Higher residuosity problem

In cryptography, most public key cryptosystems are founded on problems that are believed to be intractable.

This problem is easier to solve than integer factorization, so the assumption that this problem is hard to solve is stronger than the assumption that integer factorization is hard.

If n is an integer, then the integers modulo n form a ring.

If n = pq where p and q are primes, then the Chinese remainder theorem tells us that The units of any ring form a group under multiplication, and the group of units in

is traditionally denoted

Since p and q were assumed to be prime, the groups

are cyclic of orders p−1 and q−1 respectively.

If d is a divisor of p−1, then the set of d th powers in

form a subgroup of index d. If gcd(d,q−1) = 1, then every element in

is also a subgroup of index d. In general, if gcd(d,q−1) = g, then there are (q−1)/g d th powers in

, so the set of d th powers in

has index dg.

This is most commonly seen when d = 2, and we are considering the subgroup of quadratic residues, it is well-known that exactly one quarter of the elements in

are quadratic residues (when n is the product of two primes, as it is here).

The important point is that for any divisor d of p−1 (or q−1) the set of d th powers forms a subgroup of

Given an integer n = pq where p and q are unknown, an integer d such that d divides p−1, and an integer x < n, it is infeasible to determine whether x is a d th power (equivalently d th residue) modulo n. Notice that if p and q are known it is easy to determine whether x is a d th residue modulo n because x will be a d th residue modulo p if and only if When d = 2, this is called the quadratic residuosity problem.

The semantic security of the Benaloh cryptosystem and the Naccache–Stern cryptosystem rests on the intractability of this problem.