In cryptography, most public key cryptosystems are founded on problems that are believed to be intractable. The higher residuosity problem (also called the nth-residuosity problem) is one such problem. 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.

Mathematical background

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 unitss of any ring form a group.

From the ring isomorphism above, we have

as an isomorphism of groups. Since p and q are cyclic of orders p−1 and q−1 respectively. If d is a divisor of p−1, then the set of d form a subgroup of index d. If gcd(d,q−1) = 1, then every is a dth power, so the set of d is also a subgroup of index d. In general, if gcd(d,q−1) = g, then there are (q−1)/g d, so the set of d has index dg. This is most commonly seen when d = 2, and we are considering the subgroup of quadratic residues 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

Problem statement

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 dth power (equivalently dth residue) modulo n.

Notice that if p and q are known it is easy to determine whether x is a dth residue modulo n because x will be a dth residue modulo p if and only if

When d = 2, this is called the quadratic residuosity problem.

Applications

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

References