number theorey?

Question

Let p be any prime congruent to 1 (mod8) and let g be a genrator for Z* p
Explain why there is an element x of order 8 in Z*p
and why x^2= -x^(-2)

well.. this is an assignment i got and i sure dun understand it.. mayb u can try the link of
http://www.maths.usyd.edu.au/u/UG/SM/MATH3062/r/3062ass1.pdf
its just the question 2.. thanks a lot for anyone to help

sorry again.. maybe if someone can show me why is x^2 = -X^(-2)

and do question 2C please? thanks

Answer 1

Let's let g be a primitive root mod p (i.e. a generator
for Z_p). Then g has order p-1 mod p.
From this it follows that x = g^( (p-1)/8 ) has order 8 mod p.
Let's prove it.
First, x^8 = 1(mod p), so the order of x cannot exceed 8.
And, if the order of x were 1, 2 or 4 then
g would have order smaller than p-1.
For example if x had order 4 then x^4 = g^( (p-1)/2) would
be 1 mod p.
For example, 41 = 1(mod 8) and 6 is the least primitive root
mod 41.
So 6^5 = 27 has order 8 mod 41.

Question 2. Why does x² = -1/x² mod p?
I'll do this step by step.
Multiply both sides by x². Then It is enough to show
x^4 = -1 mod p
Look at x^8 mod p.
We know x^8 = 1 mod p
So x^8 - 1 = 0(mod p).
Factor as the difference of 2 squares:
(x^4-1)(x^4+1) = 0 mod p
Now set each factor to 0 and solve. We can
do this because the zero divisor law holds in Z_p.
We know the first factor cannot be 0 because
x has order 8 mod p
So x^4+1 = 0 mod p as required.
To continue this further(for part C.)
Look at (x-1/x)². = x² - 2 + 1/x² = -2 mod p
also (x+1/x)² = x² + 2 + 1/x² = 2 mod p.
This technique actually gives a practical way of
solving y² = 2 mod p and y² = -2(mod p).

Part C. The last 2 digits of my FID are 74.
So we must find the first prime larger than 374
which is congruent to 1 mod 8.
The first number larger than 374 which = 1 mod 8
is 377.
But 377 = 13*29.
Also 385 and 393 are both composite,
so our first suitable prime is 401.
Here the calculations become messy. Fortunately,
I have PARI to help me!
A primitive root mod 401 is 3 and an element
of order 8 is therefore 3^50 = 45 mod 401.
Also, we need the inverse of 45 mod 401
and it turns out to be 303.
When you solve your problem you will have
to solve the congruence ax = 1 mod p to find the inverse.
So to solve x² = -2 mod 401
we compute
45-303 = -258
and 303 - 45 = 258
and the squares of these are both 399 = -2 mod 401.
Now try and tackle your part C.
Hope that helps!

Answer 2

If p = 8n+1 then p-1=8n and then Z*p has elements whose orders are any divisors of p-1 and 8 is one of these.
since x^8=1 mod p we have (x^4+1)(x^4-1) = 0modp but
(x^4-1) is not 0mod p since x has order 8 by hypothesis
so (x^4+1) = 0 mod p, implies x^4 = -1 modp and multiplying
both sides by the square of the inverse of x modp we get
x^4x^(-2) = -x^(-2)modp and the left side is just x^2 so
x^2 = -x^(2) mod p.

Answer 3

let g be a genrator for Z* p
Where do you use it?

Explain why there is an element x of order 8 in Z*p
false, Z9 doesn't have any element of order 8

x^2= -x^(-2)
That doesn't make sense

Answer 4

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