Lambert cylindrical equal-area projection

The Tonelli–Shanks algorithm (referred to by Shanks as the RESSOL algorithm) is used within modular arithmetic to solve a congruence of the form

$x^2\equiv n(\mathrm{mod}p)$

where n is a quadratic residue (mod p), and p is an odd prime.

Tonelli–Shanks cannot be used for composite moduli; finding square roots modulo composite numbers is a computational problem equivalent to integer factorization.

An equivalent, but slightly more redundant version of this algorithm was developed by Alberto Tonelli in 1891. The version discussed here was developed independently by Daniel Shanks in 1973, who explained:

"My tardiness in learning of these historical references was because I had lent Volume 1 of Dickson's History to a friend and it was never returned."

The algorithm

(Note: All $\equiv$ are taken to mean $(\mathrm{mod}p)$, unless indicated otherwise).

Inputs: p, an odd prime. n, an integer which is a quadratic residue (mod p), meaning that the Legendre symbol $\left(\frac{n}{p}\right)=1$.

Outputs: R, an integer satisfying $R^2\equiv n$.

1. Factor out powers of 2 from p − 1, defining Q and S as: $p-1=Q{2}^S$ with Q odd. Note that if $S=1$, i.e. $p\equiv 3(\mathrm{mod}4)$, then solutions are given directly by $R\equiv \pm n^{\frac{p+1}{4}}$.
2. Select a z such that the Legendre symbol $\left(\frac{z}{p}\right)=-1$ (that is, z should be a quadratic non-residue modulo p), and set $c\equiv z^Q$.
3. Let $R\equiv n^{\frac{Q+1}{2}},t\equiv n^Q,M=S.$
4. Loop:
   1. If $t\equiv 1$, return R.
   2. Otherwise, find the lowest i, $0<i<M$, such that $t^{2^i}\equiv 1$; e.g. via repeated squaring.
   3. Let $b\equiv c^{2^{M-i-1}}$, and set $R\equiv Rb,t\equiv t{b}^2,c\equiv b^2$ and $M=i$.

Once you have solved the congruence with R the second solution is p − R.

Example

Solving the congruence $x^2\equiv 10(\mathrm{mod}13)$. It is clear that $13$ is odd, and since $10^{\frac{13-1}{2}}=10^6\equiv 1(\mathrm{mod}13)$, 10 is a quadratic residue (by Euler's criterion).

• Step 1: Observe $p-1=12=3\cdot 2^2$ so $Q=3$, $S=2$.

• Step 2: Take $z=2$ as the quadratic nonresidue (2 is a quadratic nonresidue since $2^{\frac{13-1}{2}}=-1(\mathrm{mod}13)$ (again, Euler's criterion)). Set $c=2^3\equiv 8(\mathrm{mod}13).$

• Step 3: $R=10^2\equiv -4,t\equiv 10^3\equiv -1(\mathrm{mod}13),M=2.$

• Step 4: Now we start the loop: $t\equiv ̸1(\mathrm{mod}13)$ so $0<i<2$; i.e. $i=1.$
  • Let $b\equiv 8^{2^{2-1-1}}\equiv 8(\mathrm{mod}13)$, so $b^2\equiv 8^2\equiv -1(\mathrm{mod}13)$.
  • Set $R=-4\cdot 8\equiv 7(\mathrm{mod}13)$. Set $t\equiv -1\cdot -1\equiv 1(\mathrm{mod}13)$, and $M=1.$
  • We restart the loop, and since $t\equiv 1(\mathrm{mod}13)$ we are done, returning $R\equiv 7(\mathrm{mod}13).$

Indeed, observe that $7^2=49\equiv 10(\mathrm{mod}13)$ and naturally also $(-7{)}^2\equiv 6^2\equiv 10(\mathrm{mod}13)$. So the algorithm yields two solutions to our congruence.

Proof

First write $p-1=Q{2}^S$. Now write $r\equiv n^{\frac{Q+1}{2}}(\mathrm{mod}p)$ and $t\equiv n^Q(\mathrm{mod}p)$, observing that $r^2\equiv nt(\mathrm{mod}p)$. This latter congruence will be true after every iteration of the algorithm's main loop. If at any point, $t\equiv 1(\mathrm{mod}p)$ then $r^2\equiv n(\mathrm{mod}p)$ and the algorithm terminates with $R\equiv \pm r(\mathrm{mod}p)$.

If $t\equiv ̸1(\mathrm{mod}p)$, then consider $z$, a quadratic non-residue of $p$. Let $c\equiv z^Q(\mathrm{mod}p)$. Then $c^{2^S}\equiv (z^Q{)}^{2^S}\equiv z^{2^{S}Q}\equiv z^{p-1}\equiv 1(\mathrm{mod}p)$ and $c^{2^{S-1}}\equiv z^{\frac{p-1}{2}}\equiv -1(\mathrm{mod}p)$, which shows that the order of $c$ is $2^S$.

