Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$. I came across this problem and i could not solve it despite many tries.
Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$.
I think the question can be solved more easily..Can anybody suggest any other method.
 A: Hint:-

$$ a^4+4b^4=(a^2+2b^2-2ab)(a^2+2b^2+2ab)$$(Sophie Germain's Identity)

A: Hint:-
We try to factorise the equation and show that each factor is greater than $2009^{182}$.
$$3^{2008}+4^{2009}=(\color{green}{3^{502}})^\color{orange}{4}+\color{orange}{4}\times\color{blue}{(4^{502})}^\color{orange}{4}.$$
Now,interestingly,this is of the form of Sophie-Germain's identity which states-

$$a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2).$$

Take,$(3^{502})=a $ and $4^{502}=b$ and resolve it into two factors and show that each factor is greater than the given value of $2009^{182}$.
A: Using $a^2+b^2 =(a+b)^2-2ab$, we have 
$3^{2008} +4^{2009}=(3^{1004})^2+(2^{2009})^2=(3^{1004}+2^{2009})^2 - 2 \cdot 2^{2009} \cdot 3^{1004}$
So we have $3^{2008} +4^{2009}=(3^{1004}+2^{2009})^2- 2^{2010} \cdot 3^{1004}$
Now, using $a^2-b^2=(a-b)(a+b)$, we have 
$3^{2008} +4^{2009}=(3^{1004}+2^{2009}+2^{1005} \cdot 3^{502})(3^{1004}+2^{2009}-2^{1005} \cdot 3^{502})$

Finally, it is not hard to see that $(3^{1004}+2^{2009}-2^{1005} \cdot 3^{502}) > 2009^{182}$, which is left as an exercise to the reader.
A: We use the standard factorisation:
x4 + 4y4 = 
(x2 + 2xy + 2y2)( x2 - 2xy + 2y2). 
We observe that for any integers x, y,
x2 + 2xy + 2y2 = (x + y)2 + y2 ≥ y2, 
x2 - 2xy + 2y2 = (x - y)2 + y2 ≥ y2, 

