Solve an exponential equation in integers

Question

Find two integers x and y such that $x^y +y^x = 94032$.

I have used

Solve[x^y + y^x == 94032, {x, y}, Integers]
Reduce[x^y + y^x == 94032, {x, y}, Integers]
FindInstance[x^y + y^x == 94032, {x, y}, Integers]

But I can not find the integers that are the solution

I would appreciate any help

Answer 1

There is always a trivial solution: $x=n-1,y=1$.

You can use brute force to find non trivial solutions.

Method 1. Calculate $x^y+y^x$ for all pairs $x < \sqrt{n}$ and $y < \sqrt{n}$. Since we are not interested in trivial solutions, we don't need to check $x > \sqrt{n}$. Then check if we have $n$ in the result list.

findNonTrivialPairs[n_]:=Select[{#1, #2, #1^#2 + #2^#1} & @@@ Tuples[#, 2] &@
    Range[Ceiling@Sqrt[n]], #[[3]] == n &];
findNonTrivialPairs[100]
(* {{2, 6, 100}, {6, 2, 100}} *)

AbsoluteTiming@findNonTrivialPairs[94032]    
(* {0.545318, {}}  <- no non-trivial solutions *)

Method 2.

We will look for pairs where $y > x$, so iterations over y start with $x$

f[{x_, y_, z_}] := {x, y + 1, x^(y + 1) + (y + 1)^x};    
g[x_, n_] := NestWhile[f, {x, x, 0}, #[[3]] < n &];     
findPairs[n_] :=Select[Map[g[#, n] &,Range[n/2]], #[[1]]^#[[2]] + #[[2]]^#[[1]] == n &];

findPairs[100]

(* {{1, 99, 100}, {2, 6, 100}} it has non-trivial solutions *)

Let's find non-trivial solutions for $100<n<200$

Rest /@ Select[findPairs /@ Range[100, 1000], Length[#] > 1 &]
(* {{{2, 6, 100}}, {{3, 4, 145}}, {{2, 7, 177}}, {{2, 8, 320}}, {{3, 5, 
   368}}, {{2, 9, 593}}, {{3, 6, 945}}} *)

For your particular $n=94032$ it seems to have no non-trivial solution (and it takes some time to calculate it):

AbsoluteTiming@findPairs[94032]
(* {108.22, {{1, 94031, 94032}}} *)

Update: We don't need to iterate up to n/2 we can iterate up to l, where l^l=n

It will significantly increase performance.

limit[n_] := Ceiling[x /. FindRoot[x^x == n, {x, 1}]];
findPairs2[n_] := 
  Select[Map[g[#, n] &,Range[limit[n]]], #[[1]]^#[[2]] + #[[2]]^#[[1]] == n &];

AbsoluteTiming@findPairs2[94032]
(* {0.401372, {{1, 94031, 94032}}} *)

Answer 2

you can actually very simply show there are only the trivial solutions:

1) note it is sufficient to consider x<=y , so 2 x^x <= 94032 or x<=6.

Then:

 Table[ {x, y /. FindRoot[ x^y + y^x == 94032 , {y, x}]}, {x, 6}]

{{1, 94031.}, {2, 16.5167}, {3, 10.4125}, {4, 8.22445}, {5, 6.99343}, {6, 6.00551}}