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**K. -C. Yang and J. -L. Lin National Tsing Hua University**

PRIMES K. -C. Yang and J. -L. Lin National Tsing Hua University

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**OUTLINE Definition And History of Prime PRIMES is in P**

Previous Researches Basic Idea and Approach Preliminary Notation The Algorithm And Verification Time Complexity Analysis Future Works

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**History History Definition**

Let p N and p > 1, p is prime if it has no positive divisor other than 1 and p. History Pythagoras (580 BC ~ 300 BC) Integer (odd, even, prime, …), Rational and Irrational number, Pythagorean Theorem… Euclid (300 BC) There are an infinite number of primes.

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**History (2) pf. Assume there are finite number of primes.**

Let p1, …, pn be all primes, and let N = p1p2…pn + 1 N is a composite number and N has a prime factor p p1, …, pn Contradiction

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**History (3) PRIMES is in P - O(logk n) for k≧1.**

How to determine if a number is prime? Sieve of Eratosthenes (240 BC) If n is composite, then n has a positive divisor less than or equal to n1/2. So to determinate whether n is prime, you can try dividing n to every m < n1/2. This is an exponential-time algorithm O(n1/2 log n). PRIMES is in P - O(logk n) for k≧1.

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**Fermat (1) Fermat’s Last Theorem (AD 1637)**

xn + yn = zn has no integer solution for n > 2 Proven by Wiles (AD 1995) Fermat’s Little Theorem (AD 1640) a N and p is prime, then ap-1 ≡1 (mod p) e.g. p = 2, a = 3, then 32 ≡ 1 (mod 2) p = 3, a = 4, then 43 ≡ 1 (mod 3) p | ap-1 - 1

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**Fermat (2) pf. of Fermat’s little theorem (by induction)**

ap-1 ≡1 (mod p) ap - a ≡ 0 (mod p) p | ap - a Assume p | ap - a, then examine (a + 1)p - (a + 1) (binomial theorem) p divides the right side, so it also divides the left side. p | (a + 1)p - (ap + 1) + (ap - a) = (a + 1)p - (a + 1) The hypothesis is true for any a.

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**Fermat (3) Time complexity – O(lg n)**

If ap-1 ≡1 (mod p) for a N , p is prime? It fails! 341 341 = 11 × 31 2340 ≡1 (mod 341) Pseudo primes: 341, 561 , 645, 1105…

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Previous Researches 1975, Miller designed a test based on Fermat Little Theorem deterministic polynomial-time algorithm – O(log4 n) Assuming Extended Riemann Hypothesis 1980, Miller’s algorithm was modified by Rabin Unconditional but randomized polynomial-time 1983, Adleman, Pomerance and Rumely deterministic in (log n)O(logloglog n) 1986, Goldwasser and Kilian randomized polynomial-time algorithm (on almost all input) 1992, G-K algorithm was modified by Adleman and Huang randomized polynomial-time algorithm on all inputs 2002, Manindra Agrawal, Neeraj Kayal, and Nitin Saxena deterministic polynomial-time O(log7.5+εn) by using algebra

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**Riemann Hypothesis (1) In 1859, proposed by Riemann Hilbert’s problems**

23 problems. The Second International Congress of Mathematicians, 1900. Three of Hilbert’s problems remain unconquered. 6. Can physics be axiomized? 8. Riemann hypothesis. 16. Develop a topology of real algebraic curves and surfaces. Partial answer by Oxenhielm, Stockholm University, 2003

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**Riemann Hypothesis (2) Riemann zeta function Trivial zero point**

-2, -4, -6, -8, … Riemann Hypothesis non trivial zero point in Reimann zeta function, σ= ½. Clay Mathematics Institute $ for the solution to this problem. ( )

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**Manindra Agrawal, Neeraj Kayal, and Nitin Saxena August 6, 2002**

PRIMES is in P Manindra Agrawal, Neeraj Kayal, and Nitin Saxena August 6, 2002

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**Basic Idea and Approach (1)**

Let aZ, nN, and (a, n) = 1. Then n is prime iff (X + a)n≡(Xn + a) (mod n) pf. If n is prime n | (X + a)n – (Xn + a) (X - a)n≡(Xn - a) (mod n) If n is composite, let q be prime, qk | n, but qk+1 | n n | (X + a)n – (Xn + a) an – a = a(an-1 – 1) ∵n | an-1 -1 (Fermat’s little thm) n | an - a (n, a) = 1 (qk, an-q) = 1

