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for what values of $n$ ,$n^4+4$ is composite number?

$n^4+4=a.b\Rightarrow (n^2-2n+2)(n^2+2n+2)=ab$

what to conclude next? Thank you

Myshkin
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    Well, if both factors are $> 1$, it's composite. So it can only be non-composite if one of the factors is $1$, that is... – Daniel Fischer Nov 26 '13 at 10:42
  • you just have to check for what values of $n$ you have $n^2-2n+2=1$ or $n^2+2n+2=1$ –  Nov 26 '13 at 10:43
  • @DanielFischer +1 :) –  Nov 26 '13 at 10:44
  • See also http://math.stackexchange.com/questions/632615/the-number-n4-4-is-never-prime-for-n1 and http://math.stackexchange.com/questions/1121407/show-that-n44-is-not-a-prime-number – Martin Sleziak Jan 27 '15 at 09:10

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Let's set the factors equal to $1$: $$n^2 - 2n + 2 = 1 \implies n^2 - 2n + 1 = 0 \implies n = 1$$

Similarly, $$n^2 + 2n + 2 = 1 \implies n^2 + 2n + 1 = 0 \implies n = -1$$

We can check and see that when $n = \pm 1$, the value of $n^4 + 4$ is indeed prime. Hence, for any other $n$, namely $n \neq \pm 1$, this will be composite.

HJ32
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