2. History
Many civilizations have
come up with secret
codes in order to pass
information
First people to give a
name to this practice were
the Greeks
3. History
Definition: Cryptography, or hidden in Greek, was the
practice and study of techniques for safe
communication in the presence of a third party.
4. History
Early cryptography was solely concerned with converting
messages into unreadable groups of figures to protect
the message’s content during the time the message was
being carried from one place to another (New World
2007).
5. Process
The process of cryptography begins with the encryption.
The creator of this encryption shares how to decode the
message with the member he would like to send the
message too.
6. Key Encapsulation
Class of encryption techniques designed to secure
symmetric cryptographic key material for transmission
using asymmetric (public-key) algorithms.
Symmetric-key algorithms have the same cryptographic
keys for both encryption and decryption
7. Key Encapsulation
Public key systems are clumsy to use in transmitting long
messages.
Instead they are often used to exchange symmetric keys,
which are usually short.
This symmetric key is then used to encrypt the longer
message
8. Example
During WWII, the German’s used a shift when they would
crypt a message.
This is an example of a symmetric key.
9. Process
Since the first World War, cryptography and the process
of encryption have become very difficult.
Today, encryption and decryption are heavily involved
with mathematical theory.
These algorithms are hard to break and almost impossible
for an inexperienced adversary to figure out.
10. RSA
The RSA or the Rivest, Shamir, Adleman was the first
practicable public-key (asymmetric) cryptosystem.
Asymmetric cryptosystem – requires two separate keys
(public and private)
Today it is used in order to secure data transmission.
11. RSA
In this system, the encryption key is public while the
decryption key is kept a secret.
Therefore the key doesn’t need to be transmitted.
Instead of sending how to decrypt the message all you
have to send is the cyphertext itself.
If intercepted with a symmetric system the way to
decrypt the system is included with the cyphertext.
12. RSA
This kind of algorithm is based on the practical difficulty of
factoring the product of two large prime numbers.
In order to create an encryption, the user must base the
public key off of two large prime numbers.
If the prime numbers are large enough, factoring them
would take even a computer a lengthy period of time.
13. Example
Instead of factoring a large prime number like 44,345,523
to find 2 prime numbers.
We can multiply two prime numbers together in order to
create a public key.
15. RSA
The prime numbers must be kept a secret
Since the encryption key is public anyone can encrypt a
message, however if the public key is large enough, only
someone who knows the prime numbers can decode
the message.
16. Operation
First, create a key.
Choose two distinct prime
numbers p and q
These should be of similar digit
length and take the Euler’s
Totient Function (phi function) of
the two numbers.
Screenshot
Euler’s Totient function – is an arithmetic function that counts the
totatives of y, that is, the positive integers less than or equal to y
that are relatively prime to y
18. Lemma
Lemma 1: If p is a positive prime number, then Φ(p) = p – 1.
Proof: If p is prime, then since its only positive divisors are 1 and p,
all of the integers 1, 2, 3, … , p – 1 are relatively prime to p.
19. Operation
By multiplying p and q together we receive and integer y.
Now we take the Euler Totient function of y.
21. Lemma
Lemma 2: For p and q distinct primes, Φ(pq) = Φ(p) Φ(q) = (p -
1)(q - 1).
Proof: There are pq - 1 natural numbers smaller than pq including
p, 2p, 3p, 4p, …, (q - 1)p and q, 2q, 3q, 4q, …, (p - 1)q, all of which
have a factor other than 1 in common with pq.
Since there are (q - 1) + (p - 1) of these numbers, this leaves
(pq) – 1 – (p - 1) – (q - 1) = pq – p – q + 1 = (p - 1)(q – 1) numbers
with no factors in common with pq other than 1.
This shows that Φ(pq) can be applied to both p and q at the
same time.
22. Operation
So by using this function designated Φ we can compute
Φ(y) = Φ(p) Φ(q) = (p - 1)(q - 1)
Now we choose an integer x so that 1< x <Φ(y) and the
gcd(x, Φ(y)) = 1
In other words, x and Φ(y) are coprime.
24. Lemma
Lemma 3: If p is prime and k is a positive integer, then gcd(n,
pk) = 1 if and only if p does not divide n.
Proof: First of all if p|n, then gcd(n, pk) ≠ 1 because p also
divides pk.
