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1. January 31, 2012 1
Cryptography and Network Security
Lecture 7: Public-key cryptography and RSA
Ion Petre
Department of IT, Åbo Akademi University
Spring 2012
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2. January 31, 2012 2
Some unanswered questions on symmetric
cryptosystems
Key management: changing the secret key or establishing one is
nontrivial
Change the keys two users share (should be done reasonably often)
Establish a secret key with somebody you do not know and cannot meet
in person: (e.g., visiting secure websites such as e-shops)
This could be done via a trusted Key Distribution Center (details in a
future lecture)
Can (or should) we really trust the KDC?
“What good would it do after all to develop impenetrable cryptosystems, if
their users were forced to share their keys with a KDC that could be
compromised by either burglary or subpoena?” – Diffie, 1988
Digital signatures
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3. January 31, 2012 3
A breakthrough idea
Rather than having a secret key that the two users must share, each
users has two keys
One key is secret and he is the only one who knows it
The other key is public and anyone who wishes to send him a
message uses that key to encrypt the message
Diffie and Hellman first (publicly) introduced the idea in 1976 – this was
radically different than all previous efforts
NSA claims to have known it since mid-1960s!
Communications-Electronic Security Group (the British counterpart of
NSA) documented the idea in a classified report in 1970
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4. January 31, 2012 4
A word of warning
Public-key cryptography complements rather than replaces
symmetric cryptography
There is nothing in principle to make public-key crypto more secure than
symmetric crypto
Public-key crypto does not make symmetric crypto obsolete: it has its
advantages but also its (major) drawbacks such as speed
Due to its low speed, it is mostly confined to key management and
digital signatures
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5. January 31, 2012 5
The idea of public-key cryptography
The concept was proposed in 1976 by Diffie and Hellman although
no practical way to design such a system was suggested
Each user has two keys: one encryption key that he makes public
and one decryption key that he keeps secret
Clearly, it should be computationally infeasible to determine the
decryption key given only the encryption key and the cryptographic
algorithm
Some algorithms (such as RSA) satisfy also the following useful
characteristic:
Either one of the two keys can be used for encryption – the other one
should then be used to decrypt the message
First we will investigate the concept with no reference yet to practical
design of a public-key system
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6. January 31, 2012 6
Essential steps in public-key encryption
Each user generates a pair of keys to be used for encryption and
decryption
Each user places one of the two keys in a public register and the
other key is kept private
If B wants to send a confidential message to A, B encrypts the
message using A’s public key
When A receives the message, she decrypts it using her private key
Nobody else can decrypt the message because that can only be done
using A’s private key
Deducing a private key should be infeasible
If a user wishes to change his keys – generate another pair of keys
and publish the public one: no interaction with other users is needed
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7. January 31, 2012 7
Bob sends an encrypted message to Alice
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8. January 31, 2012 8
Some notation
The public key of user A will be denoted KUA
The private key of user A will be denoted KRA
Encryption method will be a function E
Decryption method will be a function D
If B wishes to send a plain message X to A, then he sends the
cryptotext Y=E(KUA,X)
The intended receiver A will decrypt the message: D(KRA,Y)=X
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9. January 31, 2012 9
A first attack on the public-key scheme –
authenticity
Immediate attack on this scheme:
An attacker may impersonate user B: he sends a message E(KUA,X)
and claims in the message to be B – A has no guarantee this is so
This was guaranteed in classical cryptosystems simply through knowing
the key (only A and B are supposed to know the symmetric key)
The authenticity of user B can be established as follows:
B will encrypt the message using his private key: Y=E(KRB,X)
This shows the authenticity of the sender because (supposedly) he is the
only one who knows the private key
The entire encrypted message serves as a digital signature
Note: this may not be the best possible solution: ideally, digital signatures
