapsto a^2 \end{aligned}

(recall that an endomorphism is a homomorphism of an algebraic structure to itself, it is an easy exercise to verify that qpq_p is indeed an endomorphism). Observe that ker(qp)={1,1}\ker(q_p) = \{1, -1\}, because every element of the kernel is a root of the x21=(x1)(x+1)x^2 - 1 = (x - 1) \cdot (x + 1) polynomial, which cannot have more than two roots because of its degree, meaning that 11 and 1-1 are its only roots. Let Qp:=im(qp)={x2x(Z/pZ)}Q_p := \operatorname{im}(q_p) = \{x^2 \mid x \in (\mathbb{Z}/p\mathbb{Z})^*\}

be the set of all squares modulo pp. In the following lemma, we derive a formula for the exact number of squares modulo pp:

Lemma: Let pp be an odd prime. Then Qp=p12\st Q_p\st = \frac{p-1}{2}.

Proof: By the fundamental homomorphism theorem, there exists an isomorphism (Z/pZ)/ker(qp)im(qp)=Qp(\mathbb{Z}/p\mathbb{Z})^* / \ker(q_p) \cong \operatorname{im}(q_p) = Q_p

Since all cosets have equal order, Qp=im(qp)=(Z/pZ)ker(qp)=p12\st Q_p\st = \st \operatorname{im}(q_p)\st = \frac{\st (\mathbb{Z}/p\mathbb{Z})^*\st}{\st \ker(q_p)\st} = \frac{p - 1}{2}

This completes the proof.

Another major tool for characterizing squares modulo pp is the Legendre symbol defined as follows (ap):={0a0modp1a≢0modp and a is a square modulo p1otherwise\Big(\frac{a}{p}\Big) := \begin{cases} 0 & a \equiv 0 \mod p \\ 1 & a \not\equiv 0 \mod p \text{ and } a \text{ is a square modulo } p \\ -1 & \text{otherwise} \end{cases}

where aZa \in \mathbb{Z} and pp is a prime. If a(Z/pZ)a \in (\mathbb{Z}/p\mathbb{Z})^*, then we can think of (ap)\big(\frac{a}{p}\big) as of an “indicator” variable outputting 11 if b2amodpb^2 \equiv a \mod p holds for some bb and 1-1 otherwise. In fact, the Legendre symbol can be computed efficiently via the Euler’s criterion formulated in the following theorem.

Theorem (Euler): Let aZa \in \mathbb{Z} and pp is an odd prime. Then: (ap)ap12modp\Big(\frac{a}{p}\Big) \equiv a^{\frac{p-1}{2}} \mod p

Proof: If a0modpa \equiv 0 \mod p, then the statement holds trivially. We thus focus only on the case when a(Z/pZ)a \in (\mathbb{Z}/p\mathbb{Z})^*. By Fermat’s little theorem, it holds that (ap12)2ap11modp\Big(a^{\frac{p-1}{2}}\Big)^2 \equiv a^{p-1} \equiv 1 \mod p

which is equivalent to (ap12)210modp\Big(a^{\frac{p-1}{2}}\Big)^2 - 1 \equiv 0 \mod p

meaning that (ap121)(ap12+1)0modp\Big(a^{\frac{p-1}{2}} - 1\Big) \cdot \Big(a^{\frac{p-1}{2}} + 1\Big) \equiv 0 \mod p

Since we are working over the finite field Z/pZ\mathbb{Z}/p\mathbb{Z}, there cannot exist non-trivial divisors of zero, so it follows that ap12{1,1}a^{\frac{p-1}{2}} \in \{1, -1\} modulo pp. Thus, the following mapping φ:(Z/pZ)({1,1},)aap12\begin{aligned} \varphi : (\mathbb{Z}/p\mathbb{Z})^* &\rightarrow (\{1, -1\}, \cdot) \\ a &\mapsto a^{\frac{p-1}{2}} \end{aligned}

is a well-defined homomorphism of groups. Moreover, φ\varphi is actually an epimorphism (i.e, is a surjective homomorphism), which can be shown as follows:

  • 1im(φ)1 \in \operatorname{im}(\varphi), because φ(1)=1\varphi(1) = 1.

