Tonelli–Shanks algorithm

Given a non-zero ${\displaystyle n}$ and a prime ${\displaystyle p>2}$ (which will always be odd), Euler's criterion tells us that ${\displaystyle n}$ has a square root (i.e., ${\displaystyle n}$ is a quadratic residue) if and only if:

${\displaystyle n^{\frac {p-1}{2}}\equiv 1{\pmod {p}}}$.

In contrast, if a number ${\displaystyle z}$ has no square root (is a non-residue), Euler's criterion tells us that:

${\displaystyle z^{\frac {p-1}{2}}\equiv -1{\pmod {p}}}$.

It is not hard to find such ${\displaystyle z}$, because half of the integers between 1 and ${\displaystyle p-1}$ have this property. So we assume that we have access to such a non-residue.

By (normally) dividing by 2 repeatedly, we can write ${\displaystyle p-1}$ as ${\displaystyle Q2^{S}}$, where ${\displaystyle Q}$ is odd. Note that if we try

${\displaystyle R\equiv n^{\frac {Q+1}{2}}{\pmod {p}}}$,

then ${\displaystyle R^{2}\equiv n^{Q+1}=(n)(n^{Q}){\pmod {p}}}$. If ${\displaystyle t\equiv n^{Q}\equiv 1{\pmod {p}}}$, then ${\displaystyle R}$ is a square root of ${\displaystyle n}$. Otherwise, for ${\displaystyle M=S}$, we have ${\displaystyle R}$ and ${\displaystyle t}$ satisfying:

• ${\displaystyle R^{2}\equiv nt{\pmod {p}}}$; and
• ${\displaystyle t}$ is a ${\displaystyle 2^{M-1}}$-th root of 1 (because ${\displaystyle t^{2^{M-1}}=t^{2^{S-1}}\equiv n^{Q2^{S-1}}=n^{\frac {p-1}{2}}}$).

If, given a choice of ${\displaystyle R}$ and ${\displaystyle t}$ for a particular ${\displaystyle M}$ satisfying the above (where ${\displaystyle R}$ is not a square root of ${\displaystyle n}$), we can easily calculate another ${\displaystyle R}$ and ${\displaystyle t}$ for ${\displaystyle M-1}$ such that the above relations hold, then we can repeat this until ${\displaystyle t}$ becomes a ${\displaystyle 2^{0}}$-th root of 1, i.e., ${\displaystyle t=1}$. At that point ${\displaystyle R}$ is a square root of ${\displaystyle n}$.

We can check whether ${\displaystyle t}$ is a ${\displaystyle 2^{M-2}}$-th root of 1 by squaring it ${\displaystyle M-2}$ times and check whether it is 1. If it is, then we do not need to do anything, as the same choice of ${\displaystyle R}$ and ${\displaystyle t}$ works. But if it is not, ${\displaystyle t^{2^{M-2}}}$ must be -1 (because squaring it gives 1, and there can only be two square roots 1 and -1 of 1 modulo ${\displaystyle p}$).

To find a new pair of ${\displaystyle R}$ and ${\displaystyle t}$, we can multiply ${\displaystyle R}$ by a factor ${\displaystyle b}$, to be determined. Then ${\displaystyle t}$ must be multiplied by a factor ${\displaystyle b^{2}}$ to keep ${\displaystyle R^{2}\equiv nt{\pmod {p}}}$. So, when ${\displaystyle t^{2^{M-2}}}$ is -1, we need to find a factor ${\displaystyle b^{2}}$ so that ${\displaystyle tb^{2}}$ is a ${\displaystyle 2^{M-2}}$-th root of 1, or equivalently ${\displaystyle b^{2}}$ is a ${\displaystyle 2^{M-2}}$-th root of -1.

The trick here is to make use of ${\displaystyle z}$, the known non-residue. The Euler's criterion applied to ${\displaystyle z}$ shown above says that ${\displaystyle z^{Q}}$ is a ${\displaystyle 2^{S-1}}$-th root of -1. So by squaring ${\displaystyle z^{Q}}$ repeatedly, we have access to a sequence of ${\displaystyle 2^{i}}$-th root of -1. We can select the right one to serve as ${\displaystyle b}$. With a little bit of variable maintenance and trivial case compression, the algorithm below emerges naturally.

Operations and comparisons on elements of the multiplicative group of integers modulo p ${\displaystyle \mathbb {Z} /p\mathbb {Z} }$ are implicitly mod p.

Inputs:

• p, a prime
• n, an element of ${\displaystyle \mathbb {Z} /p\mathbb {Z} }$ such that solutions to the congruence r² = n exist; when this is so we say that n is a quadratic residue mod p.

