Square Roots Modulo a Prime

Primality Tests

• primes asks if a number N is a prime.

• The classic algorithm tests if k | N for k = 2, 3, . . . ,√ N .

• But it runs in Ω(2^{(log2 N)/2}) steps.

Primality Tests (concluded)

• Suppose N = P Q is a product of 2 distinct primes.

• The probability of success of the density attack (p. 484) is

≈ 2

√N when P ≈ Q.

• This probability is exponentially small in terms of the input length log₂ N.

The Fermat Test for Primality

Fermat’s “little” theorem (p. 487) suggests the following primality test for any given number N:

1: Pick a number a randomly from { 1, 2, . . . , N − 1 };

2: if a^N−1 ≡ 1 mod N then

3: return “N is composite”;

4: else

5: return “N is (probably) a prime”;

6: end if

The Fermat Test for Primality (concluded)

• Carmichael numbers are composite numbers that will pass the Fermat test for all a ∈ { 1, 2, . . . , N − 1 }.^a

– The Fermat test will return “N is a prime” for all Carmichael numbers N.

• Unfortunately, there are infinitely many Carmichael numbers.^b

• In fact, the number of Carmichael numbers less than N exceeds N^2/7 for N large enough.

• So the Fermat test is an incorrect algorithm for primes.

aCarmichael (1910). Lo (1994) mentions an investment strategy based on such numbers!

bAlford, Granville, & Pomerance (1992).

Square Roots Modulo a Prime

• Equation x² ≡ a mod p has at most two (distinct) roots by Lemma 63 (p. 492).

– The roots are called square roots.

– Numbers a with square roots and gcd(a, p) = 1 are called quadratic residues.

∗ They are

1² mod p, 2² mod p, . . . , (p − 1)² mod p.

• We shall show that a number either has two roots or has none, and testing which is the case is trivial.^a

aBut no efficient deterministic general-purpose square-root-extracting algorithms are known yet.

Euler’s Test

Lemma 68 (Euler) Let p be an odd prime and a = 0 mod p.

1. If

a^(p−1)/2 ≡ 1 mod p, then x² ≡ a mod p has two roots.

2. If

a^(p−1)/2 ≡ 1 mod p, then

a^(p−1)/2 ≡ −1 mod p and x² ≡ a mod p has no roots.

The Proof (continued)

• Let r be a primitive root of p.

• Fermat’s “little” theorem says r^p−1 ≡ 1 mod p, so r^(p−1)/2

is a square root of 1.

• In particular,

r^(p−1)/2 ≡ 1 or −1 mod p.

• But as r is a primitive root, r^(p−1)/2 ≡ 1 mod p.

• Hence r^(p−1)/2 ≡ −1 mod p.

The Proof (continued)

• Let a = r^k mod p for some k.

• Suppose a^(p−1)/2 ≡ 1 mod p.

• Then

1 ≡ a^(p−1)/2 ≡ r^k(p−1)/2 ≡ 

r^(p−1)/2 _k

≡ (−1)^k mod p.

• So k must be even.

The Proof (continued)

• Suppose a = r^2j mod p for some 1 ≤ j ≤ (p − 1)/2.

• Then

a^(p−1)/2 ≡ r^j(p−1) ≡ 1 mod p.

• The two distinct roots of a are

r^j, −r^j(≡ r^j+(p−1)/2 mod p).

– If r^j ≡ −r^j mod p, then 2r^j ≡ 0 mod p, which implies r^j ≡ 0 mod p, a contradiction as r is a primitive root.

The Proof (continued)

• As 1 ≤ j ≤ (p − 1)/2, there are (p − 1)/2 such a’s.

• Each such a ≡ r^2j mod p has 2 distinct square roots.

• The square roots of all these a’s are distinct.

– The square roots of different a’s must be different.

• Hence the set of square roots is { 1, 2, . . . , p − 1 }.

• As a result,

a = r^2j mod p, 1 ≤ j ≤ (p − 1)/2, exhaust all the quadratic residues.

