25 Mar
2016
25 Mar
'16
4:26 p.m.
We conclude from the latter that in a finite field with q elements, q odd,
T[q^2](x) = x.
Actually, maybe even T[q](x)=x in many cases, but anyhow q^2 should definitely work. More generally, 1+k*(q^2-1) will work for any integer k: T[1 + k*(q^2-1)](x) = x.
--actually, q needs to satisfy certain known modular conditions to make the Dickson polynomials be permutation polynomials. And when M=p*q we need both p and q to satisfy those conditions simultaneously. So I don't think you can just use any old primes p and q. But you have plenty of freedom of choice nevertheless. The question I'm posing is: why you'd want to bother doing this instead of just powering as in plain RSA.