Outputs:

• r in ${\displaystyle \mathbb {Z} /p\mathbb {Z} }$ such that r² = n

Algorithm:

1. By factoring out powers of 2, find Q and S such that ${\displaystyle p-1=Q2^{S}}$ with Q odd
2. Search for a z in ${\displaystyle \mathbb {Z} /p\mathbb {Z} }$ which is a quadratic non-residue
   • Half of the elements in the set will be quadratic non-residues
   • Candidates can be tested with Euler's criterion or by finding the Jacobi symbol
3. Let

   ${\displaystyle {\begin{aligned}M&\leftarrow S\\c&\leftarrow z^{Q}\\t&\leftarrow n^{Q}\\R&\leftarrow n^{\frac {Q+1}{2}}\end{aligned}}}$

4. Loop:
   • If t = 0, return r = 0
   • If t = 1, return r = R
   • Otherwise, use repeated squaring to find the least i, 0 < i < M, such that ${\displaystyle t^{2^{i}}=1}$
   • Let ${\displaystyle b\leftarrow c^{2^{M-i-1}}}$, and set

     ${\displaystyle {\begin{aligned}M&\leftarrow i\\c&\leftarrow b^{2}\\t&\leftarrow tb^{2}\\R&\leftarrow Rb\end{aligned}}}$

Once you have solved the congruence with r the second solution is ${\displaystyle -r{\pmod {p}}}$. If the least i such that ${\displaystyle t^{2^{i}}=1}$ is M, then no solution to the congruence exists, i.e. n is not a quadratic residue.

This is most useful when p ≡ 1 (mod 4).

For primes such that p ≡ 3 (mod 4), this problem has possible solutions ${\displaystyle r=\pm n^{\frac {p+1}{4}}{\pmod {p}}}$. If these satisfy ${\displaystyle r^{2}\equiv n{\pmod {p}}}$, they are the only solutions. If not, ${\displaystyle r^{2}\equiv -n{\pmod {p}}}$, n is a quadratic non-residue, and there are no solutions.

We can show that at the start of each iteration of the loop the following loop invariants hold:

• ${\displaystyle c^{2^{M-1}}=-1}$
• ${\displaystyle t^{2^{M-1}}=1}$
• ${\displaystyle R^{2}=tn}$

Initially:

• ${\displaystyle c^{2^{M-1}}=z^{Q2^{S-1}}=z^{\frac {p-1}{2}}=-1}$ (since z is a quadratic nonresidue, per Euler's criterion)
• ${\displaystyle t^{2^{M-1}}=n^{Q2^{S-1}}=n^{\frac {p-1}{2}}=1}$ (since n is a quadratic residue)
• ${\displaystyle R^{2}=n^{Q+1}=tn}$

At each iteration, with M' , c' , t' , R' the new values replacing M, c, t, R:

• ${\displaystyle c'^{2^{M'-1}}=(b^{2})^{2^{i-1}}=c^{2^{M-i}2^{i-1}}=c^{2^{M-1}}=-1}$
• ${\displaystyle t'^{2^{M'-1}}=(tb^{2})^{2^{i-1}}=t^{2^{i-1}}b^{2^{i}}=-1\cdot -1=1}$
  • ${\displaystyle t^{2^{i-1}}=-1}$ since we have that ${\displaystyle t^{2^{i}}=1}$ but ${\displaystyle t^{2^{i-1}}\neq 1}$ (i is the least value such that ${\displaystyle t^{2^{i}}=1}$)
  • ${\displaystyle b^{2^{i}}=c^{2^{M-i-1}2^{i}}=c^{2^{M-1}}=-1}$
• ${\displaystyle R'^{2}=R^{2}b^{2}=tnb^{2}=t'n}$

From ${\displaystyle t^{2^{M-1}}=1}$ and the test against t = 1 at the start of the loop, we see that we will always find an i in 0 < i < M such that ${\displaystyle t^{2^{i}}=1}$. M is strictly smaller on each iteration, and thus the algorithm is guaranteed to halt. When we hit the condition t = 1 and halt, the last loop invariant implies that R² = n.

Solving the congruence r² ≡ 5 (mod 41). 41 is prime as required and 41 ≡ 1 (mod 4). 5 is a quadratic residue by Euler's criterion: ${\displaystyle 5^{\frac {41-1}{2}}=5^{20}=1}$ (as before, operations in ${\displaystyle (\mathbb {Z} /41\mathbb {Z} )^{\times }}$ are implicitly mod 41).