The Proof (concluded)

• Suppose a = r^2j+1 mod p now.

• Then it has no square roots because all the square roots have been taken.

• Finally,

a^(p−1)/2 ≡ 

r^(p−1)/2 _2j+1

≡ (−1)^2j+1 ≡ −1 mod p.

The Legendre Symbol^a and Quadratic Residuacity Test

• By Lemma 68 (p. 554),

a^(p−1)/2 mod p = ±1 for a ≡ 0 mod p.

• For odd prime p, define the Legendre symbol (a | p) as

(a | p) =

⎧⎪

⎪⎨

⎪⎪

⎩

0 if p | a,

1 if a is a quadratic residue modulo p,

−1 if a is a quadratic nonresidue modulo p.

• It is sometimes pronounced “a over p.”

aAndrien-Marie Legendre (1752–1833).

The Legendre Symbol and Quadratic Residuacity Test (concluded)

• Euler’s test (p. 554) implies

a^(p−1)/2 ≡ (a | p) mod p for any odd prime p and any integer a.

• Note that (ab | p) = (a | p)(b | p).

Gauss’s Lemma

Lemma 69 (Gauss) Let p and q be two distinct odd primes. Then (q | p) = (−1)^m, where m is the number of residues in R = { iq mod p : 1 ≤ i ≤ (p − 1)/2 } that are greater than (p − 1)/2.

• All residues in R are distinct.

– If iq = jq mod p, then p | (j − i) or p | q.

– But neither is possible.

• No two elements of R add up to p.

– If iq + jq ≡ 0 mod p, then p | (i + j) or p | q.

– But neither is possible.

The Proof (continued)

• Replace each of the m elements a ∈ R such that a > (p − 1)/2 by p − a.

– This is equivalent to performing −a mod p.

• Call the resulting set of residues R^.

• All numbers in R^ are at most (p − 1)/2.

• In fact, R^ = { 1, 2, . . . , (p − 1)/2 } (see illustration next page).

– Otherwise, two elements of R would add up to p,^a which has been shown to be impossible.

aBecause then iq ≡ −jq mod p for some i = j.

5 1 2 3 4

6 5

1 2 3 4

6

p = 7 and q = 5.

The Proof (concluded)

• Alternatively, R^ = { ±iq mod p : 1 ≤ i ≤ (p − 1)/2 }, where exactly m of the elements have the minus sign.

• Take the product of all elements in the two representations of R^.

• So

[(p − 1)/2]! = (−1)^mq^(p−1)/2[(p − 1)/2]! mod p.

• Because gcd([(p − 1)/2]!, p) = 1, the above implies 1 = (−1)^mq^(p−1)/2 mod p.

Legendre’s Law of Quadratic Reciprocity

• Let p and q be two distinct odd primes.

• The next result says (p | q) and (q | p) are distinct if and only if both p and q are 3 mod 4.

Lemma 70 (Legendre, 1785; Gauss)

(p | q)(q | p) = (−1)^{p−12 q−12} .

aFirst stated by Euler in 1751. Legendre (1785) did not give a cor- rect proof. Gauss proved the theorem when he was 19. He gave at least 8 different proofs during his life. The 152nd proof appeared in 1963. A computer-generated formal proof was given in Russinoff (1990).

As of 2008, there had been 4 such proofs. Wiedijk (2008), “the Law of Quadratic Reciprocity is the first nontrivial theorem that a student encounters in the mathematics curriculum.”

The Proof (continued)

• Sum the elements of R^ in the previous proof in mod2.

• On one hand, this is just _(p−1)/2

i=1 i mod 2.

• On the other hand, the sum equals mp +

(p−1)/2

i=1



iq − p

iq p



mod 2

= mp +

⎛

⎝q ^(p−1)/2

i=1

i − p

(p−1)/2

i=1

iq p

⎞⎠ mod 2.

– m of the iq mod p are replaced by p − iq mod p.

– But signs are irrelevant under mod2.

– m is as in Lemma 69 (p. 562).

