# Truly Stealthy PGP

Eric Hughes hughes at ah.com
Sat Mar 5 10:27:54 PST 1994

```>This was discussed some time back on the pgp developers' list, and at that
>time the suggestion was made to add a multiple of n to m so that it covered
>a fuller range of values.  The recipient would then just take the exponent
>mod n and try that.

What I suggest is making the exponent (the encrypted session key)
completely random over the length assigned to it, since that's
visible, and just live with a slightly non-flat distribution of
exponents mod n.  It turns out that this can be made to work just
fine.

>Mathematically, call L the next multiple of 256 above n.

n is the modulus.  Divide L by n to get L = t * n + s, s in [0,n).
Assume x is random in [0,L).  The entropy of x mod n is

E = - s (t+1)/L log (t+1)/L - (N-s) t/L log t/L

Rearranging, we get:  (get out some paper, do the algebra)

E = log L/t - s(t+1)/L log( 1 + 1/t )

This makes sense, since if s is zero, E = log n, which is just the
entropy of the random distribution of [0,n).

What is the smallest value of E?  In other words, what's the upper
bound of the randomness we can lose?  It happens when when t = 1 and
when n = L/2+1.  This maximize the expression in t and maximizes s at
n-2.  This minimum value of E is

E_min = log L - ( ln 2 - 2/L ln 2 )

In other words, the most entropy we can lose is two bits.  That's
right, only two bits.  Since the entropy of the session key is the
length of the modulus, for a 1000 bit key the entropy loss is
negligible.

Therefore, my recommendation is that the session key representation be
chosen randomly over [0,2^k) and to use as an actual session key this
value mod n.  The effective entropy loss is small enough not to worry