ElGamal, message > p

Question

Assume that we have:

p = 89
g = 5
public key: 17
private key: 73

If we try to encrypt message M = 53 (M < p), then we get (c1, c2) == (55, 67) and further message decrypts well.

However, if we try to encrypt message M = 91 (M > p), then we get (c1, c2) == (44, 57) and further message decrypts failed (got "2" as the result).

There are 3 questions:

Why does it happen?
Is it possible to recover original message M if we know the fact that (m > p) used, p, g, public key and have (c1, c2)?
Is it possible to recover original message M if we know the fact that (m > p) used, p, g, public key and have several encrypted messages (c1, c2)?

score 3 · Answer 1 · answered Oct 26 '21 at 08:51

Why does it happen?

This happens because, in terms of arithmetic mod 89, the numbers $2$ and $91$ are equivalent: $91 \bmod 89 = 2$. You'll usually see this denoted as $91 \equiv 2 \pmod{89}$.

If you're new to modular arithmetic - this is just as how the hours $0$ and $12$ on a traditional clock are equivalent - it "wraps around".

Is it possible to recover original message M if we know the fact that (m > p) used, p, g, public key and have (c1, c2)?

Not in general, while still allowing decrypting messages $m < p$.

In a specific case, where you had a ciphertext and knew that the original message had been in e.g. $[2p, 3p - 1]$, you could recover it by simply adding the appropriate offset after decryption.

In practice this is not particularly useful, so as mentioned in comments above, one splits the message $m$ into $m_1, \ldots, m_n$ such that $m_i < p$. Each partial message is then encrypted individually.

ElGamal, message > p

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