# 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?

• Hint for the 1st question: $x!\mid y!$ if and only if $x\le y$. Also, factorial is an increasing function. Aug 13, 2011 at 6:56
• @Jyrki Lahtonen:That's a very precise mathematical hint:-)Thanks. Aug 13, 2011 at 7:04

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

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

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

• Oh, reading this made me feel silly! My first read through the question made me think of very complicated things. But this reminds me that everything in math is impossible, until it's trivial. Aug 13, 2011 at 6:58
• This is exactly the same way I solved it :-) Aug 13, 2011 at 7:01

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.

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 $x=2^{6}$.

• +1,the second one is so simpleeeee :-) Thanks for posting :-) Aug 13, 2011 at 6:49
• @FoolFormath: Welcome. Glad I could help :)
– user9413
Aug 13, 2011 at 6:50
• I rolled back,may be somebody might post a different approach for the first one :) Aug 13, 2011 at 6:51
• So far as I can see, you've proved it's a square for $n=12$, but not that it's a square only for $n=12$. But maybe OP doesn't need that. Aug 13, 2011 at 6:52

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.