Quote:
Originally Posted by paulunderwood
Taking the obvious case of r=1, then a Lucas test over y^24*y+5 can be transformed into a 5Euler PRP test and a EulerLucas test over f(z)=z^26/5*z+1 (for gcd(5,n)==1). This has solutions of f(z)=0 for z=(3+4*i)/5. The corresponding companion matrix is Z=[3/5,4/5;1,0], the determinant of which is 4/5. Thus if n is base 2Fermat PRP there is no need to do the base 5Euler PRP test  it is implied since det(Z)^k == det(Z^k) for all k.

My logic is flawed that base 2 Fermat PRPs can replace base 5 Euler PRPs. I mistakenly wrote the wrong matrix Z; It should be [3/5,4/5;4/5,3/5] (the determinant of which is 1). No more over egging the pudding.