🚀 go-pugleaf

I haven't seen these items mentioned but I'm not a math whiz so maybe
it's hidden in there somewhere.

You can get an encode/decode pair ( e and d ) by taking d*e mod
(p-1)*(q-1) = 1. This implies that the smallest value for d*e
is ((p-1)*(q-1))+1 but it's actually (((p-1)*(q-1))/2)+1. If fact, d*e 
must be some value  (((p-1)*(q-1)/2)*x)+1 where x is any integer >0
 
Also, for ANY value of n ( =p*q), there are at least nine values of m 
(the message) that will encrypt as m. I.e. E(m) = m and D(m) = m. Three 
of these are trivial, m=0, 1, and n-1. The other six allow factoring n. 
For some values of e, there are more cases but I'm not sure how easy it 
would be to get the factors knowing one of those. This brings up the 
question of whether the proceedure should watch for such cases. Only 
while signing a message would the sender be the one to discover key was 
compromised. I know it's not at all likely but it seems like it could 
cause problems.


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