This preview shows pages 1–2. Sign up to view the full content.
The University of Sydney
MATH2068
Number Theory and Cryptography
(http://www.maths.usyd.edu.au/u/UG/IM/MATH2068/)
Semester 2, 2008
Lecturer: R. Howlett
Tutorial 2
1.
For each natural number
n
, let
a
n
be the residue of 2
n
modulo 13. Observe
that
a
k
+1
≡
2
a
k
(mod 13) for each
k
∈
N
, and thus the
a
k
are easy to compute
recursively, starting at
a
0
= 1 and repeatedly doubling and reducing mod 13 to
get successive terms of the sequence. Compute the ﬁrst 15 or 20 terms, and then
compute
a
2008
.
Solution.
1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1 – and then it repeats. In other words,
a
0
=
a
12
=
a
24
=
···
, and
a
1
=
a
13
=
a
25
=
···
, etc. In general,
a
i
=
a
j
if and only if
i
≡
j
(mod 12). (In the terminology introduced in lectures,
ord
13
(2) = 12.) Now 2004
≡
4 (mod 12); so
a
2008
=
a
4
= 3.
2.
Repeat Question 1 with 13 replaced by 3, 5, 7, 11, 31, 47.
Solution.
For 3 the sequence is just 1, 2, 1, 2,
. . .
; that is,
a
i
= 1 if
i
is even and
a
i
= 2 if
i
is odd (and ord
3
(2) = 2). So
a
2008
= 1.
Using 5 we get 1, 2, 4, 3, 1,
. . .
. So ord
5
(2) = 4, and
a
2008
=
a
0
= 1 (since
2008
≡
0 (mod 4)).
Using 7 we get 1, 2, 4, 1,
. . .
. So ord
7
(2) = 3, and
a
2008
=
a
1
= 2 (since
2008
≡
1 (mod 3)).
Using 11 we get 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1,
. . .
. So ord
11
(2) = 10, and
a
2008
=
a
8
= 3 (since 2008
≡
8 (mod 10)).
Using 31 we get 1, 2, 4, 8, 16, 1,
. . .
. So ord
31
(2) = 5, and
a
2008
=
a
3
= 8 (since
2008
≡
3 (mod 5)).
Using 47 we get 1, 2, 4, 8, 16, 32, 17, 34, 21, 42, 37, 27, 7, 14, 28, 9, 18, 36,
25, 3, 6, 12, 24, 1,
. . .
. So ord
47
(2) = 23, and
a
2008
=
a
7
= 34 (since 2008
≡
7
(mod 23)).
Observe that in all these case, ord
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 One '08
 Howlett
 Number Theory, Cryptography

Click to edit the document details