Cryptography Reference
In-Depth Information
x k (mod p m ) for all k
for all m
1. Since x m
m , the base p expansions
m must agree through the p m− 1 term. Therefore, the
sequence of integers x m determines an expression of the form
for all x k with k
a n p n ,
n =0
where
m
1
a n p n
(mod p m )
x m
n =0
for all m . In other words, the sequence of integers determines a p -adic inte-
ger. Conversely, the partial sums of a p -adic integer determine a sequence of
integers satisfying (A.3).
Let's use these ideas to show that 1 is a square in the 5-adic integers. Let
x 1 =2,so
x 1 ≡− 1(mod .
Suppose we have defined x m such that
x 2 m ≡−
m ) .
1(mod
Let x m +1 = x m + b 5 m ,where
1 − x 2 m
2 · 5 m x m
b
(mod 5) .
Note that x 2 m ≡− 1(mod5 m ) implies that the right side of this last congru-
ence is defined mod 5. A quick calculation shows that
x 2 m +1 ≡− 1(mod
m +1 ) .
x m (mod 5 m )for
Since (A.3) is satisfied, there is a 5-adic integer x with x
all m .Moreover,
x 2
m )
≡− 1(mod
for all m . This implies that x 2 =
1.
In general, this procedure leads to the following very useful result.
THEOREM A.2 (Hensel's Lemma)
Let f ( X ) be a polynomial with coe cients that are p-adic integers and suppose
x 1 is an integer such that
f ( x 1 )
0(mod p ) .
If
f ( x 1 ) 0(mod p ) ,
Search WWH ::




Custom Search