Mathematics 360
Points (leave blank)
1
2
3
4
Midterm 2 (100 points)
5
∑
11/8/13
Name:
(clearly, please)
You may use a pocket calculator that does not have internet access. You can work on the problems
in any order you like. Show your work! All problems carry the same weight.
Calculation steps and explanation for statements made are a crucial part of a solution. Partial credit
will be given sparingly – rather complete one problem than start two only partially.
1) You are participating in a Diffie-Hellman key exchange for p = 101 and g = 2. You have chosen
the secret exponent a = 8.
a) What number should you send to your partner?
b) Your partner sends you the number 99. What is the secret key shared by your partner and you?
Explain the calculations you are doing.
2) Find a number a that satisfies the following three congruences:
a ≡ 0 (mod 3)
a ≡ 4 (mod 7)
a ≡ 2 (mod 11)
3) Let p = 701, then p is prime, p − 1 = 22 52 7 and 2 is a primitive root modulo p.
a) Determine all e such that OrderMod p (2e ) = 5. (Note that this question asks for the values of e,
not for those of 2e . You also may keep values for e in product form and do not need to multiply these
products out.)
b) For how many e will you have that OrderMod p (2e ) = 25? (You do not need to write down all
values of e, the problem just asks for a count!) Justify your answer.
4) Let p = 307, then g = 5 is a primitive root modulo p and p − 1 = 17 ⋅ 18. Determine an exponent
e such that g e ≡ 100 (mod p).
Hints: You might find the following facts useful: OrderMod p (100) = 153 = 9 ⋅ 17 and 100 = 81 ⋅ 168
with OrderMod p (81) = 17 and OrderMod p (168) = 9. Furthermore 518 ≡ 81 mod p and 517 ≡
139 (mod p), so the main calculation is a discrete logarithm of 168 for √
base 139 mod p where
OrderMod p (139) = 18, a Babystep-Giantstep calculation with step width ⌊ 18 + 1⌋ = 5. As a help,
you are given the following list of values of 139−i ⋅ 168 mod p for i = 0..4: [168, 306, 53, 261, 289]. and
139−1 ≡ 254 (mod p).
(Make sure that your final answer is a number e such that g e ≡ 100 (mod p).)
5) Show that there is no n such that n = 5 ⋅ φ(n). (Hint: Write n = 5a ⋅ b with gcd(5, b) = 1. Show
that you would need φ(5a ) = 5a /5 = 5(a−1) or φ(b) = b/5. Then show that neither of these two
conditions is possible.) Remember that
φ (∏ p ei i ) = ∏(p i − 1) ⋅ p ei i −1 .
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