The Proof (continued)

• Ignore odd multipliers to make the sum equal

m +

⎛

⎝^(p−1)/2

i=1

i −

(p−1)/2

i=1

iq p

⎞⎠ mod 2.

• Equate the above with _(p−1)/2

i=1 i modulo 2.

• Now simplify to obtain m ≡

(p−1)/2

i=1

iq p



mod 2.

The Proof (continued)

• _(p−1)/2

i=1 ^iq_p is the number of integral points below the line

y = (q/p) x for 1 ≤ x ≤ (p − 1)/2.

• Gauss’s lemma (p. 562) says (q | p) = (−1)^m.

• Repeat the proof with p and q reversed.

• Then (p | q) = (−1)^m, where m^ is the number of integral points above the line y = (q/p) x for

1 ≤ y ≤ (q − 1)/2.

The Proof (concluded)

• As a result,

(p | q)(q | p) = (−1)^m+m.

• But m + m^ is the total number of integral points in the [1, ^p−1₂ ] × [1, ^q−1₂ ] rectangle, which is

p − 1 2

q − 1 2 .

Eisenstein’s Rectangle

(p,q)

(p - 1)/2 (q - 1)/2

Above, p = 11, q = 7, m = 7, m^ = 8.

The Jacobi Symbol

• The Legendre symbol only works for odd prime moduli.

• The Jacobi symbol (a | m) extends it to cases where m is not prime.

– a is sometimes called the numerator and m the denominator.

• Trivially, (1 | m) = 1.

• Define (a | 1) = 1.

aCarl Jacobi (1804–1851).

The Jacobi Symbol (concluded)

• Let m = p¹p² · · · pk be the prime factorization of m.

• When m > 1 is odd and gcd(a, m) = 1, then

(a | m) =

k i=1

(a | p_i).

– Note that the Jacobi symbol equals ±1.

– It reduces to the Legendre symbol when m is a prime.

Properties of the Jacobi Symbol

The Jacobi symbol has the following properties when it is defined.

1. (ab | m) = (a | m)(b | m).

2. (a | m1m2) = (a | m1)(a | m2).

3. If a ≡ b mod m, then (a | m) = (b | m).

4. (−1 | m) = (−1)^(m−1)/2 (by Lemma 69 on p. 562).

5. (2 | m) = (−1)^(m2−1)/8.^a

6. If a and m are both odd, then (a | m)(m | a) = (−1)(a−1)(m−1)/4.

aBy Lemma 69 (p. 562) and some parity arguments.

Properties of the Jacobi Symbol (concluded)

• Properties 3–6 allow us to calculate the Jacobi symbol without factorization.

– It will also yield the same result as Euler’s test (p.

554) when m is an odd prime.

• This situation is similar to the Euclidean algorithm.

• Note also that (a | m) = 1/(a | m) because (a | m) = ±1.^a

aContributed by Mr. Huang, Kuan-Lin (B96902079, R00922018) on December 6, 2011.

Calculation of (2200 | 999)

(2200| 999) = (202 | 999)

= (2 | 999)(101 | 999)

= (−1)^(9992−1)/8(101| 999)

= (−1)¹²⁴⁷⁵⁰(101| 999) = (101 | 999)

= (−1)(100)(998)/4(999| 101) = (−1)²⁴⁹⁵⁰(999| 101)

= (999| 101) = (90 | 101) = (−1)^(1012−1)/8(45| 101)

= (−1)¹²⁷⁵(45| 101) = −(45 | 101)

= −(−1)(44)(100)/4(101| 45) = −(101 | 45) = −(11 | 45)

= −(−1)^(10)(44)/4(45| 11) = −(45 | 11)

= −(1 | 11) = −1.

A Result Generalizing Proposition 10.3 in the Textbook

Theorem 71 The group of set Φ(n) under multiplication mod n has a primitive root if and only if n is either 1, 2, 4, p^k, or 2p^k for some nonnegative integer k and an odd prime p.

This result is essential in the proof of the next lemma.

