[Guest]

Given that x and y are positive integers, solve x^y = y^x.

*Note: x and y are unique. i.e. x!=y

Edited June 4, 2009 by adiace

[bushindo]

Just nitpicking with this, but I'm pretty sure -1^(1/2) is not considered positive, and then curious, is it actually considered an integer?

Just to add something, there are infinite real solutions on the real and imaginary plane. Taking the log and re-arranging, you get

x/log(x) = y/log(y)

Here's a graph of x/log(x)

http://www93.wolframalpha.com/input/?i=x%2FLog[x]

Here's the equations for the solution function

http://www93.wolframalpha.com/input/?i=x%2...+%3D+y%2FLog[y]

Wolfram alpha is like the nerdy search engine, sitting in the corner at prom in weird gaudy clothes, while the popular Google is in the middle of the room talking up the ladies.

Edited June 4, 2009 by bushindo

[Guest]

Just to add something, there are infinite real solutions on the real and imaginary plane. Taking the log and re-arranging, you get

x/log(x) = y/log(y)

Here's a graph of x/log(x)

http://www93.wolframalpha.com/input/?i=x%2FLog[x]

Here's the equations for the solution function

http://www93.wolframalpha.com/input/?i=x%2...+%3D+y%2FLog[y]

Wolfram alpha is like the nerdy search engine, sitting in the corner at prom in weird gaudy clothes, while the popular Google is in the middle of the room talking up the ladies.

your equation is correct, however for a graphical solution you must solve for y to get a 'y=' equation

*if you figure out how to do that, please let me know, been a while since I was in algebra, so that could be where I'm having trouble

Edited June 4, 2009 by perk8504

[Guest]

Just to add something, there are infinite real solutions on the real and imaginary plane. Taking the log and re-arranging, you get

x/log(x) = y/log(y)

Here's a graph of x/log(x)

http://www93.wolframalpha.com/input/?i=x%2FLog[x]

Here's the equations for the solution function

http://www93.wolframalpha.com/input/?i=x%2...+%3D+y%2FLog[y]

Wolfram alpha is like the nerdy search engine, sitting in the corner at prom in weird gaudy clothes, while the popular Google is in the middle of the room talking up the ladies.

Nice work. But how can you get integers as asked in OP?

[Guest]

Hey all, some really interesting answers came by, but only one integer solution. 2,4.

For those who are trying to find another integer solution to this problem, stop!

x = y=1 is a trivial solution.

So we can assume x is greater than or equal 2 and y us greater than x.

Begin by dividing both sides by x^x:

x^y/x^x = y^x/x^x

therefore:

x^y-x = (y/x)^x

As the left hand side is an integer, the right hand side must be an integer as well integer.

y/x is an integer.

Let an integer k = y/x => y = kx.

therefore x^kx-x = k^x

therefore x^x(k-1) = k^x

therefore (x^x(k-1))^1/x = (k^x)^1/x

therefore x^k-1 = k

We have ruled out the trivial case x = y and y = kx, it is clear that k greater than or equal 2.

For k = 2, x¹ = 2 implies y = 4

Let us consider m greater than or equal 3:

When k = 3, x² = 3, and as x greater than or equal 2, => x² greater than 3.

Proof:

We shall now prove inductively that x^k-1 greater than k and thus x^k-1 not equal k for k greater than or equal 3.

If true then x^k-2 greater than k-1.

therefore x^k-1 = x.x^k-2 greater than x(k-1)

But as x greater than or equal 2, x^k-1 greater than 2k-2 greater than k, which is the expected result.

Hence x^k-1 greater than k for all values of k greater than or equal 3.

Therefore we prove that 2⁴ = 4² is the only solution in positive integers.

Edited June 5, 2009 by adiace

[Guest]

Hey all, some really interesting answers came by, but only one integer solution. 2,4.

For those who are trying to find another integer solution to this problem, stop!

Hence x^k-1 greater than k for all values of k greater than or equal 3.

Therefore we prove that 2⁴ = 4² is the only solution in positive integers.

This convicts me. Cong... Very good job.

Really I was checking numbers and couldn't find any other couple up to 100,000, where I stopped seeing your proof.

[Guest]

This convicts me. Cong... Very good job.

Really I was checking numbers and couldn't find any other couple up to 100,000, where I stopped seeing your proof.

I still disagree. What is wrong with:

x = 1

y = 2i

2i is an integer.

[Guest]

I still disagree. What is wrong with:

x = 1

y = 2i

2i is an integer.

or what about

y =1

x = 2i

[Guest]

there is also the inverse of any answer such as

y = 2

x = 4

and

y=4

x=2

therefore

2,4 is not the only answer

4,2 is also one

[Guest]

no!

20^40 and 40^20 both equal 1.1^52. there must be a PATTERN

2i is not real and inverse don't count captain obvious.

[Guest]

no!

20^40 and 40^20 both equal 1.1^52. there must be a PATTERN

2i is not real and inverse don't count captain obvious.

20^40=1.099x10^52

40^20=1.099x10^32