should be rather small so that one can preserve many of them over a
long period of time
Better schemes will be presented a couple of lectures on
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10. January 31, 2012 10
A scheme to authenticate the sender of the message
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11. January 31, 2012 11
Encryption and authenticity
Still a drawback: the scheme on the previous slide authenticate but
does not ensure security: anybody can decrypt the message using
B’s public key
One can provide both authentication and confidentiality using the
public-key scheme twice:
B encrypts X with his private key: Y=E(KRB,X)
B encrypts Y with A’s public key: Z=E(KUA,Y)
A will decrypt Z (and she is the only one capable of doing it):
Y=D(KRA,Z)
A can now get the plaintext and ensure that it comes from B (he is the
only one who knows his private key): decrypt Y using B’s public key:
X=E(KUB,Y)
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12. January 31, 2012 12
Secrecy and authentication using public-key schemes
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13. January 31, 2012 13
Applications for public-key cryptosystems
1. Encryption/decryption: sender encrypts the message with the
receiver’s public key
2. Digital signature: sender “signs” the message (or a representative
part of the message) using his private key
3. Key exchange: two sides cooperate to exchange a secret key for
later use in a secret-key cryptosystem
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14. January 31, 2012 14
Requirements for public-key cryptosystems
Generating a key pair (public key, private key) is computationally
easy
Encrypting a message using a known key (his own private or
somebody else’s public) is computationally easy
Decrypting a message using a known key (his own private or
somebody else’s public) is computationally easy
Knowing the public key, it is computationally infeasible for an
opponent to deduce the private key
Knowing the public key and a ciphertext, it is computationally
infeasible for an opponent to deduce the private key
Useful extra feature: encryption and decryption can be applied in
any order:
E( KUA, D(KRA,X) ) =D(KRA, E( KUA, X) )
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15. January 31, 2012 15
Designing a public-key cryptosystem
Computationally easy usually means polynomial-time algorithm
Computationally infeasible more difficult to define
Usually means super-polynomial-time algorithms, e.g., exponential-time algorithms
Classical complexity analysis (worst-case complexity or average-case complexity) are
worthless in cryptography – one should make sure a problem is difficult for virtually all
inputs and not just in the worse or in the average case
Public-key cryptosystems usually rely on difficult math functions rather than S-P
networks as classical cryptosystems
One-way function: easy to calculate in one direction, infeasible to calculate in the
other direction (i.e., the inverse is infeasible to compute)
Trap-door function: difficult function that becomes easy if some extra information is
known
Aim: find a trap-door one-way function for encryption – decryption will be the inverse
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16. January 31, 2012 16
RSA
One of the first proposals on implementing the concept of public-key
cryptography was that of Rivest, Shamir, Adleman – 1977: RSA
The RSA scheme is a block cipher in which the plaintext and the ciphertext
are integers between 0 and n-1 for some fixed n
Typical size for n is 1024 bits (or 309 decimal digits)
To be secure with today’s technology size should between 1024 and 2048 bits
Idea of RSA: it is a difficult math problem to factorize (large) integers
Choose p and q odd primes, n=pq
Choose integers d,e such that Med=M mod n, for all M<n
Plaintext: block of k bits, where 2k<n≤2k+1 – can be considered a number M with
M<n
Encryption: C=Me mod n
Decryption: Cd mod n = Mde mod n = M
Public key: KU={e,n}
Private key: KR={d,n}
Questions: How do we find d,e? How do we find large primes?
Answer: Number Theory!
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17. January 31, 2012 17
Motto for our introduction to Number Theory
The Devil said to Daniel Webster: "Set me a task I can't carry out, and I'll give you anything
in the world you ask for."
Daniel Webster: "Fair enough. Prove that for n greater than 2, the equation an + bn = cn has
no non-trivial solution in the integers."
They agreed on a three-day period for the labour, and the Devil disappeared.
At the end of three days, the Devil presented himself, haggard, jumpy, biting his lip. Daniel
Webster said to him, "Well, how did you do at my task? Did you prove the theorem?'
"Eh? No . . . no, I haven't proved it."
"Then I can have whatever I ask for? Money? The Presidency?'
