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Math Help - Show that n^5 + n^4 + 1 is not prime for n>1

  1. #1
    Super Member fardeen_gen's Avatar
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    Show that n^5 + n^4 + 1 is not prime for n>1

    No idea at all!
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  2. #2
    Moo
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    Quote Originally Posted by fardeen_gen View Post
    No idea at all!
    n^5+n^4+1=(x+x+1)(x^3-x+1)



    But I cheated and asked it to Maxima.
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  3. #3
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    Hello, fardeen_gen!

    Please don't hide the problem in the heading . . .


    Show that n^5 + n^4 + 1 is not prime for n > 1.
    Brilliant move, moo!

    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)

    - - . . . . . . .= . (n - 1)(n + 1) - n(n - 1)

    - - . . . . . . .= . (n - 1)(n - n + 1)

    . . (n - 1)P . = . (n - 1)(n + n + 1)(n - n + 1)


    Divide by (n - 1): . P . = . (n + n + 1)(n - n + 1)

    Thanks from bkarpuz
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  4. #4
    Super Member fardeen_gen's Avatar
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    Thank you. Is expressing the given expression in factors sufficient to prove that it is not a prime?
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  5. #5
    Moo
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    Quote Originally Posted by fardeen_gen View Post
    Thank you. Is expressing the given expression in factors sufficient to prove that it is not a prime?
    Not really, because it still can be 1 multiplied by a prime number.

    But you can see that n+n+1 > 1 for any n>1.
    For any n>1, n^3>n, and than n^3-n>0 --> n^3-n+1>1.

    So both factors are strictly > 1, which proves that it's a product of 2 numbers > 1, thus not prime.
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  6. #6
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    You can note that x^5+x^4+1 can be factored using primitive root of unities. Let \zeta be a primitive third root of unity then \zeta^5+\zeta^4+1 = \zeta^2 + \zeta + 1 = 0. Now the minimal polynomial for \zeta is \Phi_3(x) which is x^2+x+1. Thus, x^2+x+1 a factor of x^5+x^4+1.
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  7. #7
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    Just for those, who got here from google

    2nd solution:

    We have
    n5 + n4 + 1

    = n5 + n4 + n3 - n3 - n2 - n +n2 + n + 1

    = n3(n2 + n + 1) - n(n2 + n + 1) + (n2 + n + 1)

    = (n2 + n + 1)(n3 - n + 1)
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