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

1. ## 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?

2. $n$ 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 you’re done ( $n^4+4^n$ would be the product of two integers greater than one and so can’t be prime).

3. 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

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