The Jacobi Symbol and Primality Test

Lemma 72 If (M | N) ≡ M^(N−1)/2 mod N for all M ∈ Φ(N), then N is a prime. (Assume N is odd.)

• Assume N = mp, where p is an odd prime, gcd(m, p) = 1, and m > 1 (not necessarily prime).

• Let r ∈ Φ(p) such that (r | p) = −1.

• The Chinese remainder theorem says that there is an M ∈ Φ(N) such that

M = r mod p, M = 1 mod m.

aMr. Clement Hsiao (B4506061, R88526067) pointed out that the text- book’s proof for Lemma 11.8 is incorrect in January 1999 while he was a senior.

The Proof (continued)

• By the hypothesis,

M^(N−1)/2 = (M | N) = (M | p)(M | m) = −1 mod N.

• Hence

M^(N−1)/2 = −1 mod m.

• But because M = 1 mod m,

M^(N−1)/2 = 1 mod m, a contradiction.

The Proof (continued)

• Second, assume that N = p^a, where p is an odd prime and a ≥ 2.

• By Theorem 71 (p. 577), there exists a primitive root r modulo p^a.

• From the assumption, M^N−1 =

M^(N−1)/2 2

= (M|N)² = 1 mod N for all M ∈ Φ(N).

The Proof (continued)

• As r ∈ Φ(N) (prove it), we have

r^N−1 = 1 mod N.

• As r’s exponent modulo N = p^a is φ(N) = p^a−1(p − 1), p^a−1(p − 1) | (N − 1),

which implies that p | (N − 1).

• But this is impossible given that p | N.

The Proof (continued)

• Third, assume that N = mp^a, where p is an odd prime, gcd(m, p) = 1, m > 1 (not necessarily prime), and a is even.

• The proof mimics that of the second case.

• By Theorem 71 (p. 577), there exists a primitive root r modulo p^a.

• From the assumption, M^N−1 =

M^(N−1)/2 2

= (M|N)² = 1 mod N for all M ∈ Φ(N).

The Proof (continued)

• In particular,

M^N−1 = 1 mod p^a (14)

for all M ∈ Φ(N).

• The Chinese remainder theorem says that there is an M ∈ Φ(N) such that

M = r mod p^a, M = 1 mod m.

• Because M = r mod p^a and Eq. (14), r^N−1 = 1 mod p^a.

The Proof (concluded)

• As r’s exponent modulo N = p^a is φ(N) = p^a−1(p − 1), p^a−1(p − 1) | (N − 1),

which implies that p | (N − 1).

• But this is impossible given that p | N.

The Number of Witnesses to Compositeness

Theorem 73 (Solovay & Strassen, 1977) If N is an

odd composite, then (M | N) ≡ M^(N−1)/2 mod N for at most half of M ∈ Φ(N).

• By Lemma 72 (p. 578) there is at least one a ∈ Φ(N) such that (a | N) ≡ a^(N−1)/2 mod N.

• Let B = { b1, b2, . . . , b_k } ⊆ Φ(N) be the set of all

distinct residues such that (bi | N) ≡ b^(N−1)/2_i mod N.

• Let aB = { abi mod N : i = 1, 2, . . . , k }.

• Clearly, aB ⊆ Φ(N), too.

The Proof (concluded)

• | aB | = k.

– ab_i ≡ ab_j mod N implies N | a(b_i − b_j), which is

impossible because gcd(a, N) = 1 and N > | bi − bj |.

• aB ∩ B = ∅ because

(abi)^(N−1)/2 ≡ a^(N−1)/2b^(N−1)/2_i ≡ (a | N)(bi | N) ≡ (abi | N).

• Combining the above two results, we know

| B |

φ(N) ≤ | B |

| B ∪ aB | = 0.5.

1: if N is even but N = 2 then

2: return “N is composite”;

3: else if N = 2 then

4: return “N is a prime”;

5: end if

6: Pick M ∈ { 2, 3, . . . , N − 1 } randomly;

7: if gcd(M, N ) > 1 then

8: return “N is composite”;

9: else

10: if (M | N ) ≡ M^(N−1)/2 mod N then

11: return “N is (probably) a prime”;

12: else

13: return “N is composite”;

14: end if

15: end if

Analysis

• The algorithm certainly runs in polynomial time.

