Problem: If, $ 2^s (2^{s-2} \pm 1)=n!$ , prove $n, s$ are finite.
Proof 1: Given, $ 2^s (2^{s-2} \pm 1)=n! \cdots (1)$
Let, $x=(2^{s-2} \pm 1)$. We know,
$\log(n!) = n \log(n)-n + O(\log(n))$ (see $[1]$ Lemma, equation $(3.4)$ on page , 53)
$\implies \log( 2^s \times x)= n \log(n)-n + O(\log(n))$ $\implies \log( 2^s) +\log(x)= n \log(n)-n + O(\log(n))$ $\implies \log(x)= n \log(n)-n + O(\log(n))- \log( 2^s) \cdots (2)$
Here, $\log( 2^s)=s \times \log2=(n-s_2(n)) \times \log2$ $[\because s=n-s_2(n) ]$
$\therefore \log( 2^s) < n$, so, $n \log(n)-n + O(\log(n))- \log( 2^s) $ $> n \log(n)-n + O(\log(n))- n$.
Now, we can rewrite $(2)$ as- $ \log(x) > n \log(n)-2n + O(\log(n))=n\times (\log(n)-2)+ O(\log(n) > C \times n $ $\therefore \log(x) > C \times \log( 2^s) \implies x > C_1 \times 2^s \cdots (3)$.
$C, C_1$ are constants. For sufficient large value of $n$, $C>100, C_1>2$. But inequality $(3)$ contradicts $x=(2^{s-2} \pm 1)$. So, $n,s$ can not be infinite. $\square$
$\textbf{Proof 2:} $ If $n!=a \times 2^s$ and $a$ is the odd number, then $a\geq 2^{n-1}$ . Let, $P(n)$ be the proposition- $n! \geq 2^{2(n-1)}$
BASIS STEP: $P(8)$ is true as $8! > 2^{14}$.
INDUCTIVE STEP: Assume, $P(d)$ is true $(\forall d> d_0)$, we need to show that $P(d+1)$ is true. Form $P(d)$ , $d! \geq 2^{2(d-1)} \implies 4 \times d! \geq 2^{2d} $. Since, $d+1>4, \forall d>4,$ we can write-$ (d+1) > 4 \times d! \geq 2^{2d} $, so, $P(d+1)$ is also true. The inductive step is complete.
Thus, $n! \geq 2^{2(n-1)}$. But then, $a \times 2^{n-1} \geq 2^{2n-2}$ $[\because s \leq n-1]$
$\implies a \geq 2^{2n-2-n+1}=2^{n-1}$. So, $n,s$ can not be infinite. $\square$
Query: Is there any error in the above proof? If there is an error, explain the error as an answer. Also, if you have an alternative / correct proof, please post.
Related Post: Finitely many Positive Integers $n, s $ such that $n!=2^s(2^{s−2}−1)$. This post has an argument for $n!=2^s(2^{s−2}−1)$ using a result of Cameron Stewart, whereas current post concerns with the general case $n!=2^s(2^{s−2} \pm 1)$.
Reference:
- Stopple, J. , A Primer of Analytic Number Theory, Cambridge University Press.