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What is the largest power $n$ of 3 such that ${3}^n$ divides $2012^4-2011^4$? [\begin{align} & \text{My try follows : } \\ & \text{ 201}{{\text{2}}^{\text{4}}}\text{-201}{{\text{1}}^{\text{4}}}= \\ & =\left( \text{201}{{\text{2}}^{\text{2}}}\text{-201}{{\text{1}}^{\text{2}}} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\ & =\left( \text{2012-2011} \right)\left( \text{2012+2011} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\ & =\left( \text{2012+2011} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\ & =\left( \text{4023} \right)\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\ & =\text{ }{{\text{3}}^{\text{3}}}\text{*149}\left( \text{201}{{\text{2}}^{\text{2}}}\text{+201}{{\text{1}}^{\text{2}}} \right) \\ & \text{from which i conclude the answer to be 3} \\ & \text{Is my work right ? is there any simpler way ?} \\ & \text{Thanks for your help } \\ \end{align}]

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    $\begingroup$ The answer $3^3$ is correct. I can't think of a shorter way,though. Also ,there is a mistake in the working ,since you have written $2012+2011 = 2403$, which is incorrect. It should be $4023$, which happens to be $27 * 149$.So you've got the right answer in the wrong way. Also, you did not justify that $3$ does not divide $2012^2 + 2011^2$. $\endgroup$ Commented Apr 5, 2017 at 11:15
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    $\begingroup$ 2012+2011=4023 not 2403. Nevertheless, your conclusion still holds since $ 4023=27 \times 149 $ and 149 is not divisible by 3. As far as why the other factor is not divisible by 3, you have that $ 2012 \equiv 2 (mod 3) $ and $ 2011 \equiv 1 (mod 3) $ hence $ 2012^{2}+2011^{2} \equiv 4+1 \equiv 2 (mod 3) $. $\endgroup$
    – Andrei
    Commented Apr 5, 2017 at 11:18
  • $\begingroup$ @Raizen thank you yes 4023 it is a typo ; for the other factor i got it ; so clear $\endgroup$
    – user373141
    Commented Apr 5, 2017 at 11:21

2 Answers 2

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$3$ is the right answer, but you still need to show that $2012^2+2011^2$ has no factors of $3.$ But that's easy if you just reduce modulo $3$.

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Just to give another approach, note that $81\cdot25=2025$, so that working mod $81$ (from which we can draw conclusions mod $27$), we have

$$\begin{align} 2012^4-2011^4&\equiv(-13)^4-(-14)^4\\ &\equiv169^2-196^2\\ &\equiv7^2-34^2\\ &=(7-34)(7+34)\\ &=-27\cdot41\mod81\\ &\equiv0\mod27 \end{align}$$

Since $3\not\mid41$, we see that $-27\cdot41\not\equiv0$ mod $81$. Thus $27=3^3$ is the largest power of $3$ that divides $2012^4-2011^4$.

The main advantage of this approach is that it obviates the need to find the highest power of $3$ dividing $4023$ (which I have a hard time doing in my head); it also avoids the need for a side proof that $3$ doesn't divide $2012^2+2011^2$. The main disadvantage is that it requires you to guess (or know) that $81$ is the first power of $3$ that doesn't divide $2012^4-2011^4$.

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