The
Mathematics of RSA Encryption
The Quotient Remainder Theorem:
Given integers m, and n>0, there exist unique integers a and r such that
m= a∙n + r where n > r ≥ 0 .
In particular m = r mod(n) .
Lemma 1:
Let p be a prime. Let e , 0
<e< p, and d, 0 < d< d'
<p then ed' ≠ ed (mod p).
Proof:
If ed' = ed (mod p) then e(d'-d)=0 (mod p)
. That is e(d'-d)=ap But p is a prime.
Lemma 2:
Let p be a prime. For every e , 0
<e< p, there is a unique d,
0<d<p such that ed= 1 (mod p). In
particular
(p-1)!=1 (mod p).
Proof:
A counting argument. There are p-1 integers
0<d<p and from Lemma 1 ed (mod
p) is unique for each. So for some d
,ed= 1 (mod p).
To compute (p - 1)! (mod p). If p=2 then
(p - 1)!=1. for p > 2, 1
2 = 1 (mod p). and
(p - 1) 2 = 1 (mod
p).
For 1 < e < p -1 , e2 ≠1 (mod
p). If it were then (e-1)(e+1)=0 mod(p) but p
∤(e -1) and p ∤(e+1).
So
e≠d where ed=1 (mod p). So (p -
1)! is a product of integers or pairs of integers = 1 (mod p)
.
Fermat's Little Theorem:
Let p be a prime which does not divide the integer
a, then a (p-1) = 1 (mod p).
Proof:
Noting that we can assume 0 < a < p, and that
ma ≠na (mod p) for 0 <m ≠n < p ,
The set of integers
(1a , 2a,....,(p-1)a) are unique (mod p)
As in Lemma 2 , their product a (p-1) (p-1)! = 1 (mod p) . But
(p-1)!=1 (mod p) so a (p-1) =1 (mod p) .
A Special Case of the Chinese Remainder Theorem:
Let p ≠
q be 2 primes. Then for any
a and b,the system of equations
x=a (mod p)
x=b (mod q)
has a unique solution for x (mod pq).
Proof:
Let p' and q' be such that pp'=1
(mod q) and qq'=1 (mod p) . Let x = bpp' +
aqq'
(bpp' + aqq') (mod p)=aqq' (mod
p)=a
and
(bpp' + aqq') (mod q)=bpp' (mod
q)=b
To prove uniqueness check that if z satifies the equations
then (x-z) =0 mod (p) and mod (q) so is
divisable by p and q , hence =0 mod
(p) and mod (q) .
The RSA Computation:
For any a, a(p-1)(q-1)=1 mod
(pq)
Proof:
a(p-1)(q-1)=1(q-1)=1
mod(p)
and
a(p-1)(q-1)=1(p-1)=1
mod(q)
The computation now follows by uniqueness of solution.