  • 1im(φ)-1 \in \operatorname{im}(\varphi), because (Z/pZ)(\mathbb{Z}/p\mathbb{Z})^* is a cyclic group meaning that its generator gg cannot have order less than (Z/pZ)=p1\st (\mathbb{Z}/p\mathbb{Z})^*\st = p - 1. Thus, 1φ(g){1,1}=im(φ)1 \neq \varphi(g) \in \{1, -1\} = \operatorname{im}(\varphi), so it follows that φ(g)=1\varphi(g) = -1.

In other words, we have just shown that im(φ)=2\st \operatorname{im}(\varphi)\st = 2. At this point, to prove the theorem it suffices to show that ker(φ)=Qp\ker(\varphi) = Q_p. The inclusion ker(φ)Qp\ker(\varphi) \supseteq Q_p is easy: if aQpa \in Q_p, i.e., ab2modpa \equiv b^2 \mod p, then φ(a)=ap12=(b2)p12=bp1=1\varphi(a) = a^{\frac{p-1}{2}} = \big(b^2\big)^{\frac{p-1}{2}} = b^{p-1} = 1

where all computations are modulo pp and the last equality holds by Fermat’s little theorem. With ker(φ)Qp\ker(\varphi) \supseteq Q_p proven, since we are working over the finite field Z/pZ\mathbb{Z}/p\mathbb{Z} it suffices to show that ker(φ)=Qp\st \ker(\varphi)\st = \st Q_p\st in order to prove that ker(φ)=Qp\ker(\varphi) = Q_p. By the previous lemma, Qp=p12\st Q_p\st = \frac{p-1}{2}, so it suffices to prove that ker(φ)=p12\st \ker(\varphi)\st = \frac{p-1}{2}. Indeed, since by the fundamental homomorphism theorem (Z/pZ)/ker(φ)im(φ)(\mathbb{Z}/p\mathbb{Z})^* / \ker(\varphi) \cong \operatorname{im}(\varphi)

and all cosets are of equal size, (Z/pZ)ker(φ)=im(φ)=2\frac{\st (\mathbb{Z}/p\mathbb{Z})^*\st}{\st \ker(\varphi)\st} = \st \operatorname{im}(\varphi)\st = 2

Solving for ker(φ)\st \ker(\varphi)\st yields ker(φ)=(Z/pZ)ker(φ)=p12=Qp\st \ker(\varphi)\st = \frac{\st (\mathbb{Z}/p\mathbb{Z})^*\st}{\st \ker(\varphi)\st} = \frac{p-1}2{} = \st Q_p\st

This completes the proof. Having done this preparatory work, we now discuss and implement the Tonelli-Shanks algorithm.

Computing square roots modulo a prime

First observe that since pp is a prime, Z/pZ\mathbb{Z}/p\mathbb{Z} is a field and hence the polynomial x2ax^2 - a has exactly two roots. It follows that it suffices to compute just one root of aa, because it is trivial to obtain the other one by simply performing negation modulo pp. From now on, all computation will be modulo pp if not stated otherwise. We also assume that a(Z/pZ)a \in (\mathbb{Z}/p\mathbb{Z})^*, or, equivalently, that a0a \neq 0, because it is trivial to compute the square root of zero.

The core idea behind the Tonelli-Shanks algorithm is to make use of the fact that if am=1a^m = 1 for some odd mNm \in \mathbb{N}, then (am+12)2=am+1=a\Big(a^{\frac{m+1}{2}}\Big)^2 = a^{m+1} = a

and hence am+12a^{\frac{m+1}{2}} is a square root of aa. By the Euler’s theorem of the previous section above, we know that ap12=1a^{\frac{p-1}{2}} = 1 holds if and only if aa is a square modulo pp: if ap12=1a^{\frac{p-1}{2}} = -1, then the algorithm can directly identify that the given number is not a square and terminate accordingly. Thus, intuitively m=p12m = \frac{p-1}{2} is a “good” initial value to start with while searching for an odd mNm \in \mathbb{N} satisfying am=1a^m = 1. The remaining ideas are best shown by looking at how the algorithm works on a concrete example.

