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imqwerty
 one year ago
A question :)
imqwerty
 one year ago
A question :)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What the actual fudge is that?

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 @Empty @dan815 @ParthKohli @oldrin.bataku

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@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

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0@Peaches15 i knw the answer :) but m stuck a lil bit

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$3^{2008}+4^{2009}=\left(3^{502}\right)^4+4\cdot\left(4^{502}\right)^4$$now Sophie Germain's identity tells us that we can rewrite this using \(a=3^{502},b=4^{502}\) so that $$a^4+4b^4=(a^2+2ab+2b^2)(a^22ab+2b^2)$$ so clearly it has two integer divisors

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and note that \(2009=7^2\cdot41\) so \(2009^{182}=7^{364}\cdot 41^{182}\) so we know both divisors are at least \(a^22ab+2b^2\) so we just need to show that $$a^22ab+2b^2>7^{364}\cdot41^{182}$$
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