NUMBER THEORY AND GEOMETRY

Number Theory and Geometry



Number Theory and Geometry

Question 3

We know that

Having finite order m, let f be an element of ?G,??.

Since fm =e , fm (1) needs to be 1 . By the second claim, f m (1)=f(1) m , so it follows that f(1)=1 . Next, f m (n) needs to be 0 for any n?1. Take the smallest n>1 for which f(n)?0 . We have that

f m (n)=mf(1) m-1 f(n)+? F(a 1) …… F(am) = m f (n) = =0.

We understood that f(n)?0 , so we must have that m=0 , but disagree with that f has finite order. So f(n)=0 for all n>1 .

Thus we have that the only f with finite order e.

Question 4

Solution

m = 291, n = 3053, e = 17

The above message can be encrypted by the formula

C = me mod n

C = 29117 mod 3053 ….. (a)

Now using n = 3053 into prime factors = 43 x 71

Such that (p - 1) (q - 1) = 42 x 70 =2940

Now K = 1 mod r i.e. finding factors of k we get K = 17 x 173 which are relatively prime numbers.

Using the eq. (a) we get C (Cipher) = 3044 (Encrypted Message)

Question 6

Proof

We have to calculate -3.-6.-9… (p - 1) in two different ways. The first way i.e. let P =. Then

-3.-6.-9… (p - 1) = (-3.1) (-3.2)…… (-3.p) = 2pP!.

The second way is that we divide -3.-6.-9… (p - 1) into groups of numbers greater than and less or equal to p. The greater ones (I of them), are all negative odd numbers -p mod (p). Hence

-3.-6.-9… (p - 1) = (-1)lP! mod (p)

Note also that Euler's Criterion 2p = () mod (p). We have

2pP! (-1)lP! mod (p)

() 2p (-1)l mod (p)

We ...