R.S.A Encryption

R.S.A Encryption through
pell’s equation
By:- N.C.M

STEP 1
Select a secret ODD prime integer “R”

STEP 2
 Consider the Diophantine Equation:
Y2 – R X2 = 1
Let (Y0 , X0 ) be the least “positive”
integral
Solution of . Here X0,Y0 are kept secret.
1
1

STEP 3
Select two large ODD primes p,q
DEFINE:- N: = pq 2

STEP 4
 Define α:= [Y0+ φ(n)]2 – R [Xo + e]2;
Where “e” can be chosen such that
1<e< φ(n) and G.C.D ( e, φ(n)) = 1
Since G.C.D (e, φ(n))=1, there is a unique “positive” integer “d” such
that de≡1(Mod φ(n))
ASSUME
Here φ(n) = Euler’s φ function
3
d3 ≠ 1(Modφ(n))
e3 ≠ 1(Modφ(n))

STEP 5
 From (3), we have
α = Y0
2 +[φ(n)]2 + 2Yoφ(n) −R[Xo
2+e2+2X0e]
= Y0
2 −RXo
2 +[φ(n)]2 + 2Y0 φ(n)− Re2− 2X0eR
α ≡1 − Re2 − 2X0eR (Mod φ(n))
α + Re2 + 2X0eR ≡ 1 (Mod φ(n))
Multiply by d3 on both sides, of the above congruence
We get, αd3+ Rd + 2X0d2R ≡ d3 (Mod φ(n))

STEP 6
 Define:
S = αd3 + 2x0d2R + Rd
so, S ≡ d3 (Mod φ(n))

Step 7
 Represent the given message “m” in
the interval (0, n-1)

Step 8
 ASSUME G.C.D (m,n) =1

Step 9
Encryption :E ≡ mS (mod n)
≡ m +k∙φ(n) (mod n)
≡ m ∙[mφ(n)]k (mod n)
So, E ≡ m (mod n)
Public key : = S, n

Step 10
 Decryption = E (mod n)
= (m ) (mod n)
= m (mod n)
[d3e3 ≡1(mod φ(n)]
=m (mod n)

Step 10 Contd..
d3e3 = 1 +k1∙φ(n)
m = m∙[mφ(n)]k (Mod n)
= m(Mod n)
1

R.S.A Encryption

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