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Solutions to MCS 425 Exercise Set #3 — Spring, 2008, Assignments of Cryptography and System Security

University of Illinois - ChicagoCryptography and System Security

Solutions to exercises from mcs 425 course, focusing on modular arithmetic and number theory concepts such as fermat’s little theorem, euler’s theorem, quadratic residues, and the euler’s totient function. It includes solutions for exercises 12, 13, 25 from section 3.13, and additional exercises f, g, h, i.

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Solutions to MCS 425 Exercise Set #3 — Spring, 2008
Sec 3.13, exercise 12
Solution:
We need the value of 210203 mod 101.
Note 101 is prime. We proved in class that, for any prime p, u ≡ v (mod p−1) implies
au ≡ av (mod p). (This follows easily from Fermat’s Little Theorem.)
10203 ≡ 3 (mod 100), so 210203 ≡ 23 ≡ 8 (mod 101).
Sec 3.13, exercise 13
Solution:
The last two digits of 123562 are the value of 123562 (mod 100).
ϕ(100) = 100(1 −1/2)(1 −1/5) = 40. Since gcd(123,100) = 1, 12340 ≡ 1 (mod 100) by Euler’s
Theorem, so (12340)14 ≡ 123560 ≡ 1 (mod 100). It follows that
123562 ≡ 123560+ 2 ≡ 1235601232 ≡ 1232 ≡ 232 ≡ 529 ≡ 29 (mod 100).
Sec 3.13, exercise 25
Solution:
First we find integers w1 and w2 such: w1 ≡ 1 (mod 11), w1 ≡ 0 (mod 13),
w2 ≡ 0 (mod 11), w2 ≡ 1 (mod 13),
z1 = 143/11 = 13
z2 = 143/13 = 11
y1 ≡ 13−1 ≡ 2−1 ≡ 6 (mod 11)
y2 ≡ 11−1 ≡ 6 (mod 13)
w1 ≡ 13⋅6 ≡ 78 (mod 143)
w2 ≡ 11⋅6 ≡ 66 (mod 143)
a) x2 ≡ 133 (mod 143) is equivalent to the simultaneous congruences:
i) x2 ≡ 133 (mod 11), and ii) x2 ≡ 133 (mod 13).
x2 ≡ 1 (mod 11) x2 ≡ 3 (mod 13)
x ≡ ±1 (mod 11) x ≡ ±4 (mod 13) by inspection. For very large
moduli that are 5 (mod 8), the
method developed in class for this
case may be applied.
x ≡ 1 (mod 11) and x ≡ 4 (mod 13) implies x ≡ 1w1+4w2 ≡ 1⋅78 +4⋅66 ≡ 56 (mod 143)
x ≡ −1 (mod 11) and x ≡ 4 (mod 13) implies x ≡ −1w1+4w2 ≡ −1⋅78 + 4⋅66 ≡ 43 (mod 143)
The remaining two cases produce the negatives of 56 and 43.
So the solutions are x ≡ ±43, ±56 (mod 143)
pf3

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Solutions to MCS 425 Exercise Set #3 — Spring, 2008

Sec 3.13, exercise 12

Solution : We need the value of 2 10203 mod 101. Note 101 is prime. We proved in class that, for any prime p , u ≡ v (mod p −1) implies a u^ ≡ a v^ (mod p ). (This follows easily from Fermat’s Little Theorem.) 10203 ≡ 3 (mod 100), so 2 10203 ≡ 2 3 ≡ 8 (mod 101).

Sec 3.13, exercise 13

Solution : The last two digits of 123 562 are the value of 123 562 (mod 100). ϕ(100) = 100(1 − 1/2)(1 − 1/5) = 40. Since gcd(123,100) = 1, 123 40 ≡ 1 (mod 100) by Euler’s Theorem, so (123^40 )^14 ≡ 123 560 ≡ 1 (mod 100). It follows that 123 562 ≡ 123 560+2^ ≡ 123 560123 2 ≡ 123 2 ≡ 23 2 ≡ 529 ≡ 29 (mod 100).

