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Primes in P

Deterministic polynomial-time algorithm of
Agrawal, Kayal and Saxena
 Presented by Vladimir Braverman
History

Sieve of Eratosthenes 240BC
 Randomized, polynomial-time algorithm of
Miller-Rabin, 1980 (practically used)
 Adleman, Pomerance and Rumeley (APR) a
deterministic algorithm, an almost
polynomial time NloglogN , 1983
 Agrawal, Kayal and Saxena, deterministic,
polinomial-time algorithm 2002
Impact

Data security
 Number Theory
 Theory of algorithms
 Doesn’t have a practical value yet
The Heroes

Neeraj Kayal and Nitin Saxena are
PhD students (22 years old).
 Indian Institute of Technology,
Kanpur
 Agrawal was recently awarded the
Clay Research Award, one of the
coveted honors in mathematical
research.
Group Theory
Number Theory
Finite Fields
Primes in P
Polynomial identity idea
Algorithms in Number Theory
Content

Background
– Algorithms in number theory
– Rings
– Number Theory

Basic idea: Polynomial identity
 Algorithm
– Pseudo code
– Correctness
– Running time
Algorithms in number theory

Lemma 1
– 0 < a,b < n
– ab can be calculated in O( log3 (n))
– Maximal integer less than b/a can be calculated
in O(log2 (n))
– a mod b can be calculated in O(log2 (n))
– Gcd(a,b) can be calculated in O(log3 (n))
Algebra

A ring is a set together with addition and
multiplication

1. Additive associativity:
 2. Additive commutativity:
 3. Additive identity.
 4. Additive inverse
 5. Multiplicative associativity:
 6. Left and right distributivity:

A ring is therefore an Abelian group under addition and
a semigroup under multiplication.
Algebra (cont)


A polynomial f is defined to be a formal
expression of the form
where the coefficients a0, ... , an are elements of
some ring R and X is considered to be a formal
symbol.
 Two polynomials are considered to be equal if and
only if the sequences of their coefficients are
equal.
Algebra (end)

Polynomials with coefficients in R can be added
by simply adding corresponding coefficients and
multiplied using the distributive low and the rules



X a = a X for all elements a of the ring R
Xk Xl = Xk+l for all natural numbers k and l.
The set of all polynomials with coefficients in the
ring R forms itself a ring, the ring of polynomials
over R, which is denoted by R[X].
Algorithms in number theory
(cont)

Lemma 2
Let 1 < a,b, r and n = max(a,r). Than ab mod r can
be calculated in O( log2 (n) + log(b) log2(r))

Lemma 3
Let 1< a,r < n. The r coefficients of
(x-a)n mod (xr-1) in Zn[x] can be computed in
O(r2 log3 (n)) time.
Algorithms in number theory
(cont)

Lemma 4
Let n > 2 be an integer. There is an algorithm that decides
in O(√n log2 (n)) whether n is prime.
Algorithm:
– r:=2; s:=4 (s = r2)
– While s <= n
– Do If n mod r = 0 return NO


Else r:=r+1; s:=2r-1
Endif
– EndWhile
– Return Yes
Algorithms in number theory
(cont)

Lemma 4 (cont)
– The correctness follows from the fact that n is
not prime iff there is an integer r <= √n that
divides n.
– The while-loop makes at most √n iterations
each one taking (log2 n) time.
Algorithms in number theory
(end)

Lemma 5
Let n > 2 be an integer. There is an algorithm that
computes the largest prime factor of n in
O(√n log2 (n)).

Lemma 6
Let n > 2 be an integer. There is an algorithm that
decides in O(log4 (n) log (log (n)) ) whether
there exist integers a,b such that n = ab.
Number theory

Lemma 7
If p is prime number than for any i < p
(pi) = 0 mod p
Proof
(pi) = p(p-1)…(p-i+1) / i!
Since gcd(p, i!) = 1 the lemma is proved
Number theory(cont)

