Question

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

Answer 1

We use the standard factorisation:
$x^4+4y^4=(x^2+2xy+2y^2)(x^{2}2xy+2y^2)$.
We observe that for any integers x, y,
$x^2+2xy+2y^2=(x+y{)}^2+y^2\ge y^2$,
and $x^2-2xy+2y^2=(x-y{)}^2+y^2\ge y^2$,
We write $3^{2008}+4^{2009}=3^{2008}+4\left(4^{2008}\right)=(3^{502}{)}^4+4(4^{502}{)}^4$.
Taking $x=3^{502}$ and $y=4^{502}$, we se that $3^{2008}+4^{2009}=ab$, where
$a\ge (4^{502}{)}^2,b\ge (4^{502}{)}^2$.
But we have
$(4^{502}{)}^2=2^{2008}>2^{2002}=$ $(2^{11}{)}^{182}>(2009{)}^{182}$........since $2^{11}=2048>2009$.