General solutions/ Similar problems for the relationship between exponents and concatenation

Question

$$x^y=y\|x \quad x,y \in \mathbb{Z}$$ One such solution is x=5, y=2: $$5^2=25=2\|5$$

I have been trying to derive a general form for numbers that satisfy the above equality, however, the presence of the floor function (more generally, the fact that it is over the integers) makes the solution method quite difficult.

I am wondering if anybody has any ideas on how to tackle this problem, or if they have seen something similar before?

Answer 1

Thanks to Julián Aguirre for providing a critical step in this proof:

$$x^y=y\|x,\quad x,y \in \mathbb{Z}$$ Because concatenation is only defined for the non negative integers, it follows that: $$ x,y\geq 0$$ Rewriting the first equality: $$x^y=y\|x=y\times10^{\lfloor log_{10}(x) \rfloor +1 }+x $$ Because: $$y\times10^{\lfloor log_{10}(x) \rfloor +1 }+x\leq y\times 10x+x=(10y+1)x$$ It follows that: $$x^y\leq (10y+1)x$$ $$\frac{x^y}{x}\leq 10y+1$$ $$x^{y-1}\leq 10y+1$$ $$x\leq (10y+1)^{1/{y-1}}$$ And because $ x \in \mathbb{Z}$: $$x\leq \lfloor (10y+1)^{1/{y-1}} \rfloor $$ For which it can be shown that $y<8$ is a necessary condition to satisfy the inequality, else $x=1$, whereby: $$1^y<y\|1$$ Except for the case of $y=0$ which is a solution, should concatenation be defined across the natural numbers instead of the non negative integers. By the inequality above, x is bounded to a set of ordered pairs: \begin{array}{|c|c|} \hline x \leq& y \\ \hline 2&7 \\ \hline 2&6 \\ \hline 2&5 \\ \hline 3&4 \\ \hline 5&3 \\ \hline 21&2 \\ \hline \end{array} It can be shown thereafter, either via the derivative with respect to x and y or through the use of a modular that at most a single unique answer exists for each value of $x$ and $y$. Because the $5^2$ case is known, all of the options for the exponent 2 can be shown to not exist. Similarly, it can be shown that no solution exists for $2^y$. This leaves only the y=4 cases to check for which it can be shown that there are no solutions.

Answer 2

Let $d(x)$ be the number of digits of $x$. Then $$ y\,\|\,x=y\times10^{d(x)}+x\le y\,(10\,x)+x=(10\,y+1)\,x. $$ Any solution of $x^y=y\,\|\,x$ satisfies the inequality $$ x\le\bigl\lfloor(10\,y+1)^{1/(y-1)}\bigr\rfloor. $$ The right hand side id equal to $1$ if $y>7$. Clearly there are no solutions with $x=1$. We are left with only a few cases for $1\le y\le7$. Further restrictions can be imposed to fasten the calculations. For instance, $x$ must divide $y\times10^{d(x)}$.