<s>[INST] Two problems on number theory I need some ideas (preferable some tricks) for solving these two problems:

Find the largest number $n$ such that $(2004!)!$ is divisible by
  $((n!)!)!$
For which integer $n$ is $2^8 + 2^{11} + 2^n$ a perfect square?

For the second one the suggested solution is like this  : $ 2^8 + 2^{11} + 2^n = ((2^4)^2 + 2\times2^4\times2^6 + (2^ \frac{n}{2})^2 ) \Rightarrow n=12$
But I can't understand the approach,any ideas?
 [/INST] The second one is true. Because $$(2^{4} + 2^{6})^{2} = 2^{8} + 2 \cdot 2^{4} \cdot 2^{6} + 2^{12}$$ so your $n=12$.
As far as I know, the main idea is to write $2^{8}+2^{11}+2^{n}$ as $(2^4)^{2} + 2 \cdot 2^{4} \cdot x + x^{2}$. Then you will have to manipulate what $x$ is and intuition say $x=2^{6}$. 
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