Two problems on number theory

Question

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?

Answer 1

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 $m=5$, $n=12$.

Answer 2

If $n=7$ then $n!=5040\gt2004$, so....

If $n=6$ then $n!=720\lt2004$, so....

Answer 3

Regarding the first question,

If $a!\leq b!$,

then $a\leq b$.

So here,

$((n!)!)!\leq (2004!)!$, implies $(n!)!\leq 2004!$,

which further implies $n!\leq 2004$,

therefore $n\leq 6$.

P.S edit: Didn't see that the solution was already posted.

Answer 4

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 $x=2^{6}$.

Answer 5

A different way for the second question:

First consider $2^8+2^{11}+2^n$ modulo $3$. Because $2^2\equiv 1\pmod 3$, we have $$ 2^8 + 2^{11} + 2^n \equiv 2^0+2^1+2^{n\bmod 2} \equiv 2^{n\bmod 2} \bmod 3$$ and since $2^1$ is not a square modulo $3$, $n$ must be even. Similarly, $2^3\equiv 1\bmod 7$, so $$ 2^8+2^{11} + 2^n \equiv 2^2+2^2 + 2^{n\bmod 3} \equiv 1 + 2^{n\bmod 3} \bmod 7$$ The squares modulo $7$ are $0$, $1$, $2$, and $4$, so $2^{n\bmod 3}$ can only be $1$ and thus $n$ is a multiple of $3$.

Since $n$ is even, $2^n$ is a perfect square, and the next perfect square is $1+2^{n/2+1}+2^n$, which is already too large to be $2^8+2^{11}+2^n$ whenever $n/2+1>11$ -- that is, whenever $n>20$.

Putting it all together, $n$ is a multiple of $6$ that is at most $20$, and we can easily check $n=0, 6, 12, 18$ to find that only $n=12$ works.