180A HW4
Zian Deng
20211966
3.2.
If separately, hackers can probably find the meaning of the letters by studying the letters
frequency and then infer the meaning of the message, thus it is unsafe.
3.3
n=pqr
(n)=(p-1)(q-1)(r-1)
(n)=p+q+r+pq+qr+pr+1+pqr
(n)
180A HW1
Zian Deng
20211966
1.1
1235= 5*13*19
455= 5*7*13
Thus, gcd (455, 1235) = 5*13 =65
1.2
Let X= [3, 5, 7, 9.99] be the list of all odd integers between 3 and 100.
Let P= [2] be the list of prime so far.
Let P= [2, 3], and eliminate all the multiples
180 HW7
Zian Deng
20211966
6.1,
At such a situation, the graph will intersect like
which is a situation we dont like as it has a singular point
According to the definition, there will be no elliptic curves over K
6.2,
Draw the line y=x
We got the equation
180A HW2
Zian Deng
20211966
2.1
Z/nZ = cfw_0, 1, . . . , n 1
Suppose a, b Z/nZ, we have gcd(a,n)=gcd(b,n)=1, thus gcd(ab,n)=1, ab Z/nZ
Thus, this is a binary structure
For a,b,c Z/nZ, we have a*b*c=abc=a*(b*c), thus it is associative
And for every a Z/nZ,
180A HW5
Zian Deng
20211966
4.1. (3/97)=1
(3/389)= -1
(22/11)=0
(5!/7)=1
4.3. (3/p)=(-1)^[(p-1)/2]*(p/3)
(-1)^[(p-1)/2] depends on residue of p modulo 4
(p/3) depends on residue of p modulo 3
4.7. Since 213 1 is prime and 5 0 (mod 213 1), there are either
180A HW3
Zian Deng
20211966
2.12
In order to prove that Z/pZ is a field, we show that Z/pZ has an identity e and for every a Z/pZ
we have a Z/pZ that makes aa=aa= e
For all Z/pZ, we know that 1 Z/pZ as gcd(1, p)=1, and for all a Z/pZ, 1*a a*1 a (mod p)
Th
Hw6
2.
If P has finite order m, then mP=
Thus 2mP=mP+mP= (as is the identity)
As + is associative in the set of non-singular points in RP 2, 2mP=m(2P)= which means that 2P is
of finite order
Similarly, for any k of natural number kmP=m(kP)= which means kP
2.
(a)Suppose ( a + bi 5), ( x + yi 5) are two random elements in the ring
( a + bi 5)( x + yi 5)=(ax-5by)+(xbi+ayi) 5
( a + bi 5)+( x + yi 5)= (a + x)+( bi+ yi) 5
( a + bi 5)-( x + yi 5)= (a - x)+( bi- yi) 5
As the results are all in the set, the set is
HW4
Zian Deng
1.(a) Suppose |, then N()|N()
N()=34, N()=185
Which is a contradiction
Thus does not divide
/=(11-8i)/(3+5i)=(-7-79i)/34
=*(-2i)+(1-2i)
3+5i=(1-2i)*(-1+2i)+i
1-2i=(-i-2)*i+0
Thus the gcd is i
2.
(a)as a+bi|c+di, then (a+bi)(a-bi)|(a-bi)(c+d