By induction, for d in .NET Implementation Quick Response Code in .NET By induction, for d

By induction, for d generate, create qr bidimensional barcode none in .net projects ISO QR Code standard 15 . Primitive Roots Xd - 1 =. O<e~d,eld <Pe(x). the gcd of <Pm and <Pn is 1, we have Since we have already shown that for m lcm{xd - 1 : dln,O <. n < d < n} =. d/n,d<n <Pd(X). Thus,. Xn -. 1 = <Pn(x),. dln,d<n <Pd(X). as claimed. The assertion a .NET qr bidimensional barcode bout the degree of <Pn follows from the identity proven below for Euler"s phi-function: <p(d) = n dln,d>O.

This completes the proof of the theorem. Proposition: Let <p(x) be Euler"s phi-function <p(X). l~t~z;gcd(l,z)=l Then for m and n relatively prime <p(mn). = <p(m) . <p(n). <p(p"). (weak multiplicativity). For p prime and f a positive integer = (p _ 1) . p"-l cp(d). dln,d>O Proof: For the first statement, note that by unique factorization gcd(t, mn) = gcd(t, m) . gc d(t, n) In particular, t is relatively prime to mn if and only if t is relatively prime to both m and n. Recall that the gcd of m and n is the smallest positive integer expressible as rm + sn.

By Sun-Ze"s theorem, the map. f: {O, 1, ...

,m -I} x {O, qr barcode for .NET 1, ..

. ,n -I}. Z/mn 15.5 Primitive elements in finite fields: proofs f: (x,y) - rmy+ mx is a bijection. From rm + y QR Code JIS X 0510 for .NET n = 1, rm = 1 mod n 80 rm is relatively prime to n, and sn = 1 mod m 80 an is relatively prime to m.

Thus, rmy + snx has a common factor with m if and only if x does, and rmy + snx has a common factor with n if and only if y does. Thus, f also gives a bijection. {x: 1 ~ x < m,gcd(x,m). = I)} x {y: 1 ~ y < n,gc QR Code for .NET d(y,n) = I)}. - {z: 1 ~ z < mn,gcd(z,mn). is a bijection. This proves visual .net Quick Response Code that for gcd( m, n).

= I)}. <p(mn). = <p(m) . <p(n). Using unique factorization, qrcode for .NET this reduces the calculation of <pO to its evaluation on prime powers pe (p prime). This is easy.

88 an integer x in the range 1 ~ x < pe is relatively prime to pe if and only if it is not divisible by p, 80 there are. such x, 88 claimed. To obtain the formula dln,d>O <p(d) = n start with the case that n is a prime power pe, in which case Then use the weak multiplic ativity and unique factorization of divisors into their prime power factors. Let n = p~l ..

. pft be the prime factorization of n into powers of distinct primes Pi. We have.

This proves the desired identity for <p. 15.5 Primitive elements in finite fields: proofs Now we can prove that the m .net vs 2010 qr bidimensional barcode ultiplicative group P of a finite field k is a cyclic group. A generator of k X is 80metimes called a primitive root for k.

Thisproperly of P is essential. (Note that the superscript is a cross x, not an x.).

Theorem: Let k be a finite field. Then P is a cyclic group. 15 . Primitive Roots Proof: Let q be the number of elements in k. The group of units k X is a group. Since k is a field, any b:F 0 has a multiplicative inverse in k. So the order of P is q - 1. Thus, by coroll.

aries to Lagrange"s theorem, for b:F 0,. That is, any non-zero eleme Quick Response Code for .NET nt of k is a root of the polynomial f (x) = x q - 1 - 1. On the other hand, by a fundamental theorem from algebra, a polynomial with coefficients in a field k has at "most as many roots (in k) as its degree, so this polynomial has at most q - 1 roots in k.

Therefore, it has exactly q - 1 (distinct) roots in k. Let p be the characteristic of k. Certainly p cannot dividt:) q - 1, since if it did then the derivative of f(x) = x q - 1 - 1 would be zero, so gcd(f, I") = f and f would have multiple roots.

We have just noted that f has q - 1 distinct roots, so this doesn"t happen. Since the characteristic of k does not divide q - 1, we can apply the results from just above concerning cyclotomic polynomials. Thus,.

x q- 1 - 1 =. II !Pd(X). dlq-l Since x q- 1 - 1 has q - 1 QR Code for .NET roots in k, and since the !Pd"S here are relatively prime to each other, each !Pd with dlq - 1 must have a number of roots (in k) equal to its degree. Thus, !Pd for dlq - 1 has !p(d) > 0 roots in k (Euler"s phi-function).

Finally, the roots of !Pq-l(X) are those field elements b so that bq- 1 = 1 and no smaller positive power than q - 1 has this property. The primitive roots are exactly the roots of !Pq-l(X). The cyclotomic polynomial !Pq-l has !p(q - 1) roots.

Therefore, there aie !p(q - 1) > 0 primitive roots. Thus, the group k X has a III generator. That is, the group P is cyclic.

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