Solutions of Exponential Diophantine Equations $a^x=b^y$

Question

Let $\mathbb{N}$ be a set of natural numbers and $a,b,x,y \in \mathbb{N}$.

What can be said about the existence of natural nontrivial solutions $\langle a_0, b_0, x_0, y_0\rangle$ of equation $a^x=b^y$?

The restricted case of this task when $x=b, y=a$ had been solved here.

(Sorry! I din't know exactly, where to repace this comment on @Mythomorphic 's reply): Grate solutions! As far as my initial task concerned, based on @Mythomorphic consideration, if arbitrary $d, x_0, y_0 \in \mathbb{N}$ will be chosen, then as I see it, any tetrad that looks like < d^$y_0$, d^$x_0$ , $x_0, y_0$ > is a solution of given equation, isn't it? A good job, it suits me fine, thanks!

Answer 1

(This might not be a fine solution but it gives some thoughts here.)

Theorem If $a^x=b^y$ have integer solutions, then $$a^x=b^y=P^l=\left(\prod^n_{k=1}p_k^{c_k}\right)^l$$ where $\{p_k\}$ is a set of distinct primes and $\{c_k\}$ is a set of integers; $l=\text{lcm}(x,y)$.

Proof

If such solutions exist, we know $a,b$ must share the same set of $n$ prime factors $p_1,p_2\ldots p_n$.

$$a=\prod_{k=1}^n p_k^{u_k};b=\prod_{k=1}^n p_k^{w_k}$$

We know if $a^x=b^y$, then for every $k$, $xu_k=yw_k$. This is only possible if $$u_k=\frac{c_kl}{x},w_k=\frac{c_kl}{y}$$

where $l=\text{lcm}(x,y)$ and $c_k$ is an arbitary integer.

---

Extra: On the equation $x^y=y^x(x<y)$

For the special case $a=y,b=x$, by the theorem we know

$$y=a=P^{\frac lx}, x=b=P^{\frac ly}$$ So $$y=xP^{l(\frac1x-\frac1y)}\iff P=\left(\frac yx\right)^{\frac{xy}{l(y-x)}}\tag1$$

Again, by the theorem, we know $y$ is an integer multiple of $x$, i.e. there exists $t\in\Bbb{N}$ such that $y=tx$

Then

$$P=t^{\frac{tx^2}{tx(t-1)x}}\space=t^\frac1{t-1}$$

Obviously, $P$ is integer only if $t=2$.

Solving the equations and we get the only integer soluton $(2,4)$.