1. ${\displaystyle p-1=40=5\cdot 2^{3}}$ so ${\displaystyle Q\leftarrow 5}$, ${\displaystyle S\leftarrow 3}$
2. Find a value for z:
   • ${\displaystyle 2^{\frac {41-1}{2}}=1}$, so 2 is a quadratic residue by Euler's criterion.
   • ${\displaystyle 3^{\frac {41-1}{2}}=40=-1}$, so 3 is a quadratic nonresidue: set ${\displaystyle z\leftarrow 3}$
3. Set
   • ${\displaystyle M\leftarrow S=3}$
   • ${\displaystyle c\leftarrow z^{Q}=3^{5}=38}$
   • ${\displaystyle t\leftarrow n^{Q}=5^{5}=9}$
   • ${\displaystyle R\leftarrow n^{\frac {Q+1}{2}}=5^{\frac {5+1}{2}}=2}$
4. Loop:
   • First iteration:
     • ${\displaystyle t\neq 1}$, so we're not finished
     • ${\displaystyle t^{2^{1}}=40}$, ${\displaystyle t^{2^{2}}=1}$ so ${\displaystyle i\leftarrow 2}$
     • ${\displaystyle b\leftarrow c^{2^{M-i-1}}=38^{2^{3-2-1}}=38}$
     • ${\displaystyle M\leftarrow i=2}$
     • ${\displaystyle c\leftarrow b^{2}=38^{2}=9}$
     • ${\displaystyle t\leftarrow tb^{2}=9\cdot 9=40}$
     • ${\displaystyle R\leftarrow Rb=2\cdot 38=35}$
   • Second iteration:
     • ${\displaystyle t\neq 1}$, so we're still not finished
     • ${\displaystyle t^{2^{1}}=1}$ so ${\displaystyle i\leftarrow 1}$
     • ${\displaystyle b\leftarrow c^{2^{M-i-1}}=9^{2^{2-1-1}}=9}$
     • ${\displaystyle M\leftarrow i=1}$
     • ${\displaystyle c\leftarrow b^{2}=9^{2}=40}$
     • ${\displaystyle t\leftarrow tb^{2}=40\cdot 40=1}$
     • ${\displaystyle R\leftarrow Rb=35\cdot 9=28}$
   • Third iteration:
     • ${\displaystyle t=1}$, and we are finished; return ${\displaystyle r=R=28}$

Indeed, 28² ≡ 5 (mod 41) and (−28)² ≡ 13² ≡ 5 (mod 41). So the algorithm yields the two solutions to our congruence.

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

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

modular multiplications, where ${\displaystyle m}$ is the number of digits in the binary representation of ${\displaystyle p}$ and ${\displaystyle k}$ is the number of ones in the binary representation of ${\displaystyle p}$. If the required quadratic nonresidue ${\displaystyle z}$ is to be found by checking if a randomly taken number ${\displaystyle y}$ is a quadratic nonresidue, it requires (on average) ${\displaystyle 2}$ computations of the Legendre symbol.[5] The average of two computations of the Legendre symbol are explained as follows: ${\displaystyle y}$ is a quadratic residue with chance ${\displaystyle {\tfrac {\tfrac {p+1}{2}}{p}}={\tfrac {1+{\tfrac {1}{p}}}{2}}}$, which is smaller than ${\displaystyle 1}$ but ${\displaystyle \geq {\tfrac {1}{2}}}$, so we will on average need to check if a ${\displaystyle y}$ is a quadratic residue two times.

This shows essentially that the Tonelli–Shanks algorithm works very well if the modulus ${\displaystyle p}$ is random, that is, if ${\displaystyle S}$ is not particularly large with respect to the number of digits in the binary representation of ${\displaystyle p}$. As written above, Cipolla's algorithm works better than Tonelli–Shanks if (and only if) ${\displaystyle S(S-1)>8m+20}$. However, if one instead uses Sutherland's algorithm to perform the discrete logarithm computation in the 2-Sylow subgroup of ${\displaystyle \mathbb {F} _{p}}$, one may replace ${\displaystyle S(S-1)}$ with an expression that is asymptotically bounded by ${\displaystyle O(S\log S/\log \log S)}$.[6] Explicitly, one computes ${\displaystyle e}$ such that ${\displaystyle c^{e}\equiv n^{Q}}$ and then ${\displaystyle R\equiv c^{-e/2}n^{(Q+1)/2}}$ satisfies ${\displaystyle R^{2}\equiv n}$ (note that ${\displaystyle e}$ is a multiple of 2 because ${\displaystyle n}$ is a quadratic residue).

The algorithm requires us to find a quadratic nonresidue ${\displaystyle z}$. There is no known deterministic algorithm that runs in polynomial time for finding such a ${\displaystyle z}$. However, if the generalized Riemann hypothesis is true, there exists a quadratic nonresidue ${\displaystyle z<2\ln ^{2}{p}}$,[7] making it possible to check every ${\displaystyle z}$ up to that limit and find a suitable ${\displaystyle z}$ within polynomial time. Keep in mind, however, that this is a worst-case scenario; in general, ${\displaystyle z}$ is found in on average 2 trials as stated above.

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 and in the sieving step of the quadratic sieve.

Tonelli–Shanks can be generalized to any cyclic group (instead of ${\displaystyle (\mathbb {Z} /p\mathbb {Z} )^{\times }}$) and to kth roots for arbitrary integer k, in particular to taking the kth root of an element of a finite field.[8]

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 sped up as follows.

1. Factor out powers of 2 from p − 1, defining Q and S as: ${\displaystyle p-1=Q2^{S}}$ with Q odd.
2. Let ${\displaystyle R\leftarrow n^{\frac {Q+1}{2}},t\leftarrow n^{Q}\equiv R^{2}/n}$
3. Find ${\displaystyle b}$ from the table such that ${\displaystyle b^{2}\equiv t}$ and set ${\displaystyle R\equiv R/b}$
4. return R.

• Ivan Niven; Herbert S. Zuckerman; Hugh L. Montgomery (1991). An Introduction to the Theory of Numbers (5th ed.). Wiley. pp. 110–115. ISBN 0-471-62546-9.
• 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.
• Gagan Tara Nanda - Mathematics 115: The RESSOL Algorithm
• Gonzalo Tornaria