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4.1. GCD IN ONE VARIABLE 119
be made effective, but the results tend to be unduly pessimistic, since it
is extremely likely that all the bad primes would be clsutered just above
2M.
• In theory, the division tests could yield very large numbers if done as tests
of the remainder being zero: for example the remainder on dividing x 100
by x − 10 is 10 100 . This can be solved by a technique known as “early
abort” trial division.
Proposition 44 If h, of degree m, is a factor of f of degree n, the coefficient
of x n−m−i in the quotient is bounded by ( )
n−m 1
i lc(h) ||f||.
This is basically Corollary 22. Hence, as we are doing the trial division,
we can give up as soon as we find a coefficient in the quotient that exceeds
this bound, which is closely related to M (the difference relates to the
leading coefficient terms).
For example, if we take p = 7 in the example at the start of this chapter,
we find that the g.c.d. of A 7 and B 7 is x + 10 (it could equally well be
x + 3, but x + 10 makes the point better). Does x + 10 divide B? We
note that ||B|| ≈ 23.92. Successive terms in the quotient are 3x 5 (and 3
is a permissible coefficient), −30x 4 (and 30 < ( 5
1)
× 23.92) and 305x 3 , at
which point we observe that 305 > ( 5
2)
× 23.92 = 239.2, so this cannot be
a divisor of B. Hence 7 was definitely unlucky.
With this refinement, it is possible to state that the numbers dealt with
in this algorithm are “not much larger” than 2M, though considerable
ingenuity is needed to make this statement more precise.
If we apply this algorithm to the polynomials at the start of this section,
√ ∑8
we deduce that
i=0 a2 i = √ √ ∑6
113,
i=0 b2 i = 2√ 143, and hence corollary 8
gives a bound of
( )
√113,
2 6 2√ min
143 ≈ 510.2, (4.4)
3
so our first prime would be 1021, which is indeed of good reduction. In this
case, corollary 9 gives a bound of
( )
1 √
2 6 2 √
min 113, 143 ≈ 72.8, (4.5)
5 21
so our first prime would be 149. In general, we cannot tell which gives us the
best bound, and it is normal to take the minimum.
Open Problem 4 (Improving Landau–Mignotte for g.c.d.) A significant
factor in the Landau–Mignotte bound here, whether (4.4) or (4.5), was the
2 min(8,6) contribution from the degree of the putative g.c.d. But in fact the
exponent is at most 4, not 6, since the g.c.d. cannot have leading coefficient
divisible by 3 (since A does not). Hence the g.c.d. must have at most the degree