elgamal encryption question

given a 2048bit public key, if I want to encrypt ~70 bytes of data, what
is the largest and smallest byte size of the encrypted value that can
result?  512bytes,  514 bytes?  if the size that comes out is not constant,
 how should it be padded so that the structure is?

if I pad the data being encrypted to 256 bytes (or 255 or 257), will
that 1) improve the security of a known plaintext attack or 2) affect
the resulting size of the encrypted block?

please see datastructures.sxw page 13 for the block in question.  Current
implementations, as well as a test with bouncycastle.org's elgamal encryption
engine, shows that using a 512byte block /usually/ works, but sometimes
fails to decrypt correctly.

from handbook of applied crypto p296:

"A disadvantage of ElGamal encryption is that there is message expansion
by a factor of 2. That is, the ciphertext is twice as long as the corresponding
plaintext."

That seems to tell me that if we want the ciphertext to be 512 bytes,
 we pad the plaintext to 256 bytes.  but it also seems to say that if
the ciphertext is just 70 bytes, the plaintext will be 140 bytes, not
512. 

this needs to be resolved asap, as if a 2048bit key can legitimately
encrypt ~70 bytes of data to a resulting 514 bytes instead of 512, the
spec needs to change.

i'm going out to breakfast and will keep hacking on this when i get back.

gracias srs,
-jr



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