Euler's theorem

We define Ψ (x) as Euler functions.

• Euler theorem Description:

If (A, p) = 1 , then aψ (P) ξ1 (mod p) .

• Prove:

Warm up first:

Ⅰ. We make x1, x2, x3, ..., xs a simplified residual system for modulo p (if for any 1≤j≤s, (XJ, p) = 1 and for any a∈z, if (A, p) = 1 , then there is only one XJ that is the remainder of a to modulo p (and xI 22 is not the same).) where s =ψ (p) .

Ⅱ. If {x1,x2,x3,......, xs} is a simplified remainder of mod p, then if (k, p) = 1, {k x1,k x2,k x3,......, K XS} is also a simplified residual system for mod p , proving that:

　　　　　　　If {k x1,k x2,k x3,......, K XS} is not a simplified residual system for mod p,

Then there is at least one set of K xiξk XJ (mod p), I! = J:

　　　　　　　　　　　　∴k (XI-XJ) ξ0 (mod p)

　　　　　　　　　　　　Also ∵ (k, p) = 1, ∴ (XI-XJ) ξ0 (mod p);

　　　　　　　　　　　　this with XI, Xj∈{mod P's simplified residual system} is contradictory, so the hypothesis is not established, and if (k, p) = 1, {k x1,k x2,k x3,......, K XS} is also the simplified remainder of mod p is established.

　　　　

Let's start with the real proof:

　　　　　　We make {A1, A2, A3, ..., an} a simplified remainder of mod p because the Euler theorem (A, p) = 1, so {a a1, a a2, a A3, ..., a an} is also a simplified remainder of mod p.

∴A1 A2 A3 ...... · Anξa A1 A A2 A A3 ...... · A anξaψ (p) · A1 A2 A3 ...... · An (mod p);

　　　　　　

　　　　　

　　　　　

　　　　　∴a　 ψ (p) -1ξ0 (mod p), ∴a ψ (p) ξ1 (mod p) proof.

• Simple application of Euler's theorem:

　　　　Ask a x mod p, where (A, p) = 1.

x = S. Ψ (p) + q, q <ψ (p)

∴axΞas Ψ (p) + qΞ (Aψ (p))s ·aQΞaQΞax mod p(mod p)

　　　　

Euler's theorem