Vtyjkyuk

Here, $^yx$ means "$x$ tetrated to $y$", and is defined recursively by

$^yx = begincases
1, & textif y = 0\
x^^y-1x, & textif y > 0
endcases$

I want to know if there are non-trivial positive integer solutions to $^ba$ = $^dc$. I define a solution to be trivial if $mina, b, c, d = 1$ or $a =c$.

I have already proven that there are no solutions where $b, d = 2, 3$, and it is as follows:

Suppose that $x^x^x=y^y$ for positive integers $x, y > 1$. Then $x$ and $y$ are both perfect powers of a common integer, so let $x = p^a$, $y = p^b$ (note that $a < b$). Then we get

$p^ap^ap^a=p^bp^b$, and therefore, $ap^ap^a=bp^b$, or $fracba=fracp^ap^ap^b$

So let $b = ap^c$ for some positive integer $c$. Our equation turns to $p^ap^a = p^b+c$, or $ap^a = ap^c + c$ (note this means $a > c$). Since all variables are positive integers and $p > 1$, we know that $ap^c > c$, so $ap^c + c < 2ap^c le ap^c+1 le ap^a$, but we determined that $ap^c + c = ap^a$, so we have a contradiction.

I have not made any progress towards solving the general equal tetration case beyond here. If someone can prove that there are no solutions for larger $b$ and $d$ (which is what I suspect), this would prove that addition, multiplication, and exponentiation are the only hyper-operations $*$ where $a*b = c*d$ has non-trivial solutions among the positive integers.