So. $P(n) = n^4 + n^2 + 1$ is a polynomial. I calculated that answer is 1. But I don't understand why?
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9$(n^2-n+1)(n^2+n+1)$ – André Nicolas Apr 01 '16 at 23:04
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@AndréNicolas thank you! – sashaaero Apr 01 '16 at 23:06
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You are welcome. – André Nicolas Apr 01 '16 at 23:10
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@AndréNicolas You should post that as an answer so that the question can be marked as answered – Stella Biderman Apr 02 '16 at 00:12
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@StellaBiderman: OK, done. – André Nicolas Apr 02 '16 at 00:21
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We have $$n^4+n^2=(n^2-n+1)(n^2+n+1).\tag{1}$$ If $n\gt 1$, each term on the right-hand side of (1) is greater than $1$, so $n^4+n^2+1$ cannot be prime.
Remark: One way of "seeing" the above factorization is to note that $$n^4+n^2+1=(n^2+1)^2-n^2.$$ Now we can use the usual factorization of a difference of squares.
André Nicolas
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