"What? Oh, that—of course. But listen! If we could just prove the following two lemmas—"
—The Mathematical Magpie, Clifton Fadiman
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18. January 31, 2012 18
Notions of number theory
Fermat’s little theorem: if p is prime and a is positive integer not divisible
by p, then ap-1 ≡ 1 mod p
Corollary: For any positive integer a and prime p, ap ≡ a mod p
Comments:
This is a first step in our quest to find Med=M mod n – not quite enough though
Fermat’s little theorem provides a necessary condition for an integer p to be
prime – the condition is not sufficient
We will turn this theorem into a (probabilistic) test for primality
Example:
p=5, a=3, 35=243=3 mod 5
p=5, a=10, 105=100000=10 mod 5 = 0 mod 5
Fermat’s theorem, as useful as it will turn out to be, it does not provide us
with integers d,e we are looking for – Euler’s theorem (a refinement of
Fermat’s) does
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19. January 31, 2012 19
Euler’s totient function
Euler’s function associates to any positive integer n a number φ(n): the
number of positive integers smaller than n and relatively prime to n
Example:
φ(37)=36
φ(p)=p-1, for any prime p
φ(35)=24: {1,2,3,4,6,8,9,11,12,13,16,17,18,19,22,23,24,26,27,29,31,32,33,34}
Easy to see that for any two distinct primes p,q, φ(pq)=(p-1)(q-1)
All numbers smaller than pq are relatively primes with pq except for multiples of
p (q-1 of them) and multiples of q (p-1 of them): pq-(q-1)-(p-1)=(p-1)(q-1)
Euler’s theorem: for any relatively prime integers a,n we have
aφ(n)≡1 mod n
Corollary: For any integers a,n we have aφ(n)+1≡ a mod n
Corollary: Let p,q be two odd primes and n=pq. Then:
φ(n)=(p-1)(q-1)
For any integer m with 0<m<n, m(p-1)(q-1)+1≡m mod n
For any integers k,m with 0<m<n, mk(p-1)(q-1)+1≡m mod n
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20. January 31, 2012 20
Back to RSA
Euler’s theorem provides us the numbers d,e such that Med=M mod n
We have to choose d,e such that ed=kφ(n)+1 for some k
Equivalently, d≡e-1 mod φ(n)
To calculate the modular inverse of an interger: the extended Euclid’s algorithm!
see Lecture 5
The RSA scheme
Key generation
Choose two odd primes p,q – keep private. Compute n=pq – make public
Choose e, 1<e<φ(n) with gcd(φ(n),e)=1 – make public
Compute d≡e-1 mod φ(n) – keep private
Private key is {d,n}
Public key is {e,n}
Encryption
Plaintext: block of k bits, where 2k<n≤2k+1 – can be considered a number M with M<n
Ciphertext: C=Me mod n
Decryption:
Ciphertext: C
Plaintext: Cd mod n = Mde mod n = M
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21. January 31, 2012 21
Example
Key generation
Select primes p=17, q=11
Compute n=pq=187
Compute φ(n)=(p-1)(q-1)=160
Select e=7
Compute d: d=23 (use the extended
Euclid’s algorithm)
KU={7,187}
KR={23,187}
Encrypt M=88: 887 mod 187
887 mod 187 = [ (884 mod 187)(882 mod
187) (88 mod 187) ] = 11
Decrypt C=11: 1123 mod 187
M=1123 mod 187= [ (1116 mod 187)(114
mod 187) (112 mod 187)(11 mod 187)]
112 mod 187 =121
114 mod 187= 1212 mod 187=55
118 mod 187=552 mod 187= 33
1116 mod 187=332 mod 187=154
M=154 x 55 x 121 x 11 mod 187 = 88
RSA scheme
–Key generation
•Choose primes p,q
•Compute n=pq
•Choose e, 1<e<φ(n) with gcd(φ(n),e)=1
•Compute d≡e-1 mod φ(n)
•Private key is {d,n}
•Public key is {e,n}
–Encryption
•C=Me mod n
–Decryption:
•Cd mod n = Mde mod n = M
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22. January 31, 2012 22
Computational aspects – RSA implementation
