Question

a problem for u
How many primes are there in the form

Answer 1

\[1+n^2+n^4\]

Answer 2

n is a positive integer

Answer 3

Yeah this isn't a complete question I think...

For a polynomial, it has no factors (you said integers not complex #'s).

For a sequence, it depends on what you input for n:
A(n) = 1+n\(^2\)+n\(^4\)
A(1) = 3
A(2) = 21
A(3) = 91
A(4) = 273
A(5) = 651
etc...
\[\lim_{n \rightarrow \infty} 1+n^2+n^4 = \infty\]

Answer 4

@agentx5
what do u think of this?
\[n^4+n^2+1=n^4+2n^2+1-n^2=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)\]

Answer 5

Completing the square...

Answer 6

just n=1 gives a prime

Answer 7

Those factors do not form real roots though... (take a look at the graph for your "completing the square" version of the function as a sequence) I misspoke but that is what I think I was getting at yesterday or whenever this was.

List of of the first 1000 prime #'s in the decimal system:
http://primes.utm.edu/lists/small/1000.txt

Let's see... 91/7 = 13, most of the numbers are divisible by 3 without a remainder, but that's an assumption. Also A(0) = 1, one is a prime # too, but zero isn't consider positive or negative. So... yeah.