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p. 193 #11.
Suppose that
and
We cannot have
=
, since
otherwise
by unique factorization in the integers. Without loss of
generality assume that
=
, the case of
being similar. Then we can
quickly compute
by using the Euclidean algorithm on
and
. However, once
we know one of the proper factors of each of the n's, then by division we can
quickly find the other factor. Knowing both factors of n, we can find
in both
cases. Once we know
, we can find the modular inverse of the public
encryption exponent
e
in both the cases, yielding the decryption exponents
d
in
both cases. The RSA systems are broken.
p. 193 #14.
Since
=
, it follows that pq  (x + a)(x  a). Hence p  (x+a)(xa),
and in particular, the prime p divides one of the two factors. However, pq does not
divide x + a or x  a. Therefore p divides only one factor while q divides only the
other.
Without loss of generality, assume that x + a =
and
x  a =
.
Hence,
=
=
, or
.
We would like to find
k
and
j.
But since
, we know that there exist integers
s
and
t
for which
. Taking
and
. we get find solutions for k and j such
that
.
We know that
=
since
Now let x be the common
value of these two expressions: x =
=
.
We now show that such an x has the desired property.
We see that
= (
Hence,

.
Using the other form of x, we see that
so

Since p and q are distinct primes, it follows that

Hence,
=
For example, with
p = 71 and q = 73, then (36)(71)  (35)(73) = 1, so, taking a = 7,
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(2.1)
restart:
(14*36)*717;
%^2 mod(71*73);
35777
49
p. 195 #22.
a) Since
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 Spring '08
 Staff
 Cryptography, Integers

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