Fast modular exponentiation
Take each step in turn and discuss how
can it be implemented efficiently
For encryption and decryption we must be
able to do quick modular exponentiations
– two ideas are useful:
(ab mod n) = [(a mod n)(b mod n)] (mod n)
To compute x16 mod n we do not have to
do 15 multiplication but only 4: compute x2
mod n, x4 mod n, x8 mod n, x16 mod n
Apply this to compute quickly any
exponent, not just powers of 2
RSA scheme
–Key generation
•Choose primes p,q
•Compute n=pq
•Choose e, 1<e<φ(n) with gcd(φ(n),e)=1
•Compute d≡e-1 mod φ(n)
•Private key is {d,n}
•Public key is {e,n}
–Encryption
•C=Me mod n
–Decryption:
•Cd mod n = Mde mod n = M
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23. January 31, 2012 23
Fast modular exponentiation
Square-and-multiply algorithm
Input: n,x,b (b is in base 2 (bk-1,…,b1,b0), b≠0
Output: xb mod n
1. z=1
2. for i=k-1 downto 0
3. z=z2 mod n
4. if bi=1 then z=zx mod n
Complexity O(r3), where r=[log2n]
Example: encrypt 9726 with
KU={3533,11413}: 97263533mod 11413
3533=(1,1,0,1,1,1,0,0,1,1,0,1)
Ciphertext: 5761
RSA scheme
–Key generation
•Choose primes p,q
•Compute n=pq
•Choose e, 1<e<φ(n) with gcd(φ(n),e)=1
•Compute d≡e-1 mod φ(n)
•Private key is {d,n}
•Public key is {e,n}
–Encryption
•C=Me mod n
–Decryption:
•Cd mod n = Mde mod n = M
i bi z
11 1 9726
10 1 97262x9726=2659
9 0 26592=5634
8 1 56342x9726=9167
7 1 91672x9726=4958
6 1 49582x9726=7783
i bi z
5 0 77832=6298
4 0 62982=4629
3 1 46292x9726=10185
2 1 101852x9726=105
1 0 1052=11025
0 1 110252x9726=5761
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24. January 31, 2012 24
Computational aspects – RSA implementation
Key generation
The highlighted part in the algorithm is
easy to implement
Generate a series of random numbers
and test each against φ(n) for relative
primality
Testing whether or not two integers
are relatively prime and finding a
modular inverse can be done with the
extended Euclid’s algorithm
Very few tests are needed before a
usable e is found: the probability that
two random numbers are relatively
prime is 0.6
RSA scheme
–Key generation
•Choose primes p,q
•Compute n=pq
•Choose e, 1<e<φ(n) with gcd(φ(n),e)=1
•Compute d≡e-1 mod φ(n)
•Private key is {d,n}
•Public key is {e,n}
–Encryption
•C=Me mod n
–Decryption:
•Cd mod n = Mde mod n = M
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25. January 31, 2012 25
Computational aspects – RSA implementation
Key generation
No practical techniques to yield large
prime numbers
Procedure: generate random odd
numbers and test whether that integer is
prime
Testing whether or not an integer n is a
prime is a difficult problem (“primality is
difficult”)
There has been a long standing question
in math whether or not primality can be
tested in polynomial deterministic time
Answer (2002): YES!
Manindra Agrawal, Neeraj Kayal and Nitin
Saxena, “PRIMES is in P”, Ann. of Math.
(2), 160:2 (2004) 781--793.
Drawback: high complexity – O(log12n
f(log log n)), where f is a polynomial
RSA scheme
–Key generation
•Choose primes p,q
•Compute n=pq
•Choose e, 1<e<φ(n) with gcd(φ(n),e)=1
•Compute d≡e-1 mod φ(n)
•Private key is {d,n}
•Public key is {e,n}
–Encryption
•C=Me mod n
–Decryption:
•Cd mod n = Mde mod n = M
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26. January 31, 2012 26
Miller-Rabin primality test
Faster methods of testing primality exist – they are all probabilistic
Such an algorithm can give two answers to the question “Is n prime?”
