$x^y y^x$ in more than one way

Question

I've submitted to OEIS the sequence consisting of numbers that can be written in more than one way as $x^y y^x$ for integers $x,y$ with $1 < x \le y$. The first $6$ examples are $$ \eqalign{ 4^{16} \cdot 16^4 &= 8^8 \cdot 8^8 = 281474976710656 \cr 3^{27} \cdot 27^3 &= 9^9 \cdot 9^9 = 150094635296999121\cr 10^{80} \cdot 80^{10} &= 20^{50} \cdot 50^{20} \cr 2^{324} \cdot 324^2 &= 8^{96} \cdot 96^8 \cr 8^{144} \cdot 144^8 &= 16^{96} \cdot 96^{16} \cr 16^{144} \cdot 144^{16} &= 32^{96} \cdot 96^{32} \cr}$$

I managed to find an infinite parametric family of solutions, generalizing the first example:

$$(2^{x+1})^{2^{x-k-1}} \cdot (2^{x-k-1})^{2^{x+1}} = (2^x)^{2^x} \cdot (2^x)^{2^x}\ \text{with}\ x = (k+1) 2^{k+2} - 1$$

Questions:

1. Can you find other infinite parametric families of solutions?
2. Is there any example that can be written in three or more ways?

[EDIT] This is now sequence A367001

Answer 1

Not a full answer, but some observations regarding the examples you have found and 2 new examples:

Consider the following examples: $$ 4^{16}\cdot 16^4 = 8^8\cdot 8^8 $$ $$ 8^{144}\cdot 144^8 = 16^{96}\cdot 96^{16} $$ $$ 16^{144}\cdot 144^{16} = 32^{96}\cdot 96^{32} $$ There is a clear recurrency here, in particular they are of the form: $$ (2^{k})^{a}\cdot a^{2^k} = (2^{k+1})^b\cdot b^{2^{k+1}} $$ and I have found two more! Namely: $$ 32^{1600}\cdot 1600^{32}=64^{1280}\cdot 1280^{64} $$ $$ 64^{1600}\cdot 1600^{64} = 128^{1280}\cdot 1280^{128} $$ The pattern is now clear, and we can study it in order to generalize a possible new family of solutions. In particular I note that:

1. If $k$ is odd then $a$ and $b$ satistifes both the "consecutive" equations $$ (2^{k})^{a}\cdot a^{2^k} = (2^{k+1})^b\cdot b^{2^{k+1}} $$ $$ (2^{k+1})^{a}\cdot a^{2^{k+1}} = (2^{k+2})^b\cdot b^{2^{k+2}} $$
2. Furthermore $\frac{a}{b} = 2$ in the first equation , $\frac{a}{b} = \frac{3}{2}$ in the second and third, $\frac{a}{b} = \frac{5}{4}$ in the fourth and fifth..
3. And if $(a,b)$ is a solution to the "$k$-th and $k+1$-th equations, then they also satisfy: $$ b^2 = a\cdot (2^{k})^{(\frac{a-b}{2^{k}})}\cdot 2^{-(\frac{b}{2^{k}})} $$ $$ b^2 = a\cdot (2^{k+1})^{(\frac{a-b}{2^{k+1}})}\cdot 2^{-(\frac{b}{2^{k+1}})} $$

For example, in the last two $(a,b) = (1600,1280)$ and: $$ 1280^2 = 1600\cdot (2^{5})^{(\frac{1600-1280}{2^{5}})}\cdot 2^{-(\frac{1280}{2^{5}})} = 1638400 $$ $$ 1280^2 = 1600\cdot (2^{6})^{(\frac{1600-1280}{2^{6}})}\cdot 2^{-(\frac{1280}{2^{6}})} = 1638400 $$

Now, of course it seems that for $k = 7$ and $k = 8$, $\frac{a}{b}$ should be $\frac{9}{8}$, but zero positive integer solutions to $$ b^2 = a\cdot(2^{7})^{(\frac{a-b}{2^{7}})}\cdot 2^{-(\frac{b}{2^{7}})} $$ using $\frac{a}{b} = \frac{9}{8}$. I have also tried other fractions but I just can not found other solutions. I have also brute forced the $k=7$-th equation and didn't found any solution for $a,b < 25000$.

