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imqwerty

  • one year ago

A question :)

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  1. imqwerty
    • one year ago
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  2. anonymous
    • one year ago
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    What the actual fudge is that?

  3. imqwerty
    • one year ago
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    @ganeshie8 @Empty @dan815 @ParthKohli @oldrin.bataku

  4. anonymous
    • one year ago
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    @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
    • one year ago
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    @Peaches15 i knw the answer :) but m stuck a lil bit

  6. imqwerty
    • one year ago
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  7. anonymous
    • one year ago
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    $$3^{2008}+4^{2009}=\left(3^{502}\right)^4+4\cdot\left(4^{502}\right)^4$$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
    • one year ago
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    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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spraguer (Moderator)
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