Question

A question :)

Answer 1

What the actual fudge is that?

Answer 2

@ganeshie8 @Empty @dan815 @ParthKohli @oldrin.bataku

Answer 3

@imqwerty I ask @ganeshie8 ta help u but u have ta wait tell he text me back cause he did not answer me yet k

Answer 4

@Peaches15 i knw the answer :) but m stuck a lil bit

Answer 5

$$3^{2008}+4^{2009}=\left(3^{502}\right)^4+4\cdot\left(4^{502}\right)^4$$now Sophie Germain's identity tells us that we can rewrite this using \(a=3^{502},b=4^{502}\) so that $$a^4+4b^4=(a^2+2ab+2b^2)(a^2-2ab+2b^2)$$
so clearly it has two integer divisors

Answer 6

and note that \(2009=7^2\cdot41\) so \(2009^{182}=7^{364}\cdot 41^{182}\)

so we know both divisors are at least \(a^2-2ab+2b^2\) so we just need to show that $$a^2-2ab+2b^2>7^{364}\cdot41^{182}$$