Maths Major Assignment Solutions for Student ID 1427755

Maths Major Assignment Solutions
Student ID = 1427755
Using the student ID number we have to generate the Private and Public Keys.
Then,
Using nextprime function two prime numbers of form 4n+1 are generated
The two numbers hence obtained are,
P1=1427773 and P2=1427809
Then,
∅ (N) = (P1-1) (P2-1)
= 1427772 * 1427808
= 2038584283776
N = P1 * P2
= 1427773 * 1427809
= 2038587139357
Using the formula, ed=1Mod(∅ (N))
we get the value of,
e = 834301631117
d = 950292999365
Therefore,
Public Key (Pu) = (N,e)= (2038587139357, 834301631117)

Private Key (Pr) = (N,d)=( 2038587139357, 950292999365)
Then, from the formula,
C= Pe mod N
= 1427755^834301631117 mod 2038587139357
= 608234997909
Then, using the formula,
P= Cd mod N
= 608234997909^950292999365 mod 2038587139357
= 1427755
The value of P is 1427755 which is the required original student ID number.

Breaking the keys:
P1= 1427773
P2= 1427809
The numbers P1 and P2 are then represented as the sum of two squares.
(1082)2 + (507)2 = 1427773
(695)2 + (972)2 = 1427809
Then, two right angled triangles are drawn,
So, For the 1st triangle
C = n2 + (2m + n - 1)2
= 5072 + (2m + 507 - 1)2
(1082)2 = (2m + 507 - 1)2
2m = 1082–506
m = 288
So, (m1, n1) = (288, 507)
For 2nd triangle,
C = n2 + (2m + n - 1)2
= 6952 + (2m + 695 - 1)2
(972)2 = (2m + 695 - 1)2
2m = 972 – 694
m = 139
So, (m2, n2) = (139, 695)
1082
507
√1427773
972
695
√1427809