Show that $3^{2008} + 4^{2009}$ can be written as product of two positive integers, each of which is larger than $2009^{182}$.
To show that $3^{2008} + 4^{2009}$ is composite, I used the Sophie-Germain identity: $$[a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 − 2ab + 2b^2)]$$
So, I got: $$(3^{502})^4 + 4(4^{502})^4$$ Which gives: $$(3^{1004}+2(3^{502} \cdot 4^{502}) + 2\cdot 4^{1004})\cdot(3^{1004}-2(3^{502} \cdot 4^{502}) + 2\cdot4^{1004})$$
I don't know how to simplify this further and compare it with $2009^{182}$ and check whether my method is correct or not. If the factorization is such that both the factors are greater than $2009^{182}$, then how to prove it? If the factors are not greater than $2009^{182}$, then we're back to square $1$. How to proceed further?
$\frac{2008}{4}=502$, not $252$. In any case, it suffices to show that $$3^{1004} - 2\times3^{502}\times4^{502}+2\times4^{1004}>2009^{182}$$ Note that $2\times3^{502}\times4^{502}< 4^{1004}$, so we just need to show that $4^{1004}>2009^{182}$ (In comparison $3^{1004}$ is quite insignificant). We have $2009< 2^{11} = 2048$, so $2009^{182} < 2^{182\times 11} = 4^{1001} < 4^{1004}$ and we are done.