Example

Let p=17p = 17 and a=8a = 8. aQpa \in Q_p, because (ap)=ap12=a8=1(\frac{a}{p}) = a^{\frac{p-1}{2}} = a^8 = 1. We set m:=p12=8m := \frac{p-1}{2} = 8. Unfortunately, we cannot compute the square root of aa directly using the am+12a^{\frac{m+1}{2}} formula above, because mm is even. We thus divide mm by 22 as long as possible and hope that at some point mm will become odd such that am=1a^m = 1 still holds:

 m  m  p12=8  \frac{p-1}{2} = 8  p14=4  \frac{p-1}{4} = 4 
 am  a^m 111-1

Unfortunately, already after the first division of mm by two we get am=a4=1a^m = a^4 = -1, but we want to have one on the right-hand side. The idea is to pick some b(Z/pZ)Qpb \in (\mathbb{Z}/p\mathbb{Z})^* \setminus Q_p, because by Euler’s Theorem (see above) bp12=(bp)=1b^{\frac{p-1}{2}} = \Big(\frac{b}{p}\Big) = -1

It thus holds that 1=am1bp121=ap14bp12=(ab2)m1 = \underbrace{a^m}_{-1} \cdot \underbrace{b^{\frac{p-1}{2}}}_{-1} = a^{\frac{p-1}{4}} \cdot b^{\frac{p-1}{2}} = \big(a \cdot b^2\big)^m

Let b:=3(Z/pZ)Qpb := 3 \in (\mathbb{Z}/p\mathbb{Z})^* \setminus Q_p. We now run the algorithm recursively to compute the square root of a2:=ab2=89=4a_2 := a \cdot b^2 = 8 \cdot 9 = 4, because then we can recover the square root of aa by multiplying the square root of a2a_2 with b1b^{-1}. We repeatedly divide mm by two and recompute a2ma_2^m in the same way we did above:

 m  m  p14=4  \frac{p-1}{4} = 4  p18=2  \frac{p-1}{8} = 2 
 a2m  a_2^m 111-1

Again, already after the very first division of mm by two we obtain 1-1. We thus again write 1=a2m1bp121=a2p18bp12=(a2b4)m1 = \underbrace{a_2^m}_{-1} \cdot \underbrace{b^{\frac{p-1}{2}}}_{-1} = a_2^{\frac{p-1}{8}} \cdot b^{\frac{p-1}{2}} = \big(a_2 \cdot b^4\big)^m

and determine recursively the square root of a3:=a2b4=1a_3 := a_2 \cdot b^4 = 1:

 m  m  p18=2  \frac{p-1}{8} = 2  p116=1  \frac{p-1}{16} = 1 
 a3m  a_3^m 1111

We now finally have the desired situation that m=1m = 1 is odd and a3m=1a_3^m = 1, so we can compute one of the square roots of a3a_3: a3:=a3m+12=a31+12=a3=1\sqrt{a_3} := a_3^{\frac{m+1}{2}} = a_3^{\frac{1+1}{2}} = a_3 = 1

We now recall that a3=a2b4a_3 = a_2 \cdot b^4, so it follows that the square root of a2a_2 is a2:=a3b2=132=2\sqrt{a_2} := \sqrt{a_3} \cdot b^{-2} = 1 \cdot 3^{-2} = 2

Similarly, since a2=ab2a_2 = a \cdot b^2, it follows that a=a2b1=231=26=12\sqrt{a} = \sqrt{a_2} \cdot b^{-1} = 2 \cdot 3^{-1} = 2 \cdot 6 = 12

Hence, the square roots of a=8a = 8 modulo p=17p = 17 are ±12\pm 12, that is, 55 and 1212.

The Tonelli-Shanks algorithm

We now state the Tonelli-Shanks algorithm in general:

  1. Set m:=p12m := \frac{p-1}{2}. If am=1a^m = -1, then return “not a square”, because aa is not a square modulo pp by Euler’s theorem (see above) and ama^m is equal to the Legendre symbol.