Sec 3.13, exercise 25 Solution : First we find integers w 1 and w 2 such: w 1 ≡ 1 (mod 11), w 1 ≡ 0 (mod 13), w 2 ≡ 0 (mod 11), w 2 ≡ 1 (mod 13), z 1 = 143/11 = 13 z 2 = 143/13 = 11 y 1 ≡ 13 −^1 ≡ 2 −^1 ≡ 6 (mod 11) y 2 ≡ 11 −^1 ≡ 6 (mod 13) w 1 ≡ 13 ⋅ 6 ≡ 78 (mod 143) w 2 ≡ 11 ⋅ 6 ≡ 66 (mod 143)

a) x^2 ≡ 133 (mod 143) is equivalent to the simultaneous congruences: i) x^2 ≡ 133 (mod 11), and ii) x^2 ≡ 133 (mod 13). x^2 ≡ 1 (mod 11) x^2 ≡ 3 (mod 13) x ≡ ±1 (mod 11) x ≡ ±4 (mod 13) by inspection. For very large moduli that are 5 (mod 8), the method developed in class for this case may be applied. x ≡ 1 (mod 11) and x ≡ 4 (mod 13) implies x ≡ 1 w 1+4 w 2 ≡ 1 ⋅78 +4⋅ 66 ≡ 56 (mod 143) x ≡ −1 (mod 11) and x ≡ 4 (mod 13) implies x ≡ − 1 w 1 +4 w 2 ≡ − 1 ⋅78 +4⋅ 66 ≡ 43 (mod 143) The remaining two cases produce the negatives of 56 and 43. So the solutions are x ≡ ± 43, ± 56 (mod 143)

b) x^2 ≡ 77 (mod 143) is equivalent to the simultaneous congruences: i) x^2 ≡ 77 (mod 11), and ii) x^2 ≡ 77 (mod 13). x^2 ≡ 77 (mod 11) x^2 ≡ 77 (mod 13) x^2 ≡ 0 (mod 11) x^2 ≡ 12 (mod 13) x ≡ 0 (mod 11) x ≡ ±5 (mod 13) by inspection. See remark above x ≡ 0 (mod 11) and x ≡ 5 (mod 13) implies x ≡ 0 w 1 +5 w 2 ≡ 5 ⋅ 66 ≡ 44 (mod 143) x ≡ 0 (mod 11) and x ≡ −5 (mod 13) implies x ≡ 0 w 1 − 5 w 2 ≡ − 5 ⋅ 66 ≡ − 44 (mod 143) So the solutions are x ≡ ± 44 (mod 143)

Exercise F Show that 2 is a quadratic residue mod 103. (Note 103 is prime.) Find the square roots of 2 in modulus 103. (The exercise posted on the web site had 107 in place of 103. This was an error. 2 is not a quadratic residue mod 107.)

Solution : (103 − 1) /2 = 51. To show that 2 is a quadratic residue mod 103, we need 2 51 ≡ 1 (mod 103). 2 20 ≡ 2 (mod 103) 2 21 ≡ 2 2 ≡ 4 (mod 103) 2 22 ≡ 4 2 ≡ 16 (mod 103) 2 23 ≡ 16 2 ≡ 256 ≡ 50 (mod 103) 2 24 ≡ 50 2 ≡ 2500 ≡ 28 (mod 103) 2 25 ≡ 28 2 ≡ 784 ≡ 63 (mod 103) 51 = (110011) 2 = 2^5 + 2 4 + 2 1 + 2^0 , so 2 51 = 2 2

(^5) +2 4 +2 (^1) +2 0 = 2 2

5 2 2

4 2 2

1 2 2

0 ≡ 63 ⋅ 28 ⋅ 4 ⋅ 2 ≡ 1 (mod 103). Since 103 ≡ 3 (mod 4) and 2 is a quadratic residue mod 103, the square roots of 2 mod 103 are ± 2 (103+1)/4^ = ± 2 26 (mod 103). Now 2 26 = 2 2

4 22

3 2 2

1 ≡ 28 ⋅ 50 ⋅ 4 ≡ 5600 ≡ 38 (mod 103). So the square roots of 2 mod 53 are ± 38.

Exercise G Show that 21 is a quadratic residue mod 37. (Note 37 is prime.) Find the square roots of 21 in modulus 37.

Solution : (37 − 1) /2 = 18. To show 21 is a quadratic residue mod 37, it suffices to show that 21^18 ≡ 1 (mod 37). 2120 ≡ 21 (mod 37) 2121 ≡ 21 2 ≡ 34 (mod 37) 2122 ≡ 34 2 ≡ 9 (mod 37) 2123 ≡ 9 2 ≡ 7 (mod 37) 2124 ≡ 7 2 ≡ 12 (mod 37)