> > The conjecture is true. > If a(n) was not prime for some n, it was the product of two numbers b, > c both coprime to n > and strictly smaller than a(n). If m = carmichael's lambda(n), we then > have : b^(m/2) == c^(m/2) ==1 mod n) and hence > a(n)^(m/2) == 1 mod n, Impossible. >