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I understand it has something to do with factorization of large numbers being very difficult and time consuming but I think prime numbers have something to do with it too. ???

2006-10-06 12:24:26 · 4 answers · asked by tjcsonofallnations 3 in Computers & Internet Security

4 answers

The security of the RSA cryptosystem is based on two mathematical problems: the problem of factoring large numbers and the RSA problem. Full decryption of an RSA ciphertext is thought to be infeasible on the assumption that both of these problems are hard, i.e., no efficient algorithm exists for solving them.

The RSA problem is defined as the task of taking eth roots modulo a composite n: recovering a value m such that me=c mod n, where (e, n) is an RSA public key and c is an RSA ciphertext. Currently the most promising approach to solving the RSA problem is to factor the modulus n. With the ability to recover prime factors, an attacker can compute the secret exponent d from a public key (e, n), then decrypt c using the standard procedure. To accomplish this, an attacker factors n into p and q, and computes (p-1)(q-1) which allows the determination of d from e. No polynomial-time method for factoring large integers on a classical computer has yet been found, but it has not been proven that none exists.

2006-10-06 12:44:38 · answer #1 · answered by karkov48 4 · 0 0

One of the methods public key encryption systems can use is taking two very large prime numbers (100 digits or more), and multiplying them. The two factors form one key and the product forms the other key. Information can be encrypted with either key in such a way that it can only be encrypted by the other key (unless someone finds a fast way to factor large numbers).

Since the encryption is one-way, two ends of the communication channel each have a public and private key pair. They exchange the public keys with each other, and in this way they can set up encrypted communication keys even though the keys themselves can be sent in plain view.

2006-10-06 14:18:13 · answer #2 · answered by Ken H 4 · 0 0

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2016-11-26 21:54:52 · answer #3 · answered by Anonymous · 0 0

Go here...{:-{}.

2006-10-06 12:28:43 · answer #4 · answered by Anonymous · 0 0

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