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Math Help - If n>1 , show that n^4 + 4^n is never a prime?

  1. #1
    Super Member fardeen_gen's Avatar
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    If n>1 , show that n^4 + 4^n is never a prime?

    If n>1, show that n^4 + 4^n is never a prime?
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  2. #2
    Senior Member JaneBennet's Avatar
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    n clearly has to be odd, so \tfrac{n+1}{2} is an integer.

    Then n^4+4^n=\left(n^2+2^n\right)^2-2^{n+1}n^2=\left[n^2+2^n+2^{\frac{n+1}{2}}n\right]\left[n^2+2^n-2^{\frac{n+1}{2}}n\right]

    n^2+2^n+2^{\frac{n+1}{2}}n is clearly greater than 1; if you can show that n^2+2^n-2^{\frac{n+1}{2}}n>1 then youíre done ( n^4+4^n would be the product of two integers greater than one and so canít be prime).
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  3. #3
    Senior Member JaneBennet's Avatar
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    Okay, here we go. Prove that n^2+2^n-2^{\frac{n+1}{2}}\cdot n>1 for all odd integers n>1.

    First, check the cases n=3 and n=5 separately.

    3^2+2^3-2^{\frac{3+1}{2}}\cdot3=9+8-12=5>1

    5^2+2^5-2^{\frac{5+1}{2}}\cdot5=25+32-24=33>1

    For odd integers n\ge7, I claim that 2^n>2^{\frac{n+1}{2}}\cdot n. This can be proved by a slight variation of the method of induction.

    When n=7, 2^7=128>112=2^{\frac{7+1}{2}}\cdot7.

    Suppose 2^n>2^{\frac{n+1}{2}}\cdot n for some odd integer n\ge7.

    Now consider 2^{\frac{(n+2)+1}{2}}\cdot(n+2). (NB: n+2 is the next odd integer.)

    2^{\frac{(n+2)+1}{2}}\cdot(n+2)=2\cdot2^{\frac{n+1  }{2}}\cdot(n+2)

    \color{white}.\hspace{27mm}. =2\cdot2^{\frac{n+1}{2}}\cdot n+2\cdot2^{\frac{n+1}{2}}\cdot2

    \color{white}.\hspace{27mm}. <2\cdot2^n+2\cdot2^n\color{white}\ .*

    \color{white}.\hspace{27mm}. =4\cdot2^n=2^{n+2}

    * \ 3<n\ \Rightarrow\ n+1+2<2n\ \Rightarrow\ \frac{n+1}{2}+1<n\ \Rightarrow\ 2\cdot2^{\frac{n+1}{2}}<2^n

    Hence, 2^n>2^{\frac{n+1}{2}}\cdot n for all odd integers n>5, and it follows that n^2+2^n-2^{\frac{n+1}{2}}\cdot n>n^2>1.
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