Consider Xn+X+1 
Z/3Z[X].
For which n does this polynomial have multiple factors?
No n.
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All n.
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n = 0 mod 3.
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n = 1 mod 3.
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n = 2 mod 3.
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Wrong. For n = 2, we have
X2+X+1 = (X+1)2.
Wrong.
The derivative of X3+X+1 is 1. So, by the lemma,
there are no multiple factors for n = 3.
Wrong. For n = 2, we have X2+X+1 = (X+1)2.
Correct.
As gcd(Xn+X+1,nXn-1+1) = 1 if n = 0, 1 mod 3,
and X-1 if n = 2 mod 3, only this case can lead to multiple factors.
Indeed it does:
X2+3m+X+1 is divisible by
X2+X+1 = (X-1)2.
Wrong. For n = 1, we have X1+X+1 = -X+1, a polynomial without
multiple factors.