[SOLVED] Write a sum to a product

[anemone]

Show that $3^{2008}+4^{2009}$ can be written as a product of two positive integers each of which is larger than $2009^{182}$.

 

[kaliprasad]

Spoiler: My solution

we have $3^{2008} + 4^{2009} = (3^{1004})^2 + (2^{2009})^2$
$= (3^{1004})^2 + (2^{2009})^2 + 2 * 3^{1004} * 2^{2009} - 2 * 3^{1004} * 2^{2009}$
$= ((3^{1004}) + (2^{2009}))^2 - 3^{1004} * 2^{2010}$
$= (3^{1004} + 2^{2009})^2 - (3^{502} * 2^{1005})^2$
$= (3^{1004} + 2^{2009} +3^{502} * 2^{1005}) (3^{1004} + 2^{2009} -3^{502} * 2^{1005})$

Let us consider the lower value
$(3^{1004} + 2^{2009} -3^{502} * 2^{1005})$
$=2^{2009} + 3^{502}(3^{502} - 2^{503})$

as $3^{502} = 9^{\frac{502}{2}} = 9^{251} > 8^{251} = 2^{251 * 3} = 2^{753} > 2^{502}$

so
$(3^{1004} + 2^{2009} -3^{502} * 2^{1005}) > 2^{[2009} = 2^{ 11 * 182+7} > (2^{11})^{182} = 2048 ^{182} > 2009 ^ {182}$

as lower term is $>2009^{182}$ so we are done