## imqwerty one year ago A question :)

1. imqwerty

2. anonymous

What the actual fudge is that?

3. imqwerty

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

4. anonymous

@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

5. imqwerty

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

6. imqwerty

7. anonymous

$$3^{2008}+4^{2009}=\left(3^{502}\right)^4+4\cdot\left(4^{502}\right)^4$$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

8. anonymous

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 $$a^2-2ab+2b^2>7^{364}\cdot41^{182}$$

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