<s>[INST] What is the largest power $n$ of 3 such that ${3}^n$ divides $2012^4-2011^4$? 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}] 
 [/INST] 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 $2012^4-2011^4$.
</s>