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**Basic Idea and Approach (2)**

To evaluate n coefficients, it costs time Ω(n). To shorten the number of coefficients, we use (x + a)n ≡ (xn + a) (mod xr – 1, n) If p is prime, the above congruence holds. However, some composite numbers still satisfy this congruence. For appropriate r, n must be a prime power. e.g. 33, 75, 2×3×5

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**Basic Algorithm Input n > 1**

1. If ( n = ab for some a N and b > 1), output COMPOSITE. 2. Find the smallest r such that or(n) > 4log2n. 3. If (gcd(n, a) > 1 for some a ≦ r) , output COMPOSITE. 4. If (n ≦ r), output PRIME. 5. For a = 1 to do if , output COMPOSITE. 6. Output PRIME. Notation: or(n) = d denotes the smallest positive integer d s.t. nd ≡ 1 (mod r) Notation2: ψ(r) = |k|, where k < r and (k, r) = 1

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**Preliminary Notation (1)**

Fn denotes the finite field, where n is a prime. Let n and r be prime numbers, n ≠ r. 1. The multiplicative group of any field Fn, denoted by Fn* is cyclic. 2. Let f(x) be a polynomial with integral coefficients. Then f(x)n≡ f(xn) (mod n) 3. Let h(x) be any factor of xr - 1. Let m≡mr (mod r). Then xm ≡ xmr (mod h(x)) 4. In Fn, factorizes into irreducible polynomial each of degree or(n).

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**Preliminary Notation (2)**

Let f(x) be a polynomial with integral coefficients. Then f(x)n≡ f(xn) (mod n) pf. Let f(x) = a0 + … + adxd. The coefficient ci of xi in f(x)n is n | ci unless some ij is n. In this exception case, im = 0 for all m ≠ j. i = j × ij = nj. And cnj = ajn (mod n). Therefore, cnj ≡ aj (mod n) (Fermat’s Little Theorem) f(x)n ≡ c0 + cnxn + c2nx2n + … + cndxnd (mod n) ≡ a0 + a1xn + a2x2n + … + adxnd (mod n) ≡ f(xn) (mod n) xi1 × x2i2 × … × xdid = xi1+2i2…+did cnj = ajn + n ×Δ

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**Preliminary Notation (3)**

Let h(x) be any factor of xr – 1. Let m≡mr (mod r). Then xm ≡ xmr (mod h(x)) pf. Let m = kr + mr. Now xr ≡ 1 (mod xr - 1) xkr ≡ 1 (mod xr - 1) xkr+mr ≡ xmr (mod xr - 1) xm ≡ xmr (mod xr - 1) xm ≡ xmr (mod h(x)) xr-1 | xm-xmr h(x) ×Δ | xm-xmr h(x) | xm-xmr

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**Preliminary Notation (4)**

In Fn, factorizes into irreducible polynomial each of degree or(n). Let d = or(n) and h(x) be a irreducible factor of with degree k. Fn[x]/h(x) forms a field of size nk and the multiplicative subgroup of Fn[x]/h(x) is cyclic with a generator g(x) (by fact 1). We have g(x)n ≡ g(xn) (fact 2) g(x)nd ≡ g(xnd) g(x)nd ≡ g(x) g(x)nd-1 ≡ 1 ∵ Order of g(x) = (nk - 1), ∴(nk - 1)|(nd - 1) k | d. ∵ h(x) | (xr – 1), we also have xr ≡ 1 in Fn[x]/h(x) order of x in this field must be r (∵ r is prime). Therefore, r | (nk - 1), i.e. nk ≡ 1 (mod r) Hence, d | k. Therefore, k = d. g(xn) ≡ g(xn) g(xn)n ≡ g(xn2) g(xn2)n ≡ g(xn3) … g(x)nd ≡ g(xnd) pn ≡ 1 (mod r) xnd ≡ x1 (mod h(x)) (by fact 3) g(xnd) ≡ g(x)

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Algorithm Input n > 1 1. If ( a N and b > 1 s.t. n = ab ), output COMPOSITE. 2. Find the smallest r such that or(n) > 4log2n. 3. If ( a ≦ r s.t. 1 < gcd(n, a) < n ) , output COMPOSITE. 4. If (n ≦ r), output PRIME. 5. For a = 1 to do if , output COMPOSITE. 6. Output PRIME. Notation: (n, r) = 1, or(n) = d denotes the smallest positive integer d s.t. nd ≡ 1 (mod r) Notation2: ψ(r) = |k|, where k < r and (k, r) = 1

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**Correctness (1) Lemma. If n is prime, the algorithm returns PRIME. pf.**

1. Step 1 and Step 3 can never return COMPOSITE. n≠ab (a, n) = 1 or n a ≦ r 2. Step 5 also can not return COMPOSITE. If p is prime, (x + a)n ≡ (xn + a) (mod xr – 1, n) holds It returns PRIME either in Step 4 or Step 6.

