No idea at all!
Hello, fardeen_gen!
Please don't hide the problem in the heading . . .
Brilliant move, moo!Show that n^5 + n^4 + 1 is not prime for n > 1.
I found a way to factor it. .(It helped that I already knew the factors!)
Let: .P .= .n^5 + n^4 + 1
Multiply by (n - 1):
. . (n - 1)P . = . (n - 1)(n^5 + n^4 + 1)
- - . . . . . . .= . n^6 - n^4 + n - 1
- - . . . . . . .= . (n^6 - 1) - (n^4 - n)
- - . . . . . . .= . (n³ - 1)(n³ + 1) - n(n³ - 1)
- - . . . . . . .= . (n³ - 1)(n³ - n + 1)
. . (n - 1)P . = . (n - 1)(n² + n + 1)(n³ - n + 1)
Divide by (n - 1): . P . = . (n² + n + 1)(n³ - n + 1)
You can note that $\displaystyle x^5+x^4+1$ can be factored using primitive root of unities. Let $\displaystyle \zeta$ be a primitive third root of unity then $\displaystyle \zeta^5+\zeta^4+1 = \zeta^2 + \zeta + 1 = 0$. Now the minimal polynomial for $\displaystyle \zeta$ is $\displaystyle \Phi_3(x) $ which is $\displaystyle x^2+x+1$. Thus, $\displaystyle x^2+x+1$ a factor of $\displaystyle x^5+x^4+1$.