EDIT 1: Dividing the consecutive equations ($b^2 = \dots$) we obtain that for odd $k$: $$ (1-k)a+kb = 0 $$ where $(a,b)$ is the solution to the $k$-th equation. For example $(1-5)\cdot 1600 + 5\cdot 1280 = 0$.

EDIT 2 Using EDIT1 we know that the solution $(a,b)$ of the $7$-th equation has to satisfy: $$ (1-7)a+7b = 0 \implies 7b-6a = 0 $$ and on the other hand: $$ b^2 = a\cdot(2^{7})^{(\frac{a-b}{2^{7}})}\cdot 2^{-(\frac{b}{2^{7}})} = a\cdot 2^{\frac{7}{128}a-\frac{8}{128}b} $$ which implies that $\frac{7}{128}a-\frac{8}{128}b = n\in\mathbb{N}$. The system of equations: $$ \begin{cases} 7b-6a = 0 \\ 7a-8b = 128n \end{cases} $$ has solutions given by $a = 896n$, $b = 768n$. Substituing back we obtain: $$ (768n)^2 = 896n \cdot 2^{n} $$ which has.. no integer solutions. Very disappointing, but maybe there are no more solution of this form :(

The following code in Mathematica

f[k_] := Reduce[{-(k - 1)* x + k* y == 0, k* x - 1 (k + 1)*y == 2^k *n}, {x, y, n}, Integers]

finds for odd values of $k$ the possible value of $a$ and $b$ as a function of $n$. Substituing back we can check with the script below if there exists $n\in\mathbb{N}$ that satisfy the equation.

NSolve[(b)^2 == a*2^n, {n}, PositiveInteger]

where instead of $(a,b)$ one should substitute the values obtained before.

EDIT 3

The solutions of the previous system are of the form: $$ (a,b) = (nk 2^k, n(k-1) 2^k) $$ for every odd $k$. Using the following code we can see that it seems that there are no integer solution to the equation $b^2 = a2^n$ for $k \geq 7$:

g[k_] := NSolve[((k - 1)*2^k*n)^2 == k*2^k*n*2^n, {n}];
Table[g[2*k + 1][[3]], {k, 1, 100}]

Answer 2

The question essentially asks for solution to the following equation: $$ a^b b^a = x^y y^x $$, with conditions $a\neq x \cap b\neq y$ or $ a\neq y \cap x\neq b $.

This can be solved as a pullback $\{(a,b,x,y)\in N^4|a^b b^a=x^y y^x,(a\neq x \cap b\neq y) \cup (a\neq y \cap x\neq b)\}$.

This is essentially infinite filtering operation on $N^4$.

Given non-constructive nature of the problem, solving this requires tools that work with non-constructive problems, i.e. axiom of choice is your best bet. Thus we'll construct epimorphism and it's inverse to find infinite number of buckets and make a choice in each of the buckets to solve this problem. Choice function is obviously: $$ch()=(\exists (a,b,x,y): a^b b^a = x^y y^x \cap ((a\neq x \cap b \neq y) \cup (a\neq y \cap x\neq b))) \cup \bot$$.

Applying and evaluating the choice function against infinite number of buckets in $N^4$ takes infinite amount of time, so it's not good approach, so the normal substitution approach for (a,b,x,y) does not work in this problem. (substitution is the inverse of $\exists$ and $\forall$ in adjoint functors.) It's worth mentioning though, since we can verify found solutions with it.

I have to think this further to figure out how to extract the (a,b,x,y) numeric values from the filter...

EDIT: I figured out that instead of mapping where the equation's sides are equal, we need to map the area where they're not equal. So we need to create distance metric for the $\bot$ in the choice function. Basically distance function $$d(a,b,x,y) = a^b b^a - x^y y^x$$ and then we can jump larger distances in the huge space while trying to move between different values of a,b,x,y.