Prime Numbers

4.1 The n − 1 test 177
Also, (4.4) implies that
a1c1 + a1tF = a 2 1 + c4 − t. (4.7)
With the notation of condition (2), we have from (4.7) that
a1u + a1tv − c4v
u
c1
= a1v − +(a1c1 + a1tF )
F v F
v c4v
−
F F
u
c1
= a1v − +(a
v F
2 1 + c4 − t) v c4v
−
F F
u
c1
= a1v − +(a
v F
2 1 − t) v
. (4.8)
F
Note that (4.5), (4.6), and t ≥ 6 imply that
|a 2 1 − t| < max{a 2
1
n
1,t}≤max
9 F 3
2 , n
F 3
First suppose that u/v = c1/F . Then (4.8) and (4.9) imply that
a1u + a1tv − c4v
= |a
F
2 1 − t| v
F
1
n
<
6 F 3
2 v
F
≤ 1
n
6 F 3
2 . (4.9)
2
n2 F
< ·
6F 7
√ =
n n3/2 1
≤
6F 5 6 .
(4.10)
If u/v = c1/F ,letu ′ /v ′ be the next continued fraction convergent to c1/F
after u/v, sothat
v<
F 2
√ n ≤ v ′ ,
u
c1
− ≤
v F
1
√
n
≤ .
vv ′ vF 2
Thus, from (4.5), (4.8), and the calculation in (4.10),
a1u + a1tv − c4v
√
n 1 n3/2 1 1
≤ a1v + < + ≤
F vF 2 6 3F 5 6 2 .
Let d = a1u + a1tv, sothat|d − c4v/F| < 1/2, which implies that d =
⌊c4v/F +1/2⌋. Multiplying (4.7) by a1v, wehave
and using −a1tv = ua1 − d, weget
va 3 1 − c1va 2 1 − a 2 1tvF − a1tv + c4a1v =0,
va 3 1 +(uF − c1v)a 2 1 +(c4v − dF + u)a1 − d =0.
Hence (2) does not hold after all, which proves that if n is composite, then
either (1) or (2) does not hold.
Now suppose n is prime. If t ∈{0, 1, 2, 3, 4, 5} and (c1+tF ) 2 −4c4+4t = u 2 ,
with u integral, then
n =(c4− t)F 2 +(c1 + tF )F +1
c1 + tF + u c1 + tF − u
=
F +1
F +1 .
2
2