Primitive Roots and the Units Modulo n · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Finding a generator for the units

Mathematics · Number Theory · ages 20-21
Name ______________________   Date ____________

  1. What is the order of 2 modulo 5? That is, what is the smallest positive k with 2^k ≡ 1 (mod 5)?

    Answer: ______________

  2. What is the order of 2 modulo 5? That is, the smallest positive k with 2 to the k equal to 1 mod 5.

    Answer: ______________

  3. What is the order of 2 modulo 7?

    • 6
    • 3
    • 2

  4. Leo says 2 is a primitive root modulo 5 because its powers produce every nonzero remainder.

    Circle one:   True   False

  5. Mia says every modulus has a primitive root. Is Mia right?

    Circle one:   True   False

  6. Mia says every modulus has a primitive root.

    Circle one:   True   False

  7. What is the order of 3 modulo 7?

    Answer: ______________

  8. What is the order of 2 modulo 7?

    • 3
    • 6
    • 7
    • 2

  9. Is 6 a primitive root modulo 7?

    • No, its order is only 2
    • Yes, it generates all units
    • No, it shares a factor with 7

  10. Which of these numbers is a primitive root modulo 7?

    • 3
    • 2
    • 4
    • 6
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Answer key

For grown-ups. Fold this page away before handing over the rest.

Finding a generator for the units W1-mt_KwTDckkMze-s1

  1. 4 · The powers go 2, 4, 3, 1, so the first power returning to 1 is the 4th. The order of 2 mod 5 is 4, which equals phi(5).
  2. 4 · The powers go 2, 4, 3, 1, so 1 first returns at the 4th.
  3. 3 · The powers go 2, 4, 1, so 1 first returns at exponent 3.
  4. True · The powers 2, 4, 3, 1 cover every nonzero remainder mod 5.
  5. False · Mia is wrong. Powers exist only for odd primes, twice an odd prime, 4, and prime powers of odd primes. Mod 8 has none: every odd square is 1 mod 8.
  6. False · Mod 8 has none: every odd square is 1 there while phi of 8 is 4.
  7. 6 · The powers hit all six units before returning to 1.
  8. 3 · The powers go 2, 4, 1, so 2^3 is the first return to 1. The order of 2 mod 7 is 3, which divides phi(7) = 6.
  9. No, its order is only 2 · 6 squared is 36, which is 1 modulo 7, so it repeats far too early.
  10. 3 · Only 3 works: its order is 6 = phi(7). The orders of 2 and 4 are 3, and the order of 6 is 2, so none of them generate all units.
Worksheet · LightMySky