SOLUTION: Determine all integers n such that n^4-4n^3+15n^2-30n+27 is a prime number

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Question 794840: Determine all integers n such that n^4-4n^3+15n^2-30n+27 is a prime number
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
n%5E4-4n%5E3%2B15n%5E2-30n%2B27%3E0 for all real values of n,
so there are no factors of the form %28x-a%29,
but I can still factor that polynomial into the product of two quadratic polynomials:
n%5E4-4n%5E3%2B15n%5E2-30n%2B27=%28n%5E2%2Bbn%2Bc%29%2A%28n%5E2%2Bdn%2Be%29
Multiplying, I get

I need to find b, c, d, and e such that
ce=27, so the set c%2Ce is {1,27}, or {-1,-27}, or {3,9}, or {-3,-9}
b%2Bd=-4
c%2Bbd%2Be=15 and
eb%2Bcd=-30
Since c and e are both odd, c%2Be will be even,
and bd must be odd to make c%2Bbd%2Be odd, let alone c%2Bbd%2Be=15
To make bd odd, b and d must both be odd.
They have to add up to -4 too to make b%2Bd=-4
There are many options, but With the set {b,d}={-1,-3} I could get
b%2Bd=-1-3=-4 and
bd=3, which with {c,e}={3,9},
added to c%2Be=3%2B9 to make c%2Bbd%2Be=3%2B9%2B3=15
Trying the options left for the order of the numbers in each set,
eb%2Bcd=-30 could be made pairing them as 9%28-3%29%2B3%28-1%29=+-27-3=-30
with e=9, b=-3, c=3, and d=-1.
So n%5E4-4n%5E3%2B15n%5E2-30n%2B27=%28n%5E2-3n%2B3%29%2A%28n%5E2-n%2B9%29
With n being an integer, the factors
%28n%5E2-3n%2B3%29 and %28n%5E2-n%2B9%29 are integers.
For their product to be a prime number, one factor must be 1,
and the other factor must be a prime number.

Making the first factor equal to 1, we get
n%5E2-3n%2B3=1-->n%5E2-3n%2B3-1=0-->n%5E2-3n%2B2=0-->%28n-2%29%28n-1%29=0-->system%28n=2%2C%22or%22%2Cn=1%29
n=2-->n%5E2-n%2B9=2%5E2-2%2B9=4-2%2B9=11 leads us to
the solution highlight%28n=2%29,
which makes n%5E4-4n%5E3%2B15n%5E2-30n%2B27=%28n%5E2-3n%2B3%29%2A%28n%5E2-n%2B9%29=1%2A11=11
n=1 also makes n%5E2-3n%2B3=1, but it makes
n%5E2-n%2B9=1%5E2-1%2B9=1-1%2B9=9 and 9 is not a prime number.

Making the second factor equal to 1, we get
n%5E2-n%2B9=1-->n%5E2-n%2B9-1=0-->n%5E2-n%2B8=0, which has no real solutions.

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