Prime Numbers

290 Chapter 6 SUBEXPONENTIAL FACTORING ALGORITHMS
is large (say, k>lg |m|), then it is reasonable to suppose that m actually is a
square.
We wish to use this idea with the algebraic integers a − bα, and the
following result allows us to do so via ordinary Legendre symbols.
Lemma 6.2.4. Let f(x) be a monic, irreducible polynomial in Z[x] and let
α be a root of f in the complex numbers. Suppose q is an odd prime number
and s is an integer with f(s) ≡ 0(modq) and f ′ (s) ≡ 0(modq). LetSbe a
set of coprime integer pairs (a, b) such that q does not divide any a − bs for
(a, b) ∈S and f ′ 2 (α) (a − bα) is a square in Z[α]. Then
(a,b)∈S
(a,b)∈S
a − bs
=1. (6.9)
q
Proof. Consider the homomorphism φq from Z[α] toZq where φq(α) isthe
residue class s (mod q). We have f ′ 2 (α) (a,b)∈S (a − bα) = γ2 for some
γ ∈ Z[α]. By the hypothesis, φq(γ2 ) ≡ f ′ 2 (s) (a,b)∈S (a − bs) ≡ 0(modq).
Then φq(γ 2 ) 2
φq(γ) ′ 2
f (s)
q = q = 1 and q =1,sothat
(a,b)∈S
(a − bs)
=1,
q
which implies that (6.9) holds. ✷
So again we have a necessary condition for squareness, while we are still
searching for a sufficient condition. But we are nearly there. As we have seen,
one might heuristically argue that if k is sufficiently large and if q1,...,qk are
odd primes that divide no N(a − bα) for (a, b) ∈Sandifwehavesj∈R(qj) for j =1,...,k,wheref ′ (sj) ≡ 0(modqj), then
v(a − bα) ≡ 0 (mod2)
and
(a,b)∈S
imply that
(a,b)∈S
(a,b)∈S
a − bsj
=1forj =1,...,k
qj
(a − bα) =γ 2 for some γ ∈I.
And how large is sufficiently large? Again, since the dimensions of obstructions
(1), (2), (3) are all small, k need not be very large at all. We shall choose the
polynomial f(x) so that the degree d satisfies d2d2