Finding integer solutions to x^y+y^x=100

[himanshu121]

How one can find the solution to $$x^y+y^x=100$$ 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 $$x$$ and $$y$$ 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 $$(-2,0)$$.

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

 

[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 $$log n$$ values for $$x^y+y^x=n$$ it's really not so bad.

Consider that if $$x=y$$ then you have
[tex]2x^x=n[tex] so<br /> $$x ln x = ln \frac{n}{2}$$<br /> <br /> So, you'd really only have to check up to $$x=8$$ or so for $$n=1000000$$<br /> <br /> 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.<br /> <br /> Unless you've got something very specific in mind, that's already a pretty good solution.<br /> <br /> I suppose it would be good if you want to ask questions like:<br /> Are there n with arbitrarily many solutions?<br /> or something similar.[/tex][/tex]