Analytical Ability

[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.
x = 4[tex] no solution.
[tex]x = 5[tex] no solution.
[tex]x = 6 \rightarrow y=2
Now, since y is monotone decreasing in the next solution, y\leq1 so it's
x=99 \rightarrpw y=1
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 log n values for x^y+y^x=n it's really not so bad.

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

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

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.