imqwerty
  • imqwerty
A question :)
Mathematics
katieb
  • katieb
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imqwerty
  • imqwerty
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anonymous
  • anonymous
What the actual fudge is that?
imqwerty
  • imqwerty

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