AKS PRIMALITY TEST:

First we describe a characterization of prime numbers that will provide the conceptual mathematical foundation for our polynomial time algorithm.

Lemma 3.1:  Let a∈ Z, n ∈ N, such that (a, n) = 1. Then n is prime iff (x+a)ⁿ ≡ xⁿ +a (mod n).

Proof:

By the Binomial theorem we have:

If n is prime then   is divisible by n according to the binomial theorem. By Fermat's little theorem, we have aⁿ  ≡ a  (mod n) and hence the equivalence in the above equation holds.

If n is composite, then let q  be a prime divisor of n with q^s | n . The coefficient of x^n-q in the binomial expansion of (x + a)ⁿ is  a^q . The numerator is divisible by q^s but not by q^s+1. The denominator is divisible by q. Hence   a^q≠ 0(mod n). Since (a,n) = 1, implies (a,q^s) = 1, implies (a^q, q^s) = 1, implies  a^q ≠ 0(mod n).

Therefore (x+a)ⁿ  ≠ xⁿ + a (mod n)

The above identity suggests a simple method for testing the primality of an integer n. We can choose an integer a such that (a, n) = 1 and calculate f(x) = (x + a)ⁿ - (xⁿ + a). If this function is equal to 0 (mod n) then n is prime, else n is composite. Although this is certainly a valid primality test, it is horribly inefficient as it involves the computation of n coefficients. The trick however is in choosing a suitable integer a. The simplest method for reducing the number of coefficients that need to be computed is to evaluate f(x) modulo n and modulo some polynomial of small degree, say (x^r - 1).

Although it is clear that all primes p satisfy (x + a)^p - (x^p + a) ≡ 0 mod (p, x^r -1), some composite numbers may satisfy this equation for all values of a and r. It turns out that for a judiciously chosen r, if the above identity is satisfied for several values of a, then n can be shown to be a prime power. The number of a's and the appropriate value of r are bounded by  log(n). Therefore we have just described a deterministic polynomial time primality testing algorithm.