Thread View: ucb.math
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Started by either Jana or John
Mon, 26 Jul 1999 00:00
Help - how do I solve this?
Author: either Jana or John
Date: Mon, 26 Jul 1999 00:00
Date: Mon, 26 Jul 1999 00:00
40 lines
1421 bytes
1421 bytes
How do I solve this expression? It is part of the RSA public key encryption algorithm. The expression is e*d mod L = 1 in other words e*d = k*L+1 How do I find d? I'll back up now and show you what e and L are: p and q are prime numbers. L = (p - 1)*(q - 1) e is any number except for factors of L N = p * q To encrypt data m, m^e mod N = encrypted data c To decrypt data m = c^d mod N So, you see, I need d to decrypt data. Example: p = 19, q + 23, N = 437, L = 396, e = 13, d = 61 (but I don't know how d was calculated - that's what I need to know) Encrypt the number 180 180^13 mod 437 = 208229648656711680000000000000 mod 437 = 329 Decrypt 329 329^61 mod 437 = (a very long number I won't bother to type) mod 437 = 180 Any help gratefully appreciated. - j
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