p. 193 #11.
We cannot have
by unique factorization in the integers. Without loss of
generality assume that
, the case of
being similar. Then we can
by using the Euclidean algorithm on
. 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
cases. Once we know
, we can find the modular inverse of the public
in both the cases, yielding the decryption exponents
both cases. The RSA systems are broken.
p. 193 #14.
, it follows that pq | (x + a)(x - a). Hence p | (x+a)(x-a),
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
Without loss of generality, assume that x + a =
x - a =
We would like to find
, we know that there exist integers
. we get find solutions for k and j such
We know that
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
Using the other form of x, we see that
Since p and q are distinct primes, it follows that
For example, with
p = 71 and q = 73, then (36)(71) - (35)(73) = 1, so, taking a = 7,