<s>[INST] 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.
 [/INST] $log_{10}(3^{4^5}+4^{5^6}) = 9407.1874$ then it has $9408$ digits. Since you express this number as a product of two numbers then each of those has $4704$ digits.
$log_{10}(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}) = 4703.5937$ as we can see both numbers have $4704$ digits then we conclude that :
$$3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6} > 10^{2009}$$ $$3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6} > 10^{2009}$$
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