The second option.x^p^p is not clear, at least to me.
(2^3)^4 = 2^12 while 2^(3^4) = 2^81
I'm not sure, but how is this helpful? I used this theorem and it only makes x≡x (mod p), and for f(x) shows that it has no roots.Is not xpp=x∀x∈Zp by Fermat's little theorem?
It makes xpp−x=0 in Zp, and 0 is divisible by anything, isn't it?I'm not sure, but how is this helpful? I used this theorem and it only makes x≡x (mod p), and for f(x) shows that it has no roots.
I guess it is, just seems too simple and thus incorrect. Thank you a lot!It makes xpp−x=0 in Zp, and 0 is divisible by anything, isn't it?
But again, aren't we saying that every element in Z_p is a root for x^(p^p)-x, but what's the point, if we have the fact that x^p - x - 1 have no roots in Z_p, because if we take a^p - a -1 = 0 in Z_p we get -1 = 0.It makes xpp−x=0 in Zp, and 0 is divisible by anything, isn't it?
I think misinterpreted your post. I was thinking about divisibility of expressions for any x∈Zp. But this would not even require any Fermat theorems since Zp is a field, hence every element is divisible by any other non-zero element.But again, aren't we saying that every element in Z_p is a root for x^(p^p)-x, but what's the point, if we have the fact that x^p - x - 1 have no roots in Z_p, because if we take a^p - a -1 = 0 in Z_p we get -1 = 0.
Me too, that's why I'm seeking for help...but I don't know how to prove it.