My approach -Equation involving ceiling and floor

Question

Find all rational solutions to ${x^{\lceil x \rceil}}^{\lfloor x \rfloor}=\frac{512}{25}$

Notice that $\frac{512}{25}=\frac{2^9}{5^2}$. I tried to use the definitions of the floor and ceiling functions, but how does that help in the exponents Please help me.

My approach -

I let $x=a \frac{p}{q}$ so that we have to solve ${25(qa+p)}^{a+1}^a={512q}^{a+1}^a$ now, we should have $q \mid 25$ and afer a little bash, $q=5$ and $a=1$ works, which implies $25(5+p)^2=512$ which has no solutions?

Answer 1

It's clear that $x$ is not an integer (otherwise $\frac{512}{25} = 20.84$ would also be an integer). Let $x = a + \frac mn$ with $0 < \frac mn < 1$. So our equation becomes $${{(a+\frac mn)}^{(a+1)}}^a = \frac{512}{25}.$$

If $a \geq 2$, the expression on the left is greater than $a^{{(a+1)}^a} \ge2^{3^2} = 2^9 > 20.84$. If $a \leq -1$, we can rewrite $(a+1)^a$ as $\frac1{{(a+1)}^{-a}}$. So we're either taking an even root of $a+\frac mn$ which is imaginary, or we're taking an odd root, which is negative, and since $\frac{512}{25}$ is a positive real number, we have no solutions for $a \leq -1$. We're left with two cases: $a = 0$ and $a = 1$.

1. For $ a= 0$, $(\frac mn)^{(0+1)^0} = \frac mn = \frac{512}{25} > 1$, a contradiction.

2. For a = 1, $(1+\frac mn)^{(1+1)^1} = (1+\frac mn)^2 =\frac{512}{25} \implies 1+\frac mn = \frac{16\sqrt2}5 \implies \sqrt2 = \frac5{16}(1+\frac mn) \in \mathbb Q$, which is known to be false.

Therefore there are no solutions to this equation.

Answer 2

Rational solutions to $x^{\lceil x \rceil^{\lfloor x \rfloor}} = \frac{512}{25}$

Here is an interactive graph in Desmos visualizing the problem. Because the function in the problem increases so rapidly, the graph shows the $x$-axis on a linear scale $x\in\{ \ldots, -2, -1, 0, 1, 2, \ldots \}$), and the $y$-axis on a $\log(\log(y))$ scale $y\in \{ \ldots, 10^{10^{-2}}, 10^{-1}, 1, 10, 10^2, \ldots \}$

For convenience of notation I will define the functions: $$ \newcommand{\floor}[1]{\left\lfloor#1\right\rfloor} \newcommand{\ceil}[1]{\left\lceil#1\right\rceil} \newcommand{\abs}[1]{\left|#1\right|} g(x) = \ceil{x}^{\floor{x}} \\ f(x) = x^{\ceil{x}^{\floor{x}}} = x^{g(x)} $$ First, we can rule out the case $x < 0$:

• If $x \in (-1, 0)$, then $\ceil{x}=0$ and $\floor{x}=-1$, which means $g(x)=0^{-1} = 1/0$, and $f(x)$ is not defined.
• If $x=-1$, then $f(x) = -1$ which is not the solution.
• If $x \in (-2, -1)$, then $g(x)=(-1)^{-2} = 1/(-1)^2 = 1$, hence $f(x) = x$; the solution cannot be in this range either.
• If $x \in (-\infty, -2]$, then $|g(x)|\le 1/4$, hence $f(x)$ is complex. The solution cannot be in this range.

Moving right along, to the case $x \ge 0$:

Let's see what happens to $f(x)$ for different ranges of $x$ -- ignoring, for the moment, whether $x\in\mathbb{Q}$. The idea is, we want to start to understand the behavior of $f$ along the interval where a solution might be feasible.

\begin{array}{c|c|c|l} x \in\text{interval } X & g(x) = \ceil{x}^{\floor{x}} & f(x) = x^{g(x)} & \text{Image }f(X) \\\hline X = \{0\} & \ceil{x}^{\floor{x}}=0^0=1 & f(x)=0 & f(X) = \{0\} \\ X = (0, 1) & \ceil{x}^{\floor{x}}=1^0=1 & f(x)=x & f(X) = (0, 1) \\ X = \{1\} & \ceil{x}^{\floor{x}}=1^1=1 & f(x)=x & f(X) = \{1\} \\ X = (1, 2) & \ceil{x}^{\floor{x}}=2^1=2 & f(x)=x^2 & f(X) = (1, 4) \\ X = \{2\} & \ceil{x}^{\floor{x}}=2^2=4 & f(x)=x^4 & f(X) = \{16\} \\ X = (2, 3) & \ceil{x}^{\floor{x}}=3^2=9 & f(x)=x^9 & f(X) = (2^9, 3^9) = (512, 19683) \\ X = \{3\} & \ceil{x}^{\floor{x}}=3^3=27 & f(x)=x^{27} & f(X) = \{3^{27}\} \approx \{7.6\cdot 10^{12}\} \\ X = (3, 4) & \ceil{x}^{\floor{x}}=4^3=64 & f(x)=x^{64} & f(X) = (3^{64}, 4^{64}) \approx (3.4\cdot 10^{30}, 3.4\cdot 10^{38}) \\ X = \{4\} & \ceil{x}^{\floor{x}}=4^4=256 & f(x)=x^{256}& f(X) = \{4^{256}\} \approx \{1.3\cdot 10^{154}\} \\ \cdots \end{array}

The value of $f(x)$ quickly starts blowing up from here. The general pattern is one where at each successive integer value $n\in\mathbb{N}$, $f$ makes ever-larger discontinuous jumps to a faster-growing polynomial. More specifically, for $n\in\mathbb{N}$ even modestly large, $$ \lim_{x\to n^{-}} f(x) \ll f(n) \ll \lim_{x\to n^{+}} f(x) $$

We are trying to solve for $f(x) = 512/25 = 20.48$, but as we've checked basically through brute force, $512/25 \notin f(\mathbb{R})$. Therefore there are no real solutions. Since any rational solution would of course also be real, then there are also no rational solutions.

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N.b. Your exponential is not rendering correctly; you should listen to the

Double exponent: use braces to clarify

warning it gives you. Depending on whether your intention was to nest the exponent right-to-left (i.e. true nested, the way it is rendered on the first line - but looking at the code, that one also lacks clarifying braces) or left-to-right, this becomes two completely different problems: $x^{\ceil{x}^{\floor{x}}} \neq \left(x^{\ceil{x}}\right)^{\floor{x}} = x^{\ceil{x}\floor{x}}$.

With that said - even if you did change the problem, so that the exponents were nested in the other direction, there still would not be any solutions. Instead you would have: $$ g(x) = \ceil{x}\floor{x} \\ f(x) = \left(x^{\ceil{x}}\right)^{\floor{x}} = x^{\ceil{x}\floor{x}} = x^{g(x)} $$

The function still increases very quickly, if perhaps not quite so quickly as in the previous case with right-to-left nested exponentials. Nevertheless, it still skips over $512/25$ at the point $x=2$: \begin{align} f(x) &< 4 = \lim_{x\to 2^{-}} f(x) \text{ for all }{x < 2} \\ f(x) &= 16 \text{ for }{x < 2} \\ f(x) &> 64 = \lim_{x\to 2^{+}} f(x) \text{ for all }{x > 2} \end{align}