A summary of number theory algorithms

Ma Xiao theorem of sum-up fee for number theory algorithm

\ (A^{p-1} \equiv 1 \pmod{p}\space ((a,p) =1,isprime (p)) \)

• Proof: Link
  The general form is given here: Consider the remainder of any positive integer \ (A\mod p\space ((a,p) =1) \) , with \ (1,2,3,\ldots, p-1\).
  Then we take any positive integer \ (a ' ((a ', p) =1) \), the remaining system is multiplied by \ (a ' \) and then \ ( \mod p\) , the remaining system or \ (1,2,\ldots,p-1\) one of the permutations.
  Multiply these numbers to get:\ (A ' ^{p-1}\times (p-1)! \equiv (p-1)! \mod p\), on both sides with the exception \ ((p-1)!\) get:\ (a ' ^{p-1}\times 1 \ EQUIV 1 \mod p\), replace \ (a ' \) with \ (a\) .
  The inverse element is \ (A^{p-2}\times a \equiv 1 \pmod{p}\).
  Note that all the derivation is under the conditions of the above equation.

Euler's theorem

\ (A^{\phi (n)}\equiv 1 \pmod{n} ((a,n) =1) \)

• It is easy to find that the front Fermat theorem is the son theorem of Euler's theorem.
  The main reason is to add why \ (a\times x_i \mod N ((a,n) =1,x_i\leq n) \) has \ (\phi (n) \) results.
  Assuming \ (m_i=a\times x_i\), then the assumption exists \ (m_i\equiv m_j \pmod{n} (i\not =j) \), then there is:\ (A (x_j-x_i) = kn\) , but because \ ((a,n) =1\),\ (x_j-x_i<n\), it is impossible to have a solution anyway. The remaining proof is the same as above.

Expanding Euler's theorem

\ (A^b\equiv \begin{cases} a^{b\%\phi (P)}~~~~~~~~~~~gcd (a,p) =1\\ a^b~~~~~~~~~~~~~~~~~~gcd (a,p) \neq1,b<\phi (p) \ \ A^{b\%\phi (P) +\phi (p)}~~~~gcd (a,p) \neq1,b\geq\phi (p) \end{cases}~~~~~~~ (mod~p) \)

Proof: Link

Bsgs

Given \ (a,b,p\), the smallest nonnegative integer \ (x\) order \ ( x\) satisfies \ (a^x \equiv b \pmod{p}\).

First there is the Fermat theorem:\ (A^{k\mod (p-1)}\equiv a^k \pmod{p}\), so say \ (x< p-1\).
Then we make \ (M=\lceil \sqrt{p} \rceil\) ,\ (x=i\times m-j\), then there are:\[a^{i\times m-j}\equiv b \pmod{p}\]
\[a^{i\times m}\equiv b \times a^j\pmod{p}\]
So the violence in \ (0\to m\) enumeration \ (j\), deposit the hash table, and then in \ (1\to m\) enumeration I query.

A summary of number theory algorithms