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**Correctness (1) Lemma. If the algorithm returns PRIME, n is prime.**

If it returns PRIME in Step 4 then n must be prime. ∵n ≦ r , and (n, a) = 1 or n a ≦ r The remaining case: It returns PRIME in Step 6. (n, 1) = 1 (n, 2) = 1 … (n, n -1) = 1 (n, n) = n

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**Correctness (2) Find an appropriate r in Step 2.**

rt Find an appropriate r in Step 2. Lemma. There exist an r ≦ 16lg5n s.t. or(n) > 4lg2n pf. Let r1, r2, …, rt be all numbers s.t. ori(n) ≦ 4lg2n, note that t ≦ 16lg5n 1 2 3 16lg5n ∵n ≦ 2lgn Let ori(n) = k nk≡1 (mod ri) ri | nk - 1 < n1n2…n4lg2n = n8lg4n+2lg2n < n16lg4n

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**Correctness (3) lcm (r1, …, rt) |Π (ni - 1) < 216lg5n**

However, lcm (1, …, 16lg5n) > 216lg5n Therefore, t < 216lg5n r {ri | 0 ≦ i ≦ t}, but r < 16lg5n, and or(n) > 4lg2n Lemma. lcm (1, 2, …, m) ≧ 2m for m>6

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**Correctness (4) Assume n is composite. Let p be prime and p | n**

We fix p and r in the remainder sections. Set l = (X + a)n ≡ Xn + a (mod Xr - 1, n) for 1≦ a ≦ l (X + a)n ≡ Xn + a (mod Xr - 1, p) for 1≦ a ≦ l (X + a)p ≡ Xp + a (mod Xr - 1, p) for 1≦ a ≦ l ∵p is prime and (a, p) = 1

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Correctness (5) Definition. For polynomial f(X) and number m N, we say that m is introspective for f(X) if [f(X)]m ≡ f(Xm) (mod Xr – 1, p) n, p are introspective for f(X) = X + a Lemma. If m and m’ are introspective numbers for f(X) then so is m × m’ pf. [f(X)]mm’ ≡ [f(Xm)]m’ (mod Xr - 1, p) Let Y = Xm, [f(Y)]m’, [f(Y)]m’ ≡ f(Ym’) (mod Yr - 1, p) [f(Xm)]m’ ≡ f(Xmm’) (mod Xr - 1, p) [f(X)]mm’ ≡ f(Xmm’) (mod Xr - 1, p) Yr - 1 = Xmr - 1 Xr - 1 | Xmr – 1

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Correctness (6) Lemma. If m is introspective for f(X) and g(X) then so is f(X)g(X) pf. claim: [f(X)g(X)]m ≡ f(Xm)g(Xm) (mod Xr – 1, p) [f(X)]m ≡ f(Xm) (mod Xr – 1, p) [g(X)]m ≡ g(Xm) (mod Xr – 1, p) [f(X)]m[g(X)]m ≡ f(Xm)g(Xm) (mod Xr – 1, p)

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**Lemma 4.5. If m and m are introspective numbers for f(x) then so is m m.**

Lemma 4.6. If m is introspective for f(x) and g(x) then it is also introspective for f(x) g(x).

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Set Lemma 4.5 and 4.6 implies that every number in the set I is instropective for every polynomials in the set P. i,e,

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**Define G be the set of all residues of numbers in I modulo r , then G is a subgroup of**

Let |G| = t , and since or(n) > 4log2n, t > 4log2n.

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Lemma 4.7.

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**Lemma 4.8. If n is not a power of p,**

then

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**Lemma 4.9. If the algorithm returns PRIME then n is prime.**

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O(log3n) O(log7n) (log5n r’s) O(rlogn)= O(log6n) Each equation : O(rlog2n) Total : O(log10.5n)

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