Analyzing X^Y+Y^X=100: How to Solve Integer Equation

[himanshu121]

How one can find the solution to [tex]x^y+y^x=100[/tex] Where x,y belong to integers?

 

[HallsofIvy]

I can't answer how one would solve such an equation in general!

However, as soon as I saw that "100" on the right side of the equation, I thought "64+ 36= 8²+ 6²= 100".
(Think 3-4-5 right triangle.)

Hmmm, 8= 2³ so 8²= (2³)²= 2⁶.

Sure enough, 2⁶+ 6²= 100.

x= 2, y= 6 is a solution.

 

[himanshu121]

I got another solution apart from(2,6)
It can also be (x=1,y=99)

I believe there is no other way than churning the combination of numbers out of the mind

Also, we got 1 equation and two variables

 

[NateTG]

Well, let's take a quick look:

Clearly [tex]x[/tex] and [tex]y[/tex] must be greater than or equal to zero. If one is less than zero, there is a fractional part, if both are less than zero, the sum is on the interval [tex](-2,0)[/tex].

So, we have:
[tex]x=0[/tex] no solution.
[tex]x = 1 \rightarrow y=99[/tex]
[tex]x = 2 \rightarrow y=6[/tex] (Thanks Halls)
[tex]x = 3[/tex] no solution.
Since [tex]x=3[/tex] we have [tex]x^y \in {1,3,9,27,81}[/tex] but none of those work since the complements mod 100 are not powers of the appropriate exponents.
[tex]x = 4[tex] no solution.
[tex]x = 5[tex] no solution.
[tex]x = 6 \rightarrow y=2[/tex]
Now, since [tex]y[/tex] is monotone decreasing in the next solution, [tex]y\leq1[/tex] so it's
[tex]x=99 \rightarrpw y=1[/tex]
since the solutions are symetric.

Which gives you a complete list of solutions in the integers:
(1,99),(2,6),(6,2),(99,1)

 

[himanshu121]

Okay from reply of Halls and NateTG? I believe there is no general way

Anyway thanks Guys for using the symmetry and Mind

I do found the two solutions i was rather looking for framing the answers, NateTG's Explanation is good

 

[NateTG]

Originally posted by himanshu121
Okay from reply of Halls and NateTG? I believe there is no general way.

Considering that you really only have to check about [tex]log n[/tex] values for [tex]x^y+y^x=n[/tex] it's really not so bad.

Consider that if [tex]x=y[/tex] then you have
[tex]2x^x=n[tex]
so
[tex]x ln x = ln \frac{n}{2}[/tex]

So, you'd really only have to check up to [tex]x=8[/tex] or so for [tex]n=1000000[/tex]

Unless n is really big (so big that it's not practical to store it on a computer) that approach will give you all possible solutions fairily quickly.

Unless you've got something very specific in mind, that's already a pretty good solution.

I suppose it would be good if you want to ask questions like:
Are there n with arbitrarily many solutions?
or something similar.