Question

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Answer 1

1 works, you get 3
i don't think there are any others

Answer 2

since \(n^4+n^2+1\) factors

Answer 3

do you know the trick for factoring
\[n^4+n^2+1\]?

Answer 4

it is a completing the square gimmick, also called the "sopie germain" method

Answer 5

*sophie

Answer 6

more please

Answer 7

http://en.wikipedia.org/wiki/Sophie_Germain

Answer 8

you can make \(n^4+n^2+1\) in to a perfect square by adding (and therefore subtracting) \(n^2\) giving
\[n^4+n^2+1=n^4+2n^2+1-n^2\]

Answer 9

the first three terms is a perfect square, namely
\(n^4+2n^2+1=(n^2+1)^2\) and so
\[n^4+n^2+1=(n^2+1)^2-n^2\] and now you can factor as the difference of two squares

Answer 10

you get
\[(n^2+1+n)(n^2+1-n)\] in factored form
meaning this is not a prime number, since you can factor it

Answer 11

to everybody? or just to asker?