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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.

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You have $x^4+4y^4 = (x^2+2xy+2y^2)(x^2-2xy+2y^2)$, but $x^2-2xy+2y^2 = (x-y)^2+y^2\geq y^2$, where $y = 2^{\frac{5^6-1}{2}}$. Therefore, $$y^2 = (2^{10})^{\tfrac{5^6-1}{10}}>10^{\tfrac{3(5^6-1)}{10}} = 10^{2009},$$ because $2^{10} = 1024>1000 = 10^3.$

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$log_{10}(3^{4^5}+4^{5^6}) = 9407.1874$ 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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    is there any way from we can conclude it has $9408$ digits in math contests. – 2017-02-14
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    @mathlover: the usual way is to approach it by properties of the presented number. This answer only covers an analytic point of view made from a straightforward calculation in computer. – 2017-02-14
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    "Since you express this number as a product of two numbers then each of those has 4704 digits." - Are you saying the factors must be the same size?? And (-1) for a brute force approach which is clearly not the intent of the question. – 2017-02-14
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    @TMM: In my first comment I told the user that my answer covers a different point of view, obviously you don't have a computer in a math contest, but is easily done in whatever computer with a good straight forward calculator (in my case Fedora's calc). I took the decimal logarithm of both factors and they had the same amount of digits. – 2017-02-14
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$\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) = \left(2^{\frac{5^6}{2}}-3^{2^8}\right)^2 > (2^{7811}-3^{256})^2$

$2^8>3^5$ so $(2^{7811}-3^{256})^2 > (2^{7811}-2^{410})^2 >(2^{7810})^2 = 2^{15620}$

$2^{10}>10^3$ so $2^{15620}>10^{4686}>10^{2009}$