1. No, it is not
2. n is probably prime
The probability can be made arbitrarily large
Other algorithms may give precise answer but with low probability they
may take a long time to finish
Most popular primality test: Miller- Rabin, based on Fermat’s little
theorem
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27. January 31, 2012 27
Miller-Rabin primality test
Fermat’s little theorem: if p is prime and a is positive integer not divisible by p, then ap-1 ≡ 1 mod p
Idea of the Miller-Rabin test:
We need to test if the odd integer n is prime: test the equality in Fermat’s little theorem
for n and a random a
A speedup may be done so that we do not have to compute all powers of a – details bellow
n-1 is even, i.e., of the form n-1=2kq, with k>0, q odd: k and q easy to find
Choose an integer a such that 1<a<n-1
Compute modulo n the values a2jq, 0≤j≤q: aq, a2q,…, a2k-1q, a2kq
By Fermat’s theorem, if n is prime, then the last value in the sequence is 1 – the
sequence may have some other 1s, consider the first 1 in the sequence
Case 1: the first number in the sequence is 1 – then all other powers are also 1
Case 2: some number a2jq in the sequence is 1 – in this case a2j-1q = n-1 mod n
0 = (a2jq -1) mod n = (a2j-1q – 1) (a2j-1q + 1) mod n, i.e., n divides (a2j-1q – 1) or (a2j-1q + 1)
Since we took the first 1 in the sequence, it follows that n divides (a2j-1q + 1): a2j-1q = n-1 mod n
The test: if either the first element in the sequence is 1, or some other element is n-1,
then n could be prime. Otherwise n is certainly not prime
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28. January 31, 2012 28
Miller-Rabin primality test
TEST(n)
1. n-1=2kq: compute k and q
2. Select a random integer a,
1<a<n-1
3. If aqmod n=1 then return
“probably prime”
4. For j=0 to k-1 do
5. If a2jq mod n = n-1, then return
“probably prime”
6. Return “not a prime”
•Question: for how many integers a does the test fail?
•Failure: n is not prime but the algorithm return
“probably prime”
•Answer: for at most (n-1)/4 integers a with 1≤a≤n-1
•Thus, the probability of failure is at most ¼
•Practical implementation:
•Repeatedly invoke TEST(n) using random
choices for a
•If TEST(n) return at least once “not a prime”,
then n is not a prime
•If t executions of TEST(n) return “probably
prime”, then the probability that n is indeed a
prime is larger than 1-4-t
•t=10 gives probability larger than 0.999999
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29. January 31, 2012 29
Computational aspects – RSA implementation
Key generation
To choose primes p,q we generate random
numbers p,q on the desired scale of magnitude
and test the primality with Miller-Rabin
Question: How many trials should we expect to
do before we find a prime?
Distribution of primes
Prime number theorem: for any integer x, the
primes near x are spaced in average one every
log(x) integers
On average we have to test log(x) integers before
we find a prime – reject immediately even integers
and integers ending in 5
Correct rate: we need to test in average 0.4 log(x)
integers before we find a prime of the order of x
Example: if we look for a prime on the order of
magnitude 2200 we need to do in average 55 trials,
order of magnitude 21024: in average 284 trials
RSA scheme
–Key generation
•Choose primes p,q
•Compute n=pq
•Choose e, 1<e<φ(n) with
gcd(φ(n),e)=1
•Compute d≡e-1 mod φ(n)
•Private key is {d,n}
•Public key is {e,n}
–Encryption
•C=Me mod n
–Decryption:
•Cd mod n = Mde mod n = M
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30. January 31, 2012 30
Attacking RSA
Brute force attacks: try all possible private keys
As in the other cases defend using large keys:
nowadays integers between 1024 and 2048 bits
Mathematical attacks
Factor n into its two primes p,q: this is a hard problem
for large n
Challenges by RSA Labs to factorize large integers
Last solved challenge: 768 bits (2009)
Determine φ(n) directly without first determining p,q:
this math problem is equivalent to factoring
Determine d directly, without first determining φ(n): this
is believed to be at least as difficult as factoring
Suggestions for design
The larger the keys, the better but also the slower the
algorithm
Choosing p,q badly may weaken the algorithm
p,q should differ in length by only a few bits: for a 1024-bit
key, p,q should be on the order of magnitude 1075 to 10100
