MATH 135
Fall 2005
Assignment 8 Solutions
HandIn Problems
Exercise 71:
The following is known to be a simple substitution cipher. Break the code.
PCTPG
ANJHT
GDURG
NEIDV
GPEWN
LPHYJ
AXJHR
PTHPG
BTHHP
VTHHT
CIIDW
XHIGD
DEHLT
GTSXH
VJXHT
SQNIW
TUDAA
DLXCV
HXBEA
TBTIW
DS
Solution:
The given ciphertext has 102 letters.
Analyzing it we see that the letter
T
appears 10 times, and this is one the most repeated letters.
Because the letter
E
is so common in the english language, an educated guess is to try the substitution cipher
f
that
sends
E
to
T
.
This corresponds to the function
f
:
A → A
defined by
f
(
x
)
≡
x
+ 15 (mod 26), where A
≡
1, B
≡
2,
. . .
.
The inverse of this function would be
f

1
(
x
)
≡
x

15 (mod 26).
Applying
f

1
to the message we get
PCTPG
ANJHT
GDURG
NEIDV
GPEWN
LPHYJ
AXJHR
PTHPG
⇐
Ciphertext
ANEAR
LYUSE
ROFCR
YPTOG
RAPHY
WASJU
LIUSC
AESAR
⇐
Plaintext
BTHHP
VTHHT
CIIDW
XHIGD
DEHLT
GTSXH
VJXHT
SQNIW
⇐
Ciphertext
MESSA
GESSE
NTTOH
ISTRO
OPSWE
REDIS
GUISE
DBYTH
⇐
Plaintext
TUDAA
DLXCV
HXBEA
TBTIW
DS
⇐
Ciphertext
EFOLL
OWING
SIMPL
EMETH
OD
⇐
Plaintext
With the correct spacing it would be
“ an early user of cryptography was Julius Caesar messages sent to his troops
were disguised by the following simple method”
.
Exercise 711:
Given
n
=
pq
,
p > q
, and
φ
(
n
) = (
p

1)(
q

1) prove that
p
+
q
=
n

φ
(
n
) + 1
and
p

q
=
(
p
+
q
)
2

4
n.
Solution:
For the first part,
n

φ
(
n
) + 1
=
pq

(
p

1)(
q

1) + 1
=
pq

(
pq

p

q
+ 1) + 1
=
p
+
q.
For the second part,
(
p
+
q
)
2

4
n
=
p
2
+ 2
pq
+
q
2

4
pq
=
p
2

2
pq
+
q
2
=
(
p

q
)
2
.
Since
p

q >
0
,
we have
(
p
+
q
)
2

4
n
=
p

q
.
Exercise 714:
Each integer
n
is the product of two primes
p
and
q
, and the Euler phi function
φ
(
n
) = (
p

1)(
q

1). Determine the
prime factors
p
and
q
.
n
= 71531,
φ
(
n
) = 70992
Solution:
From Exercise 711,
p
+
q
=
n

φ
(
n
) + 1 = 71531

70992 + 1 = 540
p

q
=
(
p
+
q
)
2

4
n
=
540
2

4(71531) = 74
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Adding the two equations, 2
p
= 614 so
p
= 307.
Since
p

q
= 74 and
p
= 307, then
q
= 233.
Check:
(233)(307) = 71531 and (232)(306) = 70992.
Exercise 728:
Encrypt each message
M
, using the RSA public key (
e, n
) and the square and multiply algorithm.
M
= 2041,
(
e, n
) = (13
,
3599)
Solution:
To encrypt the message
M
we have to calculate [
M
13
] in
Z
3599
.
Using the square and multiply algorithm, 13 = (1101)
2
= 8 + 4 + 1 and
2041
2
≡
1638
(mod 3599)
2041
4
≡
1638
2
≡
1789
(mod 3599)
2041
8
≡
1789
2
≡
1010
(mod 3599)
2041
13
=
2041
8
·
2041
4
·
2041
≡
1010
·
1789
·
2041
(mod 3599)
≡
192
·
2041
(mod 3599)
≡
3180
(mod 3599)
So the enciphered message of
M
is
C
= 3180.
Exercise 736:
Use the Chinese Remainder Theorem to decrypt each received ciphertext
C
, using the RSA private key (
d, n
) where
n
=
pq
.
C
= 1120,
d
= 5051,
p
= 79,
q
= 131
Solution:
We calculate
C
d
modulo
p
and modulo
q
and then combine using the Chinese Remainder Theorem to get
C
d
modulo
n
.
Look first at
C
5051
modulo 79.
We can simplify the calculation by recognizing that since
C
≡
0 (mod 79), then
C
78
≡
1 (mod 79).
This means that
C
78
q
≡
1 (mod 79) for any integer
q
.
Since 5051 = 64(78) + 59, then
C
5051
≡
(
C
78
)
64
C
59
≡
C
59
(mod 79).
This drastically reduces the power which we have to work with.
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 Spring '08
 Vanderburgh
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