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.
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 $(3^{1004}+2^{2009}-2^{1005} \cdot 3^{502}) &gt; 2009^{182}$, which is left as an exercise to the reader.
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,