Question

all values of n such that n^4 - 51n^2 + 225 is a prime number with n is integer number ?

Answer 1

@rational

Answer 2

my job :
n^4 - 51n^2 + 225 = p
n^4 - 30n^2 + 225 - 21n^2 = p
(n^2 - 15)^2 - (sqrt(21)n)^2 = p
(n^2 - 15 + sqrt(21)n) ((n^2 - 15 - sqrt(21)n) = p
one of the faktor must be equal 1
n^2 - 15 + sqrt(21)n = 1 or n^2 - 15 - sqrt(21)n = 1
like that ?

Answer 3

and likes there is no n which integer satisfied

Answer 4

im thinking of factoring the given expression directly

Answer 5

Can you check if below is true :

\[ n^4 - 51n^2 + 225 = (n^2-9 n+15) (n^2+9 n+15)\]

Answer 6

ah, yes that's correct.. how you factor out that ?

Answer 7

like this
http://www.wolframalpha.com/input/?i=factor+n%5E4+-+51n%5E2+%2B+225

Answer 8

rest should be easy..

Answer 9

for \(M = ab\) to be prime, we must have :

1) \(a=1\) and \(b=p\)

or

2) \(a=p\) and \(b=1\)

Answer 10

ok, i get it
n^2 + 9n + 15 = 1
n^2 + 9n + 14 = 0
(n + 2)(n + 7) = 0
n = -2 or n = -7
and the rest there is no n which integer

Answer 11

thank you very much for your help :D

Answer 12

hey so what are the "n" values that make the given expression prime ?