• There are no false positives (for compositeness).

– When the algorithm says the number is composite, it is always correct.

Analysis (concluded)

• The probability of a false negative (again, for compositeness) is at most one half.

– Suppose the input is composite.

– By Theorem 73 (p. 585),

prob[ algorithm answers “no”| N is composite ] ≤ 0.5.

– Note that we are not referring to the probability that N is composite when the algorithm says “no.”

• So it is a Monte Carlo algorithm for compositeness.^a

aNot primes.

The Improved Density Attack for compositeness

All numbers < N

Witnesses to compositeness of

N via Jacobi Witnesses to

compositeness of N via common

factor

Randomized Complexity Classes; RP

• Let N be a polynomial-time precise NTM that runs in time p(n) and has 2 nondeterministic choices at each step.

• N is a polynomial Monte Carlo Turing machine for a language L if the following conditions hold:

– If x ∈ L, then at least half of the 2^p(n) computation paths of N on x halt with “yes” where n = | x |.

– If x ∈ L, then all computation paths halt with “no.”

• The class of all languages with polynomial Monte Carlo TMs is denoted RP (randomized polynomial time).^a

aAdleman & Manders (1977).

Comments on RP

• In analogy to Proposition 40 (p. 328), a “yes” instance of an RP problem has many certificates (witnesses).

• There are no false positives.

• If we associate nondeterministic steps with flipping fair coins, then we can phrase RP in the language of

probability.

– If x ∈ L, then N(x) halts with “yes” with probability at least 0.5 .

– If x ∈ L, then N(x) halts with “no.”

Comments on RP (concluded)

• The probability of false negatives is  ≤ 0.5.

• But any constant between 0 and 1 can replace 0.5.

– Repeat the algorithm k = −_log¹_{2 } times and answer

“no” only if all the runs answer “no.”

– The probability of false negatives becomes ^k ≤ 0.5.

Where RP Fits

• P ⊆ RP ⊆ NP.

– A deterministic TM is like a Monte Carlo TM except that all the coin flips are ignored.

– A Monte Carlo TM is an NTM with more demands on the number of accepting paths.

• compositeness ∈ RP;^a primes ∈ coRP;

primes ∈ RP.^b

– In fact, primes ∈ P.^c

• RP ∪ coRP is an alternative “plausible” notion of efficient computation.

aRabin (1976); Solovay & Strassen (1977).

bAdleman & Huang (1987).

cAgrawal, Kayal, & Saxena (2002).

ZPP

(Zero Probabilistic Polynomial)

• The class ZPP is defined as RP ∩ coRP.

• A language in ZPP has two Monte Carlo algorithms, one with no false positives (RP) and the other with no false negatives (coRP).

• If we repeatedly run both Monte Carlo algorithms, eventually one definite answer will come (unlike RP).

– A positive answer from the one without false positives.

– A negative answer from the one without false negatives.

aGill (1977).

The ZPP Algorithm (Las Vegas)

1: {Suppose L ∈ ZPP.}

2: {N¹ has no false positives, and N² has no false negatives.}

3: while true do

4: if N¹(x) = “yes” then

5: return “yes”;

6: end if

7: if N2(x) = “no” then

8: return “no”;

9: end if

10: end while

ZPP (concluded)

• The expected running time for the correct answer to emerge is polynomial.

– The probability that a run of the 2 algorithms does not generate a definite answer is 0.5 (why?).

– Let p(n) be the running time of each run of the while-loop.

– The expected running time for a definite answer is

∞ i=1

0.5ⁱip(n) = 2p(n).

• Essentially, ZPP is the class of problems that can be solved, without errors, in expected polynomial time.

Large Deviations

• Suppose you have a biased coin.

• One side has probability 0.5 +  to appear and the other 0.5 − , for some 0 <  < 0.5.

• But you do not know which is which.

