How do I proof that x^(p^p) - x is divisible by x^p - x - 1 over Zp, p - any prime number?

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Makuta

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I tried using Freshman's dream theorem and making a^p = a + 1, a^(p^p) = (a+1)^p = a^p + 1, which led to a^p + 1 = a + 2 and it's just not making any sense.
 
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x^p^p is not clear, at least to me.
(2^3)^4 = 2^12 while 2^(3^4) = 2^81
 
Makuta said:
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.
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It makes [imath]x^{p^p}-x = 0[/imath] in [imath]\mathbb Z_p[/imath], and 0 is divisible by anything, isn't it?
 
blamocur said:
It makes [imath]x^{p^p}-x = 0[/imath] in [imath]\mathbb Z_p[/imath], and 0 is divisible by anything, isn't it?
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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.
 
Makuta said:
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.
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I think misinterpreted your post. I was thinking about divisibility of expressions for any [imath]x\in \mathbb Z_p[/imath]. But this would not even require any Fermat theorems since [imath]\mathbb Z_p[/imath] is a field, hence every element is divisible by any other non-zero element.

I suspect you are talking of divisibility of polynomials over [imath]\mathbb Z_p[/imath]. This would make a more meaningful problem, but I don't know how to prove it.
 
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