Prove: $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) > 10^{2009}$ 
Show that the number $ 3^{{4}^{5}} {+} 4^{{5}^{6}}$ can be expressed as the product of two integers greater than $ 10^{2009}$

By Sophie Germain:
$ 3^{{4}^{5}} {+} 4^{{5}^{6}}=\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)\cdot \left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
Now $\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)$
So need to prove $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right)>10^{2009}$.
Please help with this, and I'm very enthusiast to know your method to solve above problem.

$\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) = \left(2^{\frac{5^6}{2}}-3^{2^8}\right)^2 > (2^{7811}-3^{256})^2$
$2^8>3^5$ so $(2^{7811}-3^{256})^2 > (2^{7811}-2^{410})^2 >(2^{7810})^2 = 2^{15620}$
$2^{10}>10^3$ so $2^{15620}>10^{4686}>10^{2009}$