Conversely, if gcd(n, pk) ≠ 1 then n and pk are not relatively
prime, i.e., they share a common factor greater than 1.
But the only factors of pk are powers of p and so p|n.
When choosing x, x must be coprime and thus the
gcd(x, Φ(y)) = 1
25. Lemma
Lemma 4: If p is prime and k is a positive integer, then
Φ(pk) = pk – pk-1.
Proof: There are precisely pk-1 integers between 1 and pk
that are divisible by p, namely {p, 2p, 3p, …, pk-1p}.
So of the pk that are positive and less than or equal to pk,
pk – pk-1 of them do not have a factor in common with pk.
This shows (pk) = pk – pk-1
Example: 5
26. Lemma
Lemma 5: For n > 2, Φ(n) is an even integer.
Proof: Divide this problem into two cases:
Case 1: n is a power of 2, i.e., n = 2k (k > 1). Then
Φ(n) = Φ(2k) = 2k – 2k-1 = 2k-1(2 – 1) = 2k-1, which is even.
Case 2: n is not a power of 2. In this case n is divisible by an odd
prime p, so n = pkm (k ≥ 1) and gcd(pk, m) = 1.
So, Φ(n) = Φ(pkm) = (pk – pk-1) Φ(m) = pk-1(p-1) Φ(m)
Which is even (because 2|(p-1)).
27. Operation
So now x is released as the public key exponent.
Now we must determine d (private key).
d ≡ x -1(mod Φ(y))
Thus, we solve for d given that dx ≡ 1 (mod Φ(y)).
And thus d is kept as the private key exponent.
29. Euler’s Theorem
Euler’s Theorem shows us that for some prime number n, if
n and a are coprime then n and a have no common
factors, n|(aΦ(n) – 1) and so, aΦ(n) ≡ 1(mod n).
30. Operation
So far now we have the public key which consists of the
modulus y and the public exponent x.
The private key however consists of the modulus y and
the private exponent d.
Thus the variables a, b, Φ(y) must be kept secret because
they are used to calculate d.
31. Operation
The next step is to create the encryption.
Therefore, give out your public key (y,x) and keep the d
value a secret.
Now we wish to send out a message called TAP.
Make TAP an integer by using ASCII notation designated
by b, such that T is 20, A is 01, and P is 16.
32. Operation
Next you produce the cipher text c corresponding to
c ≡ bx (mod y).
This is done through modular exponentiation.
Now we send the c value to the recipient
36. Euler’s Theorem
If n is a positive integer with gcd(a, n) = 1, then aΦ(n) ≡ 1(mod n).
Proof: If n =1, we wish to show that a0 ≡ 1(mod 1) which is
obviously true. So assume that n > 1 and let S = {a1, a2, …, aΦ(n)}
be the set of positive integers less than n that are relatively prime
to n.
Since gcd(a, n) = 1, and we know that aa1, aa2, …, aaΦ(n) are
congruent to a1, a2, …, aΦ(n) in some order.
aa1 ≡ a’1(mod n),
aa2 ≡ a’2(mod n),
…
aaΦ(n) ≡ a’ Φ(n)(mod n).
37. Euler’s Theorem
Taking the product of all of the congruence's yields
(aa1)(aa2)…(aaΦ(n)) ≡ a’1a’2…a’Φ(n)(mod n) = a1a2…
aΦ(n)(mod n),
or
aΦ(n)(a1a2…aΦ(n)) ≡ a1a2…aΦ(n)(mod n).
Letting x = a1a2…aΦ(n), we have aΦ(n)x ≡ x(mod n) and
gcd(x, n) = 1.
Now aΦ(n)x – x = kn for some integer k implies x(aΦ(n) – 1) = kn. Since
n|kn, n|x(aΦ(n) – 1). But since n and x have no common factors,
n|(aΦ(n) – 1) and so, aΦ(n) ≡ 1(mod n).
38. Works Cited
http://searchsecurity.techtarget.com/definition/cryptology
http://mathworld.wolfram.com/RSAEncryption.html
https://crypto.stanford.edu/~dabo/papers/RSA-survey.pdf
http://www.studentpulse.com/articles/41/a-brief-history-of-
cryptography
http://www.laits.utexas.edu/~anorman/BUS.FOR/course.mat/SSim/
history.html
MatLab
Cryptological Mathematics by Robert Edward Lewand
