Does the exponential Diophantine equation $2^nn^{-n}s^s(n-s)^{n-s}=1/9,9$ have a solution?

Question

Assuming $0^0=1$, how to prove/disprove the existence of integers $n,s$ such that $0\le s\le n\ge 1$ and $$\left({2\,s\over n}\right)^s\left({2\,(n-s)\over n}\right)^{n-s}\in\left\{{1\over 9},\,9\right\}$$ ... or, equivalently, satisfying either one of the following two equations: $$\begin{align}&\quad 2^n\,s^s\,(n-s)^{n-s}=9\cdot n^n\tag{1}\\ &\quad 9\cdot2^n\,s^s\,(n-s)^{n-s}=n^n\tag{2} \end{align}$$ Clearly there can be no solution with odd $n$, as in that case both (1) and (2) have odd RHS and even LHS; but other than that, I don't see how to proceed. Brute force searching turns up no solutions, and of course the numbers quickly become infeasible to compute. It appears that for both (1) and (2), $\min_{0\le s\le n}$|LHS-RHS| diverges as $n\to\infty$, but how to prove that? (Apologies if there's something obvious that I'm just not seeing.)

_{An equivalent of this question arose while answering another (in the proof sketched for Question #1 there): Discontinuities in the expectation of a stopping time (Bayesian coin-tossing) .)}

Answer 1

To show that there are no solutions, it suffices to show that both $n$ and $s$ are multiples of $3$, as then the factors $2^n$, $s^s$, $n^n$ and $(n-s)^{n-s}$ are all perfect cubes. The remaining factor $9$ is not a perfect cube, a contradiction.

First, if $s=0$ or $n=0$ or $n-s=0$ then, dividing out common factors, the equations simplify to $$2^n=9\qquad\text{ and }\qquad 9\cdot2^n=1,$$ which are clearly impossible, and therefore $n>s>0$.

Now starting off with the second equation, if $$9\cdot2^n\cdot s^s(n-s)^{n-s}=n^n,$$ then it is clear that $3$ divides $n$, so $n=3m$ for some positive integer $m$. It follows that the right hand side is divisible by $3^3$, and hence $s$ is divisible by $3$, and we are done.

The other case is similar; if $$2^n\cdot s^s(n-s)^{n-s}=9n^n,$$ it is clear that $3$ divides either $s$ or $n-s$, or both. In any case, because $s,n-s>0$ it follows that the left hand side is divisible by $3^3$ and so $3$ divides $n$. It follows that $3$ divides $s$ and we are done.

Answer 2

Your $2$ equations, with one requiring to be satisfied, are

$$2^n(s^s)(n-s)^{n-s} = 9(n^n) \tag{1}\label{eq1A}$$

$$9(2^n)(s^s)(n-s)^{n-s} = n^n \tag{2}\label{eq2A}$$

Using the assumption $0^0 = 1$, then with $s = 0$ or $s = n$, \eqref{eq1A} becomes $2^n(n^n) = 9(n^n) \; \to \; 2^n = 1$ and \eqref{eq2A} becomes $9(2^n)(n^n) = n^n \; \to \; 9(2^n) = 1$, with neither being possible. Thus, $1 \le s \le n - 1$, and $1 \le n - s \le n - 1$, with $n \ge 2$.

Next, as you stated, $n$ must be even, so for some positive integer $m$,

$$n = 2m \tag{3}\label{eq3A}$$

Thus, \eqref{eq1A} becomes

$$\begin{equation}\begin{aligned} 2^n(s^s)(2m-s)^{n-s} & = 9((2m)^n) \\ 2^n(s^s)(2m-s)^{n-s} & = 9(2^n)(m^n) \\ s^s(2m-s)^{n-s} & = 9(m^n) \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Next, let

$$d = \gcd(s, m), \; s = de, \; m = df, \; \gcd(e, f) = 1 \tag{5}\label{eq5A}$$

Using \eqref{eq5A} in \eqref{eq4A} gives

$$\begin{equation}\begin{aligned} (de)^s(2df - de)^{n-s} & = 9((df)^n) \\ (d^s)(e^s)(d^{n-s})(2f - e)^{n-s} & = 9(d^n)(f^n) \\ (d^n)(e^s)(2f - e)^{n-s} & = 9(d^n)(f^n) \\ (e^s)(2f - e)^{n-s} & = 9(f^n) \end{aligned}\end{equation}\tag{6}\label{eq6A}$$

This shows $f$ must divide the left side. Since \eqref{eq5A} gives $\gcd(e,f) = 1$, this means $f \mid (2f - e)^{n-s}$, which is only possible if $f = 1$. Since $2f - e \gt 0$, then $e = 1$. However, \eqref{eq6A} then becomes $1 = 9$, which is not possible.

A similar argument can be applied with \eqref{eq2A} to also get a contradiction, which I'll leave to you to do.