Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1. Show that $3^{2n+1}-4^{n+1}+6^n$ is never prime for natural n except 1. I tried factoring this expression but couldn't get very far. It is simple to show for even n but odd n was more difficult, at least for me.
 A: You can factor it as $(3^n-2^n)(3^{n+1}+4*2^n)=3^{2n+1}-3*6^n+4*6^n-4*4^n$. Here we juggle between $(ab)^n=a^nb^n$.
Since we can factorise it, to have a prime we need one of these factors to be 1, which only happens when $n$ is one, i.e the first term is $(3-2)$ and the second is 17. Note that the second term can't be 1 as it's an addition of two positive quantities.
A: The first step is to make explicit the algebraic dependencies among $\,3^{2n},4^n,6^n.\,$ Clearly all can be expressed in terms of the multiplicatively independent basis $\, x = 3^n\,$ and $\,y = 2^n\,$ as follows
$$\begin{align} 
&3^{2n} = (3^n)^2 = x^2\\
&4^n =\ (2^n)^2 = y^2\\
&6^n = 3^n 2^n = xy\end{align}$$ 
Rewriting our expression in these terms we obtain
$$\begin{align}  &3\cdot 3^{2n} - 4\cdot 4^n + 6^n \\
=\ & 3\cdot  x^2 - 4\cdot  y^2 + xy\end{align}$$
Next we can factor this polynomial $\, f(x) = 3\,x^2 + y\, x - 4\,y^2 \ $ using the AC-method
$$ {\begin{eqnarray}
f \, &\,=\,& \ \ \,  3 x^2+\ y\ x\,\ -\ \ 4y^2\\
\Rightarrow\,\ 3f\, &\,=\,&\!\,\ (3x)^2\! +y(3x)-12y^2\\
         &\,=\,& \ \ \   {X^2}+\, y\  X\,\ -\ 12y^2,\,\ \ X\, =\, 3x\\
         &\,=\,&  \ \ (X+4y)\ (X-\,3y)\\
         &\,=\,& \  (3x+4y)\,(3x-3y)\\
\Rightarrow\ \  f\:=\: {3^{-1}}\,(3f)\,  &\,=\,&   \,  (3x+4y)\ (x-y)\\
\end{eqnarray}}$$
A: $$3 \cdot 3^{2n}-4\cdot 2^{2n}+2^n\cdot  3^n$$ has the form
$$3x^2-4y^2+xy=(x-y)(3x+4y)$$
