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l = 1 mod pe. in .NET Implementation qr-codes in .NET l = 1 mod pe.

l = 1 mod pe. using barcode integrated for .net vs 2010 control to generate, create qr code 2d barcode image in .net vs 2010 applications. .NET CF 15.7 Primitive roots in Z/pe with p lx, then let"s show that 91 is already a primitive root mod. pe for any e ~ 1. By Lagrange"s theorem, the order of 91 in Z/pex is a divisor.

of rp(pe) = (p_l)pe-l. Since p - 1 is the smallest positive exponent l 80 that 9f = 1 mod p, p - 1 divides the order of 91 in Z/pex (from our discussion of cyclic subgroups). Thus, the order of 91 is in the list.

p - 1, (P - I)p, (P - I)p2 .net framework QR Code JIS X 0510 , ..

. , (P _ I)pe-l ". Thus, the question is to find the smallest positive l so that We are assuming that gf-l = 1 +px with,p lx, so the question is to find the smallest positive l so that (1 + px)P = 1 mod pe , From the proposition, the smallest positive l with this property is l = e - 1. That is, we have proven that 91 is a primitive root mod pe for every e ~ 1. Now suppose that gf-l = 1 +px with pix.

Then consider. 9 = (1. + P)91. Certainly 9 is still a primitive root mod p, because 9 = 91 mod p. And we compute + p)p-l = 1 + (p ~ l)p + (p;. = + pp-l !)pp-2 1 p. ((p ~ 1) + (p; l)p ~ (p; I)p2 + ...) = 1 + + I)p2 + ... + ~ ...

.. Since 1 (P-I) =p-l we see that y=p-l modp so ply. Thus,. gP-l = 1 + p)91)P-l = (I + py){1 + px) = 1 + p{y + x + pxy). Since pix, we have y+x+pxy=ymodp 15 . Primitive Roots In particular, ply + x + p visual .net QR Code ISO/IEC18004 xy. Thus, by adjusting the primitive root a bit, we have returned to the first case above, that gP-l is of the form gP-l = 1 + pz with p lz.

In that case we already saw that such 9 is a primitive root mod pe for any e ~ 1. This finishes the proof of existence of primitive roots in Zjpe for p an odd. Corollary: (of proof) In f act, for an integer 9 which is a primitive root mod p, either 9 is a primitive root mod pe and mod 2pe for all e ~ 1, or else (1 + p)g is. In particular, if gP-l # 1 mod p2, then 9 is a primitive root mod pe and mod 2pe for all e ~ 1. Otherwise, (1 + p)g is.

. 15.8 Counting primitive roots After proving existence of primitive roots, it is at least equally interesting to have an idea how many there are. Theorem: If Zjn has a primitive root, then there are exactly. cp( cp(n)). primitive roots mod n. (Ye s, that is Euler"s phi of Euler"s phi of n.) For example, there are cp( cp(p)) = cp(p - 1) primitive roots modulo a prime p, and there are.

cp(cp(pe)) = cp(p _ 1) . ( .net vs 2010 QR Code JIS X 0510 p _ l)pe-2.

primitive roots mod pe for an odd prime p and e Proof: The hypothesis that qr-codes for .NET Zjn has a primitive root is that the multiplicative group Zjn X is cyclic. That is, for some element 9 (the "primitive root").

Zinx = (g). Of course, the order Igl o QR Code 2d barcode for .NET f 9 must be the order cp(n) of Zinx. From general discussion of cyclic subgroups, we know that.

is a complete list of all the different elements of (g). And from general properties of cyclic groups k order of 9 order of 9 = gcd{k,lgl) So the generators for (g) are exactly the elements. yk with 1 $ k < Igl and k relatively prime to Igl By definition of Euler"s c p-function, there are cp(lgl) of these. Thus, since Igl = cp{n), there are cp{cp(n)) primitive roots. III.

15.9 Non..existence of primitive roots Corollary: For an odd prim e p, the fraction 1{)(P-l}/p of the elements of Z/yX consists of primitive roots.. Proof: From the theorem just proven the ratio of primitive roots to a.ll elements as claimed. many primitive roots modulo pe. Remark: Thus, there are relatively 15.9 Non-existence of primitive roots For generic integers n, th ere is no primitive root in Zin. Theorem: If n is not 2,4, nor of the forms pe, 2pe for p an odd prime (and e a positive integer), then there is no primitive root modulon..

b = 1 + 2x for integer x. Then Proof: First, let"s look at Z/2 e with e (1 + 2X}2. ~ 3. Any b E Z/2ex can be written as = 1 + 4~ + 4x2 = 1 + 4x(x + I). The peculiar feature here VS .NET QR Code ISO/IEC18004 is that for /mY integer x, the expression x(x+l} is divisible by 2. Indeed, if x is even surely x(x + I} is even, and if x is odd then x + 1 is even and x(x + I} is again even.

Thus, (1 + 2x}2.
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