Lemma 8
If p is prime number than for any a,
ap = a mod p
Proof (induction)

a=1
 a -> a+1
(a+1)p =  (pi) ai
(a+1)p = ap+1 = a+1 mod p
Polynomial identity

Theorem 1
Let a, n be integers
– If n is a prime number, then (x-a)n = xn-a in
the ring Zn[x]
– If gcd(a, n) =1 and n is not a prime number,
then (x-a)n <> xn-a in the ring Zn[x]
Proof of Theorem 1
(x-a)n =  (ni) xi(-a)n-I.
1. Prime case
–
–
2.
If n is prime, (ni) = 0 mod n, according to Lemma 7
Therefore
(x-a)n = xn – an = xn – a mod n,
according to Lemma 8
n is not a prime and gcd(n,a) = 1
q – prime factor of n, qk | n, k – maximal
gcd (a,q) = 1 and gcd (an-q, qk) = 1
Proof of Theorem 1(Cont)
Lemma: qk doesn’t divide (nq)
Proof: (by contradiction)
 Suppose, (nq) = b qk. Than,
n(n-1)…(n-q+1)/q! = b qk.
n = (q-1)! b qk+1 /(n-1)…(n-q+1).
 For each 0 < j < q , q doesn’t
divide (n-j). Indeed, since q | n and (n-j) = 0 mod
q we have j = 0 mod q.
 Therefore, (q-1)! b /(n-1)…(n-q+1) is an
integer and qk+1 | n that contradicts with maximal
property of k
Proof of Theorem 1(Cont)





(x-a)p The coefficient of xp is (nq)(-1)n-qan-q
If it is divisible by n, than (nq)an-q = bn
Hence, (nq)an-q/qk = bn / qk
Right-hand side is integer and
gcd(an-q,qk) = 1
Therefore qk | (nq) that contradicts Lemma
We proved that (x-a)n <> xn-a in the ring Zn[x]
Improved polynomial identity

Theorem 2
Let n be an integer and let q, r be prime numbers and
1. gcd (m,n) = 1 for all m < r
2. q divides r-1
3. q ≥ 2√r log(n) + 2
4. n(r-1)/q ≠ 1 mod r
5. (x-a)a = (xn-a) mod (xr-1) in Zn[x] for
a < 2√r log(n) + 1
Then n is a power of prime number.

Lemma 9
There exist constants 0 < c1 < c2 such that for any
large n, there exists a prime number r
1. c1 log6(n) < r < c2 log6(n)
2. r-1 has a prime factor q, q ≥ 2√r log(n) + 2
3. n(r-1)/q ≠ 1 mod r
Algorithm Prime(n)



r := 2; found := false;
While r < n and found = false
Do if gcd(r,n) ≠ 1 then return NO
–
–
If r is a prime and r > 2
Then q := largest prime factor of r-1



–
–

If q ≥ 2√r log(n) + 2 and n(r-1)/q ≠ 1 mod r
Then found = true
Endif
Endif
If found = false then r++ endif
Endwhile
Algorithm Prime(n) (Cont)
For a:= 1 to 2√r log(n) + 1
 Do if (x-a)a ≠ (xn-a) mod (xr-1) in Zn[x]

–
–





Then return NO
Endif
Endfor
If n = ab for some integers a,b ≥ 2
Then return NO
Else return YES
Endif
Correctness of the algorithm
(Sketch of the proof )

If n is prime,




Algorithm doesn’t return NO during While loop,
since gcd(n,r) = 1
By Theorem 1, (x-a)a ≠ (xn-a) mod (xr-1) in Zn[x]
Thus, the algorithm doesn’t return NO during For
loop
N is not in the form ab, therefore YES is returned
If n is not a prime, then, according to
Theorem 2, the algorithm returns NO
Running time
(Sketch of the proof )

While loop
Action
Time
Reason
gcd
O(log3(n))
r is prime
O(√r log2(r))
Lemma 4
largest prime factor of r-1
O(√r log2(r))
Lemma 5
n(r-1)/q mod r
O(log2(n) + log3(r))
Lemma 2
One while-loop iteration
O(log2(n) + (√r log2(r))
Number of while iterations
O(log6(n))
Lemma 9
While-loop
O( log9(n) log2(log(n)) )
r=O(log6(n))
Lemma 1
Running time(Cont)

For loop
Action
Time
Coefficient of (x-a)n mod (xr-1)
in Zn[x]
Note that the same time for
(xn -a)
O(r2 log3 (n))
For-loop
O(log19(n))

Testing if n is perfect power
– O(log4(n) log(log(n))) , by Lemma 6
Reason
Lemma 3
r=O(log6(n)),
Lemma 9
References

Agrawal, Kayal and Saxena. Primes in P
 M. Smid. Primality testing in polynomial
time
 F. Bornemann. PRIMES Is in P: A
breakthrough for “Everyman”
END