p-1 and q-1 should both contain a large prime factor
gcd(p-1,q-1) should be small
d should be larger than n1/4
RSA scheme
–Key generation
•Choose primes p,q
•Compute n=pq
•Choose e with gcd(φ(n),e)=1
•Compute d≡e-1 mod φ(n)
•Private key is {d,n}
•Public key is {e,n}
–Encryption
•C=Me mod n
–Decryption:
•Cd mod n = Mde mod n = M
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31. January 31, 2012 31
Attacks on RSA
Timing attacks: determine a private key by keeping track
of how long a computer takes to decipher a message
(ciphertext-only attack) – this is essentially an attack on the
fast exponentiation algorithm but can be adapted for any
other algorithm
Whenever a bit is 1 the algorithm has more computations to
do and takes more time
Countermeasures:
Ensure that all exponentiations take the same time before
returning a result: degrade performance of the algorithm
Add some random delay: if there is not enough noise the
attack succeeds
Blinding: multiply the ciphertext by a random number before
performing exponentiation – in this way the attacker does not
know the input to the exponentiation algorithm. (implemented
in the commercial products from RSA Data Security Inc.)
Decryption M=Cd mod n is modified as follows:
Generate a secret random number r between 0 and n-1
Compute C’=C(re) mod n where e is the public exponent
Compute M’=(C’d) mod n with the ordinary exponentiation
Compute M=M’r-1 mod n
Reported performance penalty: 2 to 10%
Square-and-multiply algorithm
– Input: n,x,b
(b is in base 2 (bk-1 ,…,b1,b0)
– Output: xb mod n
1. z=1
2. for i=k-1 downto 0
3. z=z2 mod n
4. if bi=1 then z=zx mod n
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32. January 31, 2012 32
Pseudo-random number generators
Essential in RSA (and elsewhere) to be able to generate pseudo-
random numbers
A sequence of numbers is random if they have uniform distribution and
are independent (no value can be deduced from the others)
We generally use algorithmic techniques to generate such numbers –
they will not be independent and thus not random
The whole point is to make them “look” random, i.e., make them pass many
test of randomness
Three tests to be used in evaluating a pseudo-random number
generator
The function should be full-period generating function: generate all numbers
in its range before repeating
The generated sequence should appear random: pass many statistical tests
The function should implement efficiently with 32-bit arithmetic
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33. January 31, 2012 33
Pseudo-random number generators
The most widely used technique is the linear congruential method (Lehmer
1951)
Xn+1=(aXn+c) mod m
One should be very careful in choosing constants a, c, m: a=c=1 is bad
choice!
Value of m should be as large as possible: usually close to 231, very often
chosen to be the prime number 231-1; in this case one can take c=0
There are very few good choices for a: for m= 231-1 only a handful of choices
are advisable – very often used is a=75=16807
Xn+1=16807 Xn mod (231-1)
Using this in cryptography needs extra care:
If the attacker finds one single value, then he will be able to compute all
subsequent values
Idea: restart the sequence often, using the clock as seed (initial value)
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34. January 31, 2012 34
Cryptographically generated pseudo-random numbers
Idea: use cryptographic primitives to generate
pseudo-random numbers
One possibility: Use a counter and encrypt each
value for the counter, e.g., with DES – the
cryptotext will be the key
Stronger version: instead of a counter use a
PRNG (pseudo-random number generator)
Technique can be made stronger using a more
sophisticated scheme and 3DES, see ANSI
X9.17 PRNG
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35. January 31, 2012 35
Another speed-up in RSA implementation
Operations modulo big integers become more time-consuming as
the integers grows bigger
Efficient implementation: use Chinese Remainder Theorem (CRT)
In its simplest formulation, CRT essentially says that if n=pq, then
instead of addition/difference/multiplication modulo n one can perform
the same modulo p and modulo q and then compute the result mod n
Big advantage because the modules are much smaller
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