<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] A way to do the second problem is the following. Check small values of $n<8$ by hand (nothing there). Then assume that $n\ge8$. Now $2^8+2^{11}+2^n=2^8(1+8+2^{n-8})$ is a perfect square, iff the latter factor $9+2^{n-8}$ is a perfect square also. But 
$$
9+2^{n-8}=m^2\Leftrightarrow 2^{n-8}=m^2-9=(m-3)(m+3).
$$
So by the unique factorization both $m-3$ and $m+3$ must be powers of two. The difference between these two factors is 6, and the differences between powers of 2 are larger than 6 unless both powers are at most 8. The only solution is thus $m=5$, $n=12$.
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