Question 3

ThePerfectHacker · #1 $Old$ October 23rd, 2006, 10:39 AM

Starting next week the way these problems are posed might change. But for know here is this week's problem.

Let, $a,b>0$ .
Find all solutions to the Diophantine equation*):
$a^b=b^a$

*)Diophantine equations are named in the honor of Diophantus of Alexandria, they are regular equations except we can only use integers.

TD! · #2 $Old$ October 24th, 2006, 09:30 AM

Trivial solutions: x = y.
Non-trivial: (2,4) and (4,2).

Proof: trivial

Just kidding, but I'll leave that to someone who wants to spend time writing it.
Starting hint: take logarithms, rewrite (consider f(x) = ln(x)/x), check increasing/decreasing...

topsquark · #3 $Old$ October 25th, 2006, 12:02 PM

Quote:

Originally Posted by TD! $View Post$

Trivial solutions: x = y.
Non-trivial: (2,4) and (4,2).

Proof: trivial

Just kidding, but I'll leave that to someone who wants to spend time writing it.
Starting hint: take logarithms, rewrite (consider f(x) = ln(x)/x), check increasing/decreasing...

Just for the record, TD didn't find all the solutions. But a slight adjustment of his method will. (Figuring out what he missed will probably drive him crazy until he sees the answer.

)

-Dan

topsquark · #4 $Old$ October 25th, 2006, 12:19 PM

Okay, I'll give a hint. I didn't see the other solutions either until I tried TD's proof. There's a subtle hint in his proof that can lead to the extra solutions. His proof isn't flawed, it simply contains an unspoken assumption that he made about the solutions.

-Dan

TD! · #5 $Old$ October 25th, 2006, 12:23 PM

I'm not sure what you're referring to, but if it's the negative version of the non-trivial solutions I gave (which can't be found this way because I suggested taking the logarithm), then I'd like to point out that ThePerfectHacker mentioned a,b > 0. If you're talking about non-equal, positive integer solutions other than those I gave; then I'm interested

CaptainBlack · #6 $Old$ October 25th, 2006, 12:28 PM

Quote:

Originally Posted by TD! $View Post$

I'm not sure what you're referring to, but if it's the negative version of the non-trivial solutions I gave (which can't be found this way because I suggested taking the logarithm), then I'd like to point out that ThePerfectHacker mentioned a,b > 0. If you're talking about non-equal, positive integer solutions other than those I gave; then I'm interested

I think he is referring to the assumption that almost everyone made in
solving Question 1, and then when the assumption was pointed out
no one answered the more general question. If I'm making sense ...?

RonL

topsquark · #7 $Old$ October 25th, 2006, 12:42 PM

Quote:

Originally Posted by TD! $View Post$

I'm not sure what you're referring to, but if it's the negative version of the non-trivial solutions I gave (which can't be found this way because I suggested taking the logarithm), then I'd like to point out that ThePerfectHacker mentioned a,b > 0. If you're talking about non-equal, positive integer solutions other than those I gave; then I'm interested

Ah I missed the a, b > 0. I wonder why he excluded negative integers? Yes, that was why I mentioned your proof, because the domain of ln(x) is positive only.

-Dan

TD! · #8 $Old$ October 25th, 2006, 12:43 PM

Good, now I won't have to go crazy

topsquark · #9 $Old$ October 25th, 2006, 12:45 PM

Quote:

Originally Posted by CaptainBlack $View Post$

I think he is referring to the assumption that almost everyone made in
solving Question 1, and then when the assumption was pointed out
no one answered the more general question. If I'm making sense ...?

RonL

You are making sense, but in this case TPH specifically mentioned that a, b are integers. (Positive integers, though I didn't see that statement originally.)

-Dan

ThePerfectHacker · #10 $Old$ October 30th, 2006, 06:08 AM

This problem was developed by me, however I have been able to find it somewhere on the internet that says Euler solved it. Though I do not believe in that because it is way too simple for Euler, he would not concern himself with something like that. In addition, I discussed this with a professor and he said he liked to give this problem in his number theory class.

Unlike most diophantine equations this one solves nicely and easily. The reason why I am not in the favor of TD!'s method is because he is introducing the real numbers into this problem, while it only deals with positive integers, though I am sure you can make it work. The standard classical way to appraoch diophantine equation is to work with divisibility between these expressions, which is the approach that will be used here.
---
Let us assume that $a,b$ are postive integers such as,
$a^b=b^a$ .
By trichtonomy $a>b,a<b,a=b$ .
Let us first work with $a>b$ .
Let, $\gcd(a,b)=d$ , then we can write,
$a=a'd$ and $b=b'd$ where $\gcd(a',b')=1$ .
Thus, the diophantine equation becomes,
$(a'd)^b=(b'd)^a$ .
Thus, we have,
$(a')^bd^b=(b')^ad^a$
Thus, (note that $a>b$

)
$(a')^b=(b')^ad^{a-b}$
Note, the right hand side is divisible by $b'$ thus the left hand side is divisible by $b'$ .
But, $\gcd(a',b')=1$ thus, $b'=1$ .
This condition can only happen when,
$\gcd(a,b)=b$ .
That means $b|a$ if and only if $a=bk$ for some $k>1$ .
Substituting this into our diophantine equation,
$(bk)^b=b^{bk}$
Thus,
$(bk)^b=(b^k)^b$
If, $b=1$ then $a^b\not = b^a$ .
Thus, $b\not = 1$ thus,
$bk=b^k$
Dividing both sides by $b$ ,
$k=b^{k-1}$
~~~
It should seem clear the right hand side is larger then the left hand side because it is an exponential, but we need to determine when this inequality occurs.
~~~
If $b=2$ (since $b>1$ ) then,
$k=2^{k-1}$
Since $k>1$ we go through the values of $k$ and see what happens,
$2=2^1$
$3<2^2$
$3<2^4$
.....
We show by induction that the right hand side is larger for $k>2$ ,
$k<2^{k-1}$
$k+1<2k<2^k$
Thus, we have equality only for $k=2,b=2$ in that case, $a=bk=(2)(2)=4$ .

