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Thread: 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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    $\displaystyle n$ clearly has to be odd, so $\displaystyle \tfrac{n+1}{2}$ is an integer.

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

    $\displaystyle n^2+2^n+2^{\frac{n+1}{2}}n$ is clearly greater than 1; if you can show that $\displaystyle n^2+2^n-2^{\frac{n+1}{2}}n>1$ then youíre done ($\displaystyle n^4+4^n$ 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 $\displaystyle n^2+2^n-2^{\frac{n+1}{2}}\cdot n>1$ for all odd integers $\displaystyle n>1$.

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

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

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

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

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

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

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

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

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

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

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

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

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