  2. Otherwise we have that am=1a^m = 1. Set k:=1k := 1, so that m=p12km = \frac{p-1}{2^k}

  3. At this point we have the invariant that am=1a^m = 1 for m=p12km = \frac{p-1}{2^k}. While mm is even and am=1a^m = 1, set k:=k+1k := k + 1 and update m:=p12km := \frac{p-1}{2^k}.

  4. If am=1a^m = 1, then the square root of aa is am+12a^{\frac{m+1}{2}} by the above discussion, since mm is odd.

  5. Otherwise it holds that am=1a^m = -1 for m=p12km = \frac{p-1}{2^k}, where k2k \ge 2. Choose an arbitrary non-square b(Z/pZ)Qpb \in (\mathbb{Z}/p\mathbb{Z})^* \setminus Q_p by repeatedly picking b(Z/pZ)b \in (\mathbb{Z}/p\mathbb{Z})^* uniformly at random until (bp)=bp12=1(\frac{b}{p}) = b^{\frac{p-1}{2}} = -1.

  6. Run the same algorithm recursively from step 3 with the variable aa set to a:=ab2k1a' := a \cdot b^{2^{k-1}} and other variables left unchanged. This establishes the invariant of step 3, because

1=am1bp121=ap12k1bp121=ap12k(b2k1)p12k=(ab2k1)p12k=(ab2k1a)m\begin{aligned} 1 = \underbrace{a^m}_{-1} \cdot \underbrace{b^{\frac{p-1}{2}}}_{-1} = \underbrace{a^{\frac{p-1}{2^k}}}_{-1} \cdot \underbrace{b^{\frac{p-1}{2}}}_{-1} &= a^{\frac{p-1}{2^k}} \cdot \Big(b^{2^{k-1}}\Big)^{\frac{p-1}{2^k}} \\ &= \Big(a \cdot b^{2^{k-1}}\Big)^{\frac{p-1}{2^k}} = \Big(\underbrace{a \cdot b^{2^{k-1}}}_{a'}\Big)^m \end{aligned}
  • Let a\sqrt{a'} be the square root of aa' obtained from the recursive call. We are interested in computing a\sqrt{a} such that (a)2=a(\sqrt{a})^2 = a (interpret a\sqrt{a'} and a\sqrt{a} as new variables, not as operations). Observe that
(a)2=a=ab2k1=(a)2b2k1\big(\sqrt{a'}\big)^2 = a' = a \cdot b^{2^{k-1}} = (\sqrt{a})^2 \cdot b^{2^{k-1}}
  • Bringing (a)2(\sqrt{a})^2 to the left side yields
(a)2=(a)2b2k1=(ab2k2)2(\sqrt{a})^2 = \big(\sqrt{a'}\big)^2 \cdot b^{-2^{k-1}} = \Big(\sqrt{a'} \cdot b^{-2^{k-2}}\Big)^2
  • Note that we have used the fact that k2k \ge 2 here. It follows that ab2k2\sqrt{a'} \cdot b^{-2^{k-2}} is a square root of aa modulo pp.

We now argue why this algorithm always terminates and is correct. Observe that mm is changed throughout the algorithm in a way which ensures that am{1,1}a^m \in \{1, -1\} always holds. This is the case, because 1,11, -1 are the only square roots of 11 modulo pp and in every step we divide the exponent mm only by two. The case when am=1a^m = -1 is reduced to the case when am=1a^m = 1 but for a different aa, such that the square root for the original aa can be recovered from the square root of the new aa. Finally, observe that the algorithm either terminates directly because am=1a^m = 1 holds for an odd mm, or kk gets increased by at least one and hence mm gets divided by a power of two greater than one. This guarantees termination, because iterative division of mm by two will at some point yield an odd number whereupon the algorithm will terminate.

We now make a few additional remarks about the Tonelli-Shanks algorithm.

  • If p12\frac{p-1}{2} is odd or, equivalently, if p3mod4p \equiv 3 \mod 4, then the algorithm will not make any recursive calls and output the square root am+12=ap+14a^{\frac{m+1}{2}} = a^{\frac{p+1}{4}} of aa.

