Download Chap4 Exponential Inverses

Exponential Inverses
Finding modular inverses is good enough for
decoding simple modular cryptography.
However, in RSA encryption consists of
exponentiating modulo N, i.e. m e mod N.
We want to find a different exponent d
based on e and N which will give us back m,
i.e. we want m de mod N =m. In other
words, we want an exponential inverse for
e modulo N.
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Exponential Inverses.
Prime Modulii
To tackle the general problem, start first with the
case of N a prime number. Exponentiation
modulo a prime number is well understood.
EG: Consider exponentiating 3 modulo 7:
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31 mod 7 = 3
32 mod 7 = 2
33 mod 7 = 6
34 mod 7 = 4
35 mod 7 = 5
36 mod 7 = 1
7. 37 mod 7 = 3
8. 38 mod 7 = 2
9. 39 mod 7 = 6
10.310 mod 7 = 4
11.311 mod 7 = 5
12.312 mod 7 = 1
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Exponential Inverses.
Prime Modulii
Exponentiating to the p -1 power results in 1. Therefore,
any further exponentiation results in a cycling, with
repetitions occurring every 6 exponentiations.
Fermat’s Little Theorem says that this effect
happens for all rel-prime numbers under prime
modulus:
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31 mod 7 = 3
32 mod 7 = 2
33 mod 7 = 6
34 mod 7 = 4
35 mod 7 = 5
36 mod 7 = 1
7.
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37 mod 7 = 3
38 mod 7 = 2
39 mod 7 = 6
310 mod 7 = 4
311 mod 7 = 5
312 mod 7 = 1
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Fermat’s Little Theorem
THM (FLT): Suppose that p is a prime number. If a is not
divisible by p then
a p-1  1 (mod p) .
Furthermore, all numbers satisfy
a p  a (mod p) .
EG: Compute 9100 mod 17:
p =17, so p-1 = 16. 100 = 6·16+4. Therefore,
9100=96·16+4=(916)6(9)4 . So mod 17 we have 9100 
(916)6(9)4 (mod 17)  (1)6(9)4 (mod 17)
 (81)2 (mod 17)  (-4)2 (mod 17)  16
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Exponential Inverses.
Prime Modulii
COR: If e is relatively prime to p –1, where p is prime,
then its exponential inverse modulo p exists and is
the inverse of d modulo p-1.
Proof. Supposing de  1 (mod p-1). Then for some k,
de = 1+k (p-1). So if a is any number not divisible by
p, FLT implies:
ade  a1+k(p-1) (mod p)  a (mod p)
In other words, exponentiating by de doesn’t change
numbers, modulo p, so by definition, d and e are
exponential inverses.
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Exponential Inverses.
Prime Modulii
EG: Find the exponential inverse of 3 modulo
11.
p =11, so p-1 = 10. The inverse of 3 modulo 10 is
7, which is the answer.
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