How the Euler's function is computed

The general method to evaluate the Euler's function is as follows:

1. we know that the Euler's function f (p) of a prime number P is equivalent to P (P). Then, the k power of p, that is, n = P ^ K, is easy to prove: F (N) = P ^ K-P ^ (k-1 );

 

Proof: it is known that the number of less than P ^ K has a P ^ K-1, of which the number of non-interconnectivity with P ^ K has P ^ (k-1)-1, respectively (p * 1, p * 2, P * 3, ·, p * (P ^ (k-1)-1 )). So F (n) = P ^ (k)-1-(P ^ (k-1)-1) = P ^ (k)-P ^ (k-1 ).

 

2. Assume that p and q are the numbers of the two mutex, then: F (p * q) = f (p) * F (Q );

It turns out that I kept writing a slight word here, not because it was too simple, but because I did not prove it. At first I wrote a proof, but later I found it wrong, and I never figured out how to prove it, today, I found the proof when I read the third chapter of elementary number theory at the same time, so I have to sort it out.

 

Proof: here we mainly use the simplification of the Residual Series, x1, x2 respectively through the modulo M1, M2 to simplify the residual series, then m2 * X1 + m1 * X2 simplifies the residual system by modulo m1 * M2. Because m2 * X1 + m1 * X2 uses an integer of F (m1 * m2), X1 uses an integer of F (M1), and X2 uses an integer of F (m2, therefore, F (m1 * m2) = f (M1) * F (m2 );

With the above two properties, the Euler's function is very simple. First, an integer is decomposed for a number, and then ···

The following describes a quick method:

 

Any positive integer can be uniquely expressed as follows:
K = p1 ^ A1 * P2 ^ A2 *...... * PI ^ AI; (Form of decomposition prime factor) can be introduced: E (K) = (p1-1) (p2-1 )...... (Pi-1) * (P1 ^ (a1-1) (P2 ^ (a2-1 ))...... (PI ^ (ai-1 ))
= K * (p1-1) (p2-1 )...... (Pi-1)/(P1 * P2 *...... Pi );
= K * (1-1/P1) * (1-1/P2)... (1-1/PK)
PS: using the following functions in the program, you can quickly find the value of the Euler function (A is a qualitative factor of N) if (N % A = 0 & (n/a) % A = 0), E (n) = E (n/a) *; if (N % A = 0 & (n/a) %! = 0) then there are: E (n) = E (n/a) * (A-1 );