Prime Numbers

4.2 The n + 1 test 185
of both n − 1andn + 1, that is, a fully factored divisor of n 2 − 1, may be
combined into one test.
Theorem 4.2.8. Suppose f,∆ are as in (4.12) and n is a positive integer
with gcd(n, 2b) =1and
∆
1/3 = −1. Suppose n +1=FR with F>n +1
n
and (4.14) holds. Write R in base F , so that R = r1F + r0, 0 ≤ ri ≤ F − 1.
Then n is prime if and only if neither x2 +r0x−r1 nor x2 +(r0−F )x−r1 −1
has a positive integral root.
Note that in the case Rn,(F + 1)(F 2 − 1) >n,wehave1≤ c, d ≤ F − 1. Note
that
n +1
r1F + r0 = R = = cdF + d − c,
F
so that d − c ≡ r0 (mod F ). It follows that d = c + r0 or d = c + r0 − F ,that
is, d = c + r0 − iF for i = 0 or 1. Thus,
r1F + r0 = c(c + r0 − iF )F + r0 − iF,
so that r1 = c(c + r0 − iF ) − i, which implies that
c 2 +(r0 − iF )c − r1 − i =0.
But then x 2 +(r0 −iF )x−r1 −i has a positive integral root for one of i =0, 1.
This proves one direction.
Suppose now that x 2 +(r0 − iF )x − r1 − i has a positive integral root c for
one of i =0, 1. Undoing the above algebra we see that cF + 1 is a divisor of n.
But n ≡−1(modF ), so n is composite, since the hypotheses imply F>2.
✷
We can improve the n + 1 test further, requiring only F ≥ n 3/10 .The
proof is completely analogous to Theorem 4.1.6, and we leave said proof as
Exercise 4.15.
Theorem 4.2.9. Suppose n ≥ 214 and the hypotheses of Theorem 4.2.8
hold, except that n 3/10 ≤ F ≤ n 1/3 +1. Say the base-F expansion of n +1 is
c3F 3 + c2F 2 + c1F ,andletc4 = c3F + c2. Then n is prime if and only if the
following conditions hold:
(1) (c1 + tF ) 2 − 4t +4c4 is not a square for t integral, |t| ≤5,
(2) with u/v the continued fraction convergent to c1/F such that v is
maximal subject to v < F 2 / √ n and with d = ⌊c4v/F +1/2⌋, the