• How to decide which side is the more likely side—with high confidence?

• Answer: Flip the coin many times and pick the side that appeared the most times.

• Question: Can you quantify your confidence?

The Chernoff Bound

Theorem 74 (Chernoff, 1952) Suppose x¹, x², . . . , xn are independent random variables taking the values 1 and 0 with probabilities p and 1 − p, respectively. Let X = _n

i=1 x_i. Then for all 0 ≤ θ ≤ 1,

prob[X ≥ (1 + θ) pn ] ≤ e^−θ2pn/3.

• The probability that the deviate of a binomial random variable from its expected value

E[ X ] = E

 _n



i=1

x_i



= pn decreases exponentially with the deviation.

aHerman Chernoff (1923–). The bound is asymptotically optimal.

The Proof

• Let t be any positive real number.

• Then

prob[X ≥ (1 + θ) pn ] = prob[ e^tX ≥ e^{t(1+θ) pn} ].

• Markov’s inequality (p. 535) generalized to real-valued random variables says that

prob

e^tX ≥ kE[ e^tX ]

≤ 1/k.

• With k = e^{t(1+θ) pn}/E[ e^tX ], we have^a

prob[X ≥ (1 + θ) pn ] ≤ e^{−t(1+θ) pn}E[ e^tX ].

aNote that X does not appear in k. Contributed by Mr. Ao Sun (R05922147) on December 20, 2016.

The Proof (continued)

• Because X = _n

i=1 xi and xi’s are independent, E[ e^tX ] = (E[ e^tx1 ])ⁿ = [ 1 + p(e^t − 1) ]ⁿ.

• Substituting, we obtain

prob[X ≥ (1 + θ) pn ] ≤ e^{−t(1+θ) pn}[ 1 + p(e^t − 1) ]ⁿ

≤ e^{−t(1+θ) pn}e^pn(et−1) as (1 + a)ⁿ ≤ e^an for all a > 0.

The Proof (concluded)

• With the choice of t = ln(1 + θ), the above becomes prob[X ≥ (1 + θ) pn ] ≤ epn[ θ−(1+θ) ln(1+θ) ].

• The exponent expands to

−θ²

2 + θ³

6 − θ⁴

12 + · · · for 0 ≤ θ ≤ 1.

• But it is less than

−θ²

2 + θ³

6 ≤ θ²



−1

2 + θ 6

≤ θ²



−1

2 + 1 6

= −θ² 3 .

Other Variations of the Chernoff Bound

The following can be proved similarly (prove it).

Theorem 75 Given the same terms as Theorem 74 (p. 599),

prob[X ≤ (1 − θ) pn ] ≤ e^−θ2pn/2.

The following slightly looser inequalities achieve symmetry.

Theorem 76 (Karp, Luby, & Madras, 1989) Given the same terms as Theorem 74 (p. 599) except with 0 ≤ θ ≤ 2,

prob[X ≥ (1 + θ) pn ] ≤ e^−θ2pn/4, prob[X ≤ (1 − θ) pn ] ≤ e^−θ2pn/4.

Power of the Majority Rule

The next result follows from Theorem 75 (p. 603).

Corollary 77 If p = (1/2) +  for some 0 ≤  ≤ 1/2, then prob

 _n



i=1

xi ≤ n/2



≤ e^−2n/2.

• The textbook’s corollary to Lemma 11.9 seems too loose, at e^−2n/6.^a

• Our original problem (p. 598) hence demands, e.g.,

n ≈ 1.4k/² independent coin flips to guarantee making an error with probability ≤ 2^−k with the majority rule.

aSee Dubhashi & Panconesi (2012) for many Chernoff-type bounds.

BPP

(Bounded Probabilistic Polynomial)

• The class BPP contains all languages L for which there is a precise polynomial-time NTM N such that:

– If x ∈ L, then at least 3/4 of the computation paths of N on x lead to “yes.”

– If x ∈ L, then at least 3/4 of the computation paths of N on x lead to “no.”

• So N accepts or rejects by a clear majority.

aGill (1977).