Integer solutions to $x^{x-2}=y^{x-1}$

Question

Find all $x,y \in \mathbb{Z}^+ $ such that $$x^{x-2}=y^{x-1}.$$

I can only find the following solutions: $x=1,2$. Are there any other solutions?

Answer 1

The only solutions are $x=1$ and $y=$ anything, and $x = 2$ and $y = 1$.

Start by rearranging the equation: $(\frac{x}{y})^{x-1} = x$.

If x = 1 then we have: $(\frac{1}{y})^0 = 1$ which is true for any $y$.

If y = 1 then we have: $x^{x-1} = x \implies x^{x-2} = 1$, which means either $x = 1$ which is covered by the case above, or $x = 2$.

If $x \not = 1$, $y \not = 1$ and $x$ is not a multiple of $y$, then $(\frac{x}{y})^{x-1} = x$ is in the form $\text{fraction}^{\text{positive integer}} = \text{positive integer}$, which is always false.

If $x \not = 1$, $y \not = 1$ and $x$ is a multiple of $y$, then we can write $x = ky$ for some $k \in \mathbb{Z}^+$. So we get $k^{ky-1} = ky \implies k^{ky-2} = y \implies (k^k)^y = k^2y$. If $k = 1$ then $y = 1$ and $x = 1$ (covered), or if $k = 2$ then $y = 1$ and $x = 2$ (covered), so any other solutions has $k > 2$. We see then that at $y = 1$, the $\text{LHS} > \text{RHS}$. You now want to show that it's true for every $y$. You can do that by doing the following:

Let $u_y = (k^k)^y$, $v_y = k^2y$. Compute $\frac{u_y}{u_{y-1}} = k^k \geq 27$, $\frac{v_y}{v_{y-1}} = \frac{y}{y-1} = 1 + \frac{1}{y-1} \leq 2$. So $\frac{u_y}{u_{y-1}} > \frac{v_y}{v_{y-1}}$. So $u_y > v_y, \forall y$. No more solutions.

Answer 2

$\require{begingroup} \begingroup\let\geq\geqslant\let\leq\leqslant$Let $x\geq2$.
$x^{x-2}$ is an $(x-1)$-th power. Because $\gcd(x-1,x-2)=1$ this means that $x$ has to be an $(x-1)$-th power. (Looking at the prime factorisation, $n$ is an $m$th power iff every prime has exponent divisible by $m$.)
Say $x=a^{x-1}$ for some $a\geq2$.

But $$x=a^{x-1}\geq2^{x-1}=2\cdot2^{x-2}>2(x-2)$$ (See Prove that $ n < 2^{n}$ for all natural numbers $n$.)
so $$x<4:$$

• $1^{-1}=y^0$ for any $y$
• $2^0=y^1$ for $y=1$
• $3^1=y^2$ is impossible$\endgroup$

---

Edit: user3491648's answer hints at an alternative argument why $x$ is an $(x-1)$-th power. We have

$$x^{x-1}=y^{x-1}\cdot x.$$

Because $a^n\mid b^n$ implies $a\mid b$ we obtain $y\mid x$, hence $$x=\left(\frac xy\right)^{x-1}\qquad\text{with }\frac xy\in\mathbb N.$$

Answer 3

Let $x=\prod p^{a_p}$, $y=\prod p^{b_p}$. Then we need $(x-2)a_p=(x-1)b_p$ for all primes $p$. Specifically, $\gcd(x-2,x-1)=1$ implies $x-2\mid b_p$ and $x-1\mid a_p$ In fact, we conclude that for suitable $z\in\mathbb N$, we have $x=z^{x-1}$ and $y=z^{x-2}$. But for $x\ge3$ and $z\ge 2$, $x=z^{x-1}$ has no solution as $z^{x-1}\ge 2^{x-1}>x$.