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

2. 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

3. 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.

4. 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

5. 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.

6. 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

7. 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.

8. 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…

9. 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

10. 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

11. 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. ( )

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

13. Basic Idea and Approach (1)
Let aZ, nN, 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

14. 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

15. 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

16. 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).

17. 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

18. 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

19. 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)

20. 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

21. 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.

22. 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

23. 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

24. 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

25. 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

26. 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

27. 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)

28. 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).

29. 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,

30. 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.

31. Lemma 4.7.

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

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

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