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MATH 3CY3, ASSIGNMENT # 4, SOLUTIONS
Question 1: Bob, Bill and Bradd use the same encryption exponents e = 3 in their
individual cryptosystems (51, 3), (65, 3), (77, 3). Alice sends all three of them the time
m, when the four of them can meet. The encrypted messages are
31 mod 51
25 mod 65
76 mod 77.
Use the Chinese Remainder Theorem to find m.
Hint: m3 < 51 · 65 · 77.
Solution: We know the values of m3 modulo 51, 65 and 77. The Chinese Remainder
Theorem then provides m3 modulo 51 · 65 · 77, which turns out to be
m3 ≡ 1000 mod 51 · 65 · 77.
Since m < 51, we have m3 < 51 · 65 · 77, and therefore m3 = 1000.
The answer is: m = 10.
√
Question 2: Let n = 11111 and let m = [ n]. Let B = {2, 3, 5, 7, 11, 13, } be a factor
base.
a) With j running through the integers between −10 and 10 find the squares (m + j)2
modulo n, which mod n factor into powers of primes from B.
b) Find the relations of the form x2 ≡ y 2 mod n from the list in a) .
c) Use the information from b) to factor n.
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Solution: √
(a) m = [ 11111] = 105. We produce the following list of squares and their prime
factorization mod n.
952
 962

 972

 982

 992

 1002

 1012

 1022

 1032

1042
1052

=
=
=
=
=
=
=
=
=
=
=
52 · 192
210 · 32
972
22 · 74
34 · 112
24 · 54
1012
22 · 32 · 172
1032
26 · 132
32 · 52 · 72
1062
1072
1082
1092
1102
1112
1122
1132
1142
1152
=
=
=
=
=
=
=
=
=
=

125 = 53
338 = 2 · 132


553 = 7 · 79


770 = 2 · 5 · 7 · 11 

989 = 23 · 43


1210 = 2 · 5 · 112 

1433


1658 = 2 · 829


1885 = 5 · 13 · 29 

2114 = 2 · 7 · 151
Among these only the numbers 96, 98, 99, 100, 104, 105, 106, 107, 109 and 111 have
squares, whose prime factors modulo 1111 are in B.
(b) Consider 1062 · 1072 · 1112 = 22 · 54 · 112 · 132 (mod 11111).
Note that 106 · 107 · 111 = 3419 (mod 11111) and 2 · 52 · 11 · 13 = 7150 (mod 11111).
Clearly 3419 6= ±7150 (mod 11111) and 3419 + 7150 = 10569.
(c) Applying the Euclidean Algorithm,
11111 = 1 · 10569 + 542
10569 = 19 · 542 + 271
542 = 2 · 271
we obtain gcd(10569, 11111) = 271. One then sees that 11111 = 41 · 271. It can be checked
that 41 and 271 are primes. Thus, we have the factorization of n = 11111.
Question 3: Show that the polynomial
x5 + x3 + x2 + x + 1 ∈ F2 [x]
is irreducible.
Solution:
The polynomial has no roots in F2 , and therefore the only way it can factor is into a
product of an irreducible quadratic polynomial g(x) and an irreducible cubic polynomial
h(x):
3
f (x) = x5 + x3 + x2 + x + 1 = g(x) · h(x).
The only irreducible polynomial of degree 2 in F2 [x] is g(x) = x2 + x + 1.
Using Euclid’s Algorithm we obtain
x5 + x3 + x2 + x + 1 = (x3 + x2 + x + 1) · (x2 + x + 1) + x,
so that g(x) dose not divide f (x). Therefore f (x) is irreducible.
Question 4: Use the Extended Euclidean Algorithm to find the gcd of the polynomials
f (x) = x5 + x4 + x + 1
and
g(x) = x3 + 1 ∈ F3 [x],
and write it in the form
f (x)s(x) + g(x)t(x)
with
s(x), t(x) ∈ F3 [x].
Solution: The gcd is x + 1 and
x + 1 = x · f (x) + (1 + x + x2 ) · g(x).
Question 5: Consider the finite field F25 = F5 [x]/(x2 + 2).
a) Show that the order of x is 8 and that the order of x + 2 is 3.
b) Use the information from a) to show that 3 + 2x is a generator of F∗25 .
c) Compute L3+2x (1 + 2x).
Solution:
a) The possible orders of an element in F25 are the divisors of 24: 1, 2, 3, 4, 6, 8, 12, 24.
We have to show that x8 = 1 in F25 and that the powers x, x2 , x3 , x6 are 6= 1. We
compute:
x, x2 = 3, x3 = 3x, x4 = −1. This implies that x8 = 1. Since x6 = x2 · x4 = −3 = 2, we
are done: The order of x is 8.
We have
(x + 2)2 = x2 + 4x + 4 = 4x + 2, (x + 2)3 = (4x + 2)(x + 2) = 4x2 + 4 = 1,
which shows that x + 2 has order 3.
b) We note that 3 + 2x = x · (x + 2). The order of 3 + 2x is the least common multiple
of the orders of x and x + 2, hence equal to 24. This shows that 3 + 2x is a generator
of F∗25 .
c) (3 + 2x)2 = 1 + 2x, and therefore L3+2x (1 + 2x) = 2.