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Fermat's little theorem and Euler's theorem

aᵖ⁻¹ ≡ 1 (mod p), the totient, and fast modular arithmetic.

For prime p and a not divisible by p, aᵖ⁻¹ ≡ 1 (mod p); Euler generalises to a^φ(n) ≡ 1 (mod n). These collapse huge exponents: 2¹⁰⁰ mod 13 reduces via 2¹² ≡ 1. Picture it: the powers of 2 mod 13 cycling around a 12-hour clock. Think it: this is Lagrange's theorem in the group (ℤ/n)*.

Gewerkte voorbeeld · 2^100 mod 13

Number-Theory · Modular-Arithmetic

2^100 mod 13

2,\ 100,\ 13

Stap met stap

  1. 2^{100} \bmod 13

    Never compute the huge power. Reduce after every multiplication (repeated squaring).

  2. 2 \equiv 2 \pmod{13}

    Reduce the base first.

  3. 100 = 1100100_2

    Write the exponent in binary: 7 squarings at most.

  4. 2^2 \equiv 4 \pmod{13}

    Square the running power.

  5. 4^2 \equiv 3 \pmod{13}

    Square the running power.

  6. r \leftarrow r \cdot 3 \equiv 3 \pmod{13}

    This bit is 1: multiply the result by the current power.

  7. 3^2 \equiv 9 \pmod{13}

    Square the running power.

  8. 9^2 \equiv 3 \pmod{13}

    Square the running power.

  9. 3^2 \equiv 9 \pmod{13}

    Square the running power.

  10. r \leftarrow r \cdot 9 \equiv 1 \pmod{13}

    This bit is 1: multiply the result by the current power.

  11. 9^2 \equiv 3 \pmod{13}

    Square the running power.

  12. r \leftarrow r \cdot 3 \equiv 3 \pmod{13}

    This bit is 1: multiply the result by the current power.

  13. 2^{100} \equiv 3 \pmod{13}

    Done.

Openbaar die antwoord
2^{100} \bmod 13 = 3

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Simbools wat hier gebruik word

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Hoe om: Fermat's little theorem and Euler's theorem

  1. Reduce the base mod n.
  2. Reduce the exponent mod p − 1 (prime) or mod φ(n) (general): Fermat/Euler.
  3. Finish with repeated squaring on what remains.

Vrae wat mense vra

Why are primes so important?

Every integer factors into primes in exactly one way, so primes are the atoms of multiplication. Cryptography relies on that factoring being easy to state and hard to do.

How do I tell whether a big number is prime?

Trial division up to the square root works for small numbers. For large ones, probabilistic tests (Miller-Rabin) give an answer that is wrong with negligible probability, and deterministic tests (AKS) exist but are slower.

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