EECS 658_________________________PROBLEM SET #5_________________________Fall 1999

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ASSIGNED: Oct. 14, 1999. READ: Primitive Roots+Generators, Rings+Fields of Polynomials.
DUE DATE: Oct. 28, 1999. THIS WEEK: Cyclotomic, primitive roots, public-key cryptography.

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1. Factor x¹⁵-1 over the field of rationals, i.e., into cyclotomic polynomials (straightforward).
   NOTE: "over the rationals" means that all coefficients are restricted to be rational numbers.

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2. The fundamental theorem of algebra states that a polynomial of degree N has at most N zeros.
   Show that polynomial x²-1 over Z/(15) has more than two zeros! Explain how this can be true.

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3. Let p(x)=x⁴+x³+x²+x+1. FACT: p(x) is prime over GF(2)={{0,1}, +,× mod(2)}.
   So {{polynomials of degree < 4 over GF(2)}, +,× mod(p(x))} is the finite field GF(2⁴)=GF(16).
   Huh? There are 16 polynomials with coefficients 0 or 1 with degree < 4. Add and multiply
   mod(p(x)) and then reduce the coefficients mod(2) gives another such polynomial (closure).

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   The other properties required of a field can also be shown to hold (not required).
   1. Show that "x" is not a primitive element. HINT: Show that its order < 15.
   2. Show that "x+1" IS a primitive element. HINT: Show that its order is 15.
   3. Show that f(z)=z⁴+z³+1 is the minimal polynomial of "x+1".
      MEANING: This is the polynomial of smallest degree such that f(z)|_z=x+1=0.

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4. The decimal expansion of 1/p (where p is an odd prime) repeats after k digits. Examples:
   1/7=0.142857142857...(k=6), 1/11=0.0909...(k=2) and 1/13=0.0769230769230...(k=6).
   1. Show that the period k of 1/p is the order of 10 mod(p). Also show k is a factor of p-1.
   2. What is the condition for k to take on its maximum possible value of p-1 (as with p=7)?

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5. Public Key Cryptography using Euler's totient function Ø(N) (simple using this concept):
   • p and q are large prime numbers (kept secret); but their product r=pq is made public.
   • s is the sending key (made public); s must be relatively prime to p-1 and q-1.
   • t is the decoding key (kept secret); t is defined by solving st=1 mod((p-1)(q-1)).
   • m is the message; encoded using 01=A, 02=B, 03=C...26=Z (can encode other characters).

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   Send and receive a message m by transmitting and receiving the coded message M:
   • Encode the message using M=m^s mod(r). Remember r and s have been made public.
   • Decode the message using m=M^t mod(r). Remember t has been kept private.
   • POINT: If r could be factored into p and q, then t could quickly be found from s.
   • But factoring the product of large prime numbers is very difficult (but not impossible!)

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   Now explain why this works, and try a simple but illustrative example using Matlab:
   1. Explain why m can be decoded from M. HINT: Ø(pq)=(p-1)(q-1) and Euler's theorem.
   2. Let r=2146189 and s=257. Decode the message M=1494449 (isn't that a cute number?).
   3. I will give you p in class Thursday; otherwise, you must break the code by factoring r!

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   Even for this small example, you need some help in computing m^s and M^t:
   • Noting 257=2⁸, you could compute M from m by squaring and reducing mod(r) 8 times.
   • For the numbers in this problem, you can compute m from M by doing something similar.
   • Do NOT let the numbers get too large, or Matlab will give incorrect answers (roundoff).

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