Similarly we have $t^{2^S}\equiv 1(\mathrm{mod}p)$, so the order of $t$ divides $2^S$. Suppose the order of $t$ is $2^{S^{\prime }}$. Since $n$ is a square modulo $p$, $t\equiv n^Q(\mathrm{mod}p)$ is also a square, and hence $S^{\prime }\le S-1$.

Now we set $b\equiv c^{2^{S-S^{\prime }-1}}(\mathrm{mod}p)$ and with this $r^{\prime }\equiv br(\mathrm{mod}p)$, $c^{\prime }\equiv b^2(\mathrm{mod}p)$ and $t^{\prime }\equiv c^{\prime }t(\mathrm{mod}p)$. As before, $r{\text{'}}^2\equiv n{t}^{\prime }(\mathrm{mod}p)$ holds; however with this construction both $t$ and $c^{\prime }$ have order $2^{S^{\prime }}$. This implies that $t^{\prime }$ has order $2^{S^{″}}$ with $S^{″}<S^{\prime }$.

If $S^{″}=0$ then $t^{\prime }\equiv 1(\mathrm{mod}p)$, and the algorithm stops, returning $R\equiv \pm r^{\prime }(\mathrm{mod}p)$. Else, we restart the loop with analogous definitions of $b^{\prime }$, $r^{″}$, $c^{″}$ and $t^{″}$ until we arrive at an $S^{(j{)}^{\prime }}$ that equals 0. Since the sequence of S is strictly decreasing the algorithm terminates.

Speed of the algorithm

The Tonelli–Shanks algorithm requires (on average over all possible input (quadratic residues and quadratic nonresidues))

$2m+2k+\frac{S(S-1)}{4}+\frac{1}{2^{S-1}}-9$

modular multiplications, where $m$ is the number of digits in the binary representation of $p$ and $k$ is the number of ones in the binary representation of $p$. If the required quadratic nonresidue $z$ is to be found by checking if a randomly taken number $y$ is a quadratic nonresidue, it requires (on average) $2$ computations of the Legendre symbol. The average of two computations of the Legendre symbol are explained as follows: $y$ is a quadratic residue with chance $\frac{\frac{p+1}{2}}{p}=\frac{1+\frac{1}{p}}{2}$, which is smaller than $1$ but $\ge \frac{1}{2}$, so we will on average need to check if a $y$ is a quadratic residue two times.

This shows essentially that the Tonelli–Shanks algorithm works very well if the modulus $p$ is random, that is, if $S$ is not particularly large with respect to the number of digits in the binary representation of $p$. Cipolla's algorithm works better than Tonelli–Shanks if (and only if) $S(S-1)>8m+20$.

The algorithm requires us to find a quadratic nonresidue $z$. There is no known deterministic algorithm that runs in polynomial time for finding such a $z$. However, if the generalized Riemann hypothesis is true, there exists a quadratic nonresidue $z<2{\ln }^{2}p$, making it possible to check every $z$ up to that limit and find a suitable $z$ within polynomial time. Keep in mind, however, that this is a worst case scenario; in general, $z$ is found in on average 2 trials as stated above.

Uses

The Tonelli–Shanks algorithm can (naturally) be used for any process in which square roots modulo a prime are necessary. For example, it can be used for finding points on elliptic curves. It is also useful for the computations in the Rabin cryptosystem.

Generalizations

Tonelli–Shanks can be generalized to any cyclic group (instead of $\mathbb{\mathbb{Z}}/p{\mathbb{\mathbb{Z}}}^{*}$) and to kth roots for arbitrary integer k, in particular to taking the kth root of an element of a finite field .

If many square-roots must be done in the same cyclic group and S is not too large, a table of square-roots of the elements of 2-power order can be prepared in advance and the algorithm simplified and speeded up as follows.

1. Factor out powers of 2 from p − 1, defining Q and S as: $p-1=Q{2}^S$ with Q odd.
2. Let $R\equiv n^{\frac{Q+1}{2}},t\equiv n^Q\equiv R^2/n$
3. Find $b$ from the table such that $b^2\equiv t$ and set $R\equiv R/b$
4. return R.

Notes

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References

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Pages 110–115 describe the algorithm and explain the group theory behind it.

• Daniel Shanks. Five Number Theoretic Algorithms. Proceedings of the Second Manitoba Conference on Numerical Mathematics. Pp. 51–70. 1973.
• Alberto Tonelli, Bemerkung über die Auflösung quadratischer Congruenzen. Nachrichten von der Königlichen Gesellschaft der Wissenschaften und der Georg-Augusts-Universität zu Göttingen. Pp. 344–346. 1891. [1]
• Gagan Tara Nanda - Mathematics 115: The RESSOL Algorithm [2]

External links

• Implementation in C# http://shankstonelli.blogspot.com/2010/12/shanks-tonelli-algorithm-in-c.html
• Implementation in Python http://eli.thegreenplace.net/2009/03/07/computing-modular-square-roots-in-python