Prime Numbers

422 Chapter 8 THE UBIQUITY OF PRIME NUMBERS
Using appropriate banks of unitary operators, it turns out that if q>n,
and x be a chosen residue (mod n), then one can also form the state
ψ ′ = 1
2d/2 2 d −1
a=0
| x a mod n 〉,
again as a superposition. The difference now is that if we ask for the probability
that the entire register be found in state | b 〉, that probability is zero unless
b is an a-th power residue modulo n.
We end this very brief conceptual sketch by noting that the sovereign of all
divide-and-conquer algorithms, namely the FFT, can be given a concise QTM
form. It turns out that by employing unitary operators, all of them pairwise
as above, in a specific order, one can create the state
ψ ′′ = 1 q−1
√ e
q
2πiac/q | c 〉,
a=0
and this allows for many interesting algorithms to go through on QTMs—
at least in principle—with polynomial-time complexity. For the moment, we
remark that addition, multiplication, division, modular powering and FFT
can all be done in time O(d α ), where d is the number of qbits in each of
(finitely many) registers and α is some appropriate power. The aforementioned
references have all the details for these fundamental operations. Though
nobody has carried out the actual QTM arithmetic—only a few atomic sites
have been built so far in laboratories—the literature descriptions are clear:
We expect nature to be able to perform massive parallelism on d-bit integers,
in time only a power of d.
8.5.2 The Shor quantum algorithm for factoring
Just as we so briefly overviewed the QTM concept, we now also briefly discuss
some of the new quantum algorithms that pertain to number-theoretical
problems. It is an astute observation in [Shor 1994, 1999] that one may factor
n by finding the exponent orders of random integers (mod n) via the following
proposition.
Proposition 8.5.1. Suppose the odd integer n>1 has exactly k distinct
prime factors. For a randomly chosen member y of Z ∗ n with multiplicative
order r, the probability that r is even and that y r/2 ≡−1(modn) is at least
1 − 1/2 k−1 .
(See Exercise 8.22, for a slightly stronger result.) The implication of this
proposition is that one can—at least in principle—factor n by finding “a few”
integers y with corresponding (even) orders r. For having done that, we look
at
gcd(y r/2 − 1,n)