  • The loop in step 3 of the algorithm can be optimized by replacing it with a binary search algorithm for finding an mm such that am=1a^m = -1. If during such binary search, we have am=1a^m = 1, then we continue searching for smaller values of mm. If am{1,1}a^m \notin \{1, -1\}, then this means that we have divided mm too many times, so it follows that binary search is to be continued for larger values of mm. Although this optimization is beautiful theoretically, in practice it may only give a very tiny performance gain, if any.

  • The non-square bb used in the algorithm needs to be found and computed only once. During every recursive call, the same value of bb can be reused to deal with the am=1a^m = -1 situation. This implies that b1b^{-1} needs to be computed only once, so the computation of the ab2k2\sqrt{a'} \cdot b^{-2^{k-2}} value can be optimized.

  • As regards the expected time needed to find a non-square bb via guessing, since exactly half of the elements of (Z/pZ)(\mathbb{Z}/p\mathbb{Z})^* are non-squares, two is the expected number of times we need to generate bb uniformly at random and to test whether it is a non-square or not.

  • The expected running time of the algorithm is O(log(p)2)O(\log(p)^2), if we assume that multiplication modulo pp is a constant-time operation.

Implementation in Python

In this section I would like to show how the Tonelli-Shanks algorithm discussed above can be implemented in Python. We will need the following utility algorithms:

I will reuse the implementation of the extended Euclidean algorithm I described and proved correct in this post. We are now ready to dive into the code:

# file: "tonellishanks.py"
import random


def power_modulo(a: int, b: int, n: int) -> int:
    """ Computes a ^ b mod n """
    result = 1
    while b != 0:
        if b % 2 == 1:
            # b odd
            result = (result * a) % n
        a = (a * a) % n
        b >>= 1
    return result


def extended_gcd(a: int, b: int) -> (int, int, int):
    # optional check
    if a == 0:
        return b, 0, 1

    # without this check the first iteration will divide by zero
    if b == 0:
        return a, 1, 0

    un_prev = 1
    vn_prev = 0
    un_cur = 0
    vn_cur = 1

    while True:
        qn = a // b
        new_r = a % b
        a = b
        b = new_r

        if b == 0:
            return a, un_cur, vn_cur

        # Update coefficients
        un_new = un_prev - qn * un_cur
        vn_new = vn_prev - qn * vn_cur

        # Shift coefficients
        un_prev = un_cur
        vn_prev = vn_cur
        un_cur = un_new
        vn_cur = vn_new


def inverse_modulo(a: int, n: int) -> int:
    _, b, _ = extended_gcd(a, n)
    return b % n


def legendre_symbol(a: int, p: int, /) -> int:
    return power_modulo(a, (p - 1) >> 1, p)


def _choose_b(p: int, /) -> int:
    b = 2
    while legendre_symbol(b, p) == 1:
        b = random.randrange(2, p)
    return b


def _tonelli_shanks_recursive(a: int, k: int, p: int, b: int, b_inverse: int, /):
    """
    Computes a square root of a modulo prime p
    :param a: the number to take the square root of
    :param k: positive integer, such that a^m = 1 (mod p) where m = (p-1)/(2^k)
    :param p: odd prime p modulo which we are working
    :param b: an arbitrary non-square modulo p
    :param b_inverse: the inverse of b modulo p, i.e., b * b_inverse = 1 (mod p)
    :return: one of the square roots of a modulo p (the other can be obtained via negation modulo p)
    """

    m = (p - 1) >> k
    a_m = 1

    while m % 2 == 0 and a_m == 1:
        m >>= 1
        k += 1
        a_m = power_modulo(a, m, p)

    if a_m == p - 1:
        # a^m = -1 (mod p)
        b_power = 1 << (k - 1)
        b_power_half = 1 << (k - 2)
        a_next = (a * power_modulo(b, b_power, p)) % p
        a_next_root = _tonelli_shanks_recursive(a_next, k, p, b, b_inverse)
        a_root = a_next_root * power_modulo(b_inverse, b_power_half, p)
        return a_root % p

    # we now handle the case when m is odd
    # this case is easy, a^((m+1)/2) is a square root of a
    return power_modulo(a, (m + 1) >> 1, p)


def tonelli_shanks(a: int, p: int, /) -> int | None:
    """
    Computes a square root of a modulo prime p
    :param a: the number to take the square root of
    :param p: odd prime p modulo which we are working
    :return: one of the square roots of a modulo p (the other can be obtained via negation modulo p)
    """

    if legendre_symbol(a, p) != 1:
        # a is not not a square modulo p
        return None

    b = _choose_b(p)
    b_inverse = inverse_modulo(b, p)
    return _tonelli_shanks_recursive(a, 1, p, b, b_inverse)

Recall that performing a left bitwise shift (<<) is essentially a fast way to multiply the left operand with two raised to the right operand value’s power. Similarly, a right bitwise shift (>>) divides the left operand by two raised to the right operand value’s power.

Running the implementation on the example of the previous section, that is, for a=8a = 8 and p=17p = 17 yields the expected correct roots:

$ python3 -i tonellishanks.py
>>> tonelli_shanks(8, 17)
5
>>> tonelli_shanks(8, 17)
12

Note: The above implementation is a simplified version of my thoroughly tested implementation of the algorithm available in this GitHub repository.

I find the problem of computing square roots modulo a prime especially interesting because it is in some sense very “close” to problems where there are very good reasons to believe that they are hard in the complexity-theoretic sense. For example, the same problem is not known to be efficiently solvable if we are working modulo a Blum number, that is, if we are interested in solving b2amodpqb^2 \equiv a \mod p \cdot q

for bb, where pqp \neq q are prime numbers such that pq3mod4p \equiv q \equiv 3 \mod 4. Moreover, this modified problem is at least as hard as the problem of factoring Blum numbers, because given an oracle for computing the four square roots for aa, one could factor a given Blum number n=pqn = p \cdot q efficiently as follows:

  1. Choose a number x(Z/nZ)x \in (\mathbb{Z}/n\mathbb{Z})^* uniformly at random (can be implemented by simply choosing a random number xx between 22 and n1n - 1 inclusively, because if x(Z/nZ)x \notin (\mathbb{Z}/n\mathbb{Z})^*, then we have already effectively guessed the factor 1<gcd(x,n)<n1 < \gcd(x, n) < n of nn)

  2. Use the (hypothetical) oracle to compute some square root rr of x2x^2 modulo nn.

  3. If rxmodnr \equiv x \mod n or rxmodnr \equiv -x \mod n, then we have failed to factor nn.

  4. Otherwise it holds that (rx)(r+x)0modn(r - x) \cdot (r + x) \equiv 0 \mod n and the factors of nn are gcd(rx,n)\gcd(r - x, n) and gcd(r+x,n)\gcd(r + x, n).

Since x(Z/nZ)x \in (\mathbb{Z}/n\mathbb{Z})^* is chosen at random and we only pass x2x^2 to the oracle, the oracle’s answer is independent of xx. It is not hard to show that every quadratic residue modulo a Blum number has exactly 4 distinct roots (using the Chinese Remainder Theorem), two of them are xx and x-x which do not yield a factor of nn in the algorithm above. It follows that the probability of successfully factoring a number is 1/21/2, meaning that the algorithm above is a randomized polynomial-time Turing reduction. Moreover, the success probability of the above algorithm can actually be significantly amplified, but this is outside of the scope of this post. It is also worth noting that the above security reduction is useful in cryptography. For example, the Rabin cryptosystem’s security heavily relies on the assumption that computing square roots modulo a Blum number is computationally hard, which as we have shown above is implied by the assumption that factoring Blum numbers is hard, which is a plausible hardness assumption to date. If Rabin’s cryptosystem did all computations modulo a prime instead of a Blum number, then the cipher would be easily breakable, since it would then be possible to decrypt plaintexts using the Tonelli-Shanks algorithm.


Copyright © 2019 — 2022 Alexander Mayorov. All rights reserved.