Now we turn to when $b=3$ we have for the first few values,
$2<3^1$
$3<3^2$
$4<3^3$
We show this is true by induction,
$k>2$
$k<3^{k-1}$
$k+1<3k<3^k$

Now it is reasonable to say for larger bases $b$ we can no way get equality. Thus, for $b\geq 3$ and $k>1$ we have,
$k<b^{k-1}$
$k+1<bk<b^k$ --->Proved.

These inequalities show that $a=4,b=2$ is the only possible solution. We need to check and we see that,
$4^2=2^4$ .
---
When, $a<b$ without loss of generality we have $a=2,b=4$ .
---
When, $a=b$ we have the trivial solutions.

o_O · #11 $Old$ August 30th, 2008, 02:05 PM

Here's an article devoted to this: Solving the Equation x^y = y^x

mr fantastic · #12 $Old$ August 30th, 2008, 11:03 PM

Quote:

Originally Posted by ThePerfectHacker $View Post$

This problem was developed by me, however I have been able to find it somewhere on the internet that says Euler solved it. Though I do not believe in that because it is way too simple for Euler, he would not concern himself with something like that. In addition, I discussed this with a professor and he said he liked to give this problem in his number theory class.

Unlike most diophantine equations this one solves nicely and easily. The reason why I am not in the favor of TD!'s method is because he is introducing the real numbers into this problem, while it only deals with positive integers, though I am sure you can make it work. The standard classical way to appraoch diophantine equation is to work with divisibility between these expressions, which is the approach that will be used here.
---
Let us assume that $a,b$ are postive integers such as,
$a^b=b^a$ .
By trichtonomy $a>b,a<b,a=b$ .
Let us first work with $a>b$ .
Let, $\gcd(a,b)=d$ , then we can write,
$a=a'd$ and $b=b'd$ where $\gcd(a',b')=1$ .
Thus, the diophantine equation becomes,
$(a'd)^b=(b'd)^a$ .
Thus, we have,
$(a')^bd^b=(b')^ad^a$
Thus, (note that $a>b$

)
$(a')^b=(b')^ad^{a-b}$
Note, the right hand side is divisible by $b'$ thus the left hand side is divisible by $b'$ .
But, $\gcd(a',b')=1$ thus, $b'=1$ .
This condition can only happen when,
$\gcd(a,b)=b$ .
That means $b|a$ if and only if $a=bk$ for some $k>1$ .
Substituting this into our diophantine equation,
$(bk)^b=b^{bk}$
Thus,
$(bk)^b=(b^k)^b$
If, $b=1$ then $a^b\not = b^a$ .
Thus, $b\not = 1$ thus,
$bk=b^k$
Dividing both sides by $b$ ,
$k=b^{k-1}$
~~~
It should seem clear the right hand side is larger then the left hand side because it is an exponential, but we need to determine when this inequality occurs.
~~~
If $b=2$ (since $b>1$ ) then,
$k=2^{k-1}$
Since $k>1$ we go through the values of $k$ and see what happens,
$2=2^1$
$3<2^2$
$3<2^4$
.....
We show by induction that the right hand side is larger for $k>2$ ,
$k<2^{k-1}$
$k+1<2k<2^k$
Thus, we have equality only for $k=2,b=2$ in that case, $a=bk=(2)(2)=4$ .

Now we turn to when $b=3$ we have for the first few values,
$2<3^1$
$3<3^2$
$4<3^3$
We show this is true by induction,
$k>2$
$k<3^{k-1}$
$k+1<3k<3^k$

Now it is reasonable to say for larger bases $b$ we can no way get equality. Thus, for $b\geq 3$ and $k>1$ we have,
$k<b^{k-1}$
$k+1<bk<b^k$ --->Proved.

These inequalities show that $a=4,b=2$ is the only possible solution. We need to check and we see that,
$4^2=2^4$ .
---
When, $a<b$ without loss of generality we have $a=2,b=4$ .
---
When, $a=b$ we have the trivial solutions.

For the record, the non-integer solutions can be written in parametric form as

$x = \left( 1 + \frac{1}{t} \right)^{t+1}$

$y = \left( 1 + \frac{1}{t} \right)^{t}$

The behaviour of this solution is straightforward for $t \leq -1$ and $t \geq 0$ .

When $-1 < t < 0$ the behaviour is not so straightforward.

chiph588@ · #13 $Old$ October 23rd, 2008, 10:19 PM

How did you obtain this?

mr fantastic · #14 $Old$ October 24th, 2008, 03:59 AM

Quote:

Originally Posted by mr fantastic $View Post$

For the record, the non-integer solutions can be written in parametric form as

$x = \left( 1 + \frac{1}{t} \right)^{t+1}$

$y = \left( 1 + \frac{1}{t} \right)^{t}$

The behaviour of this solution is straightforward for $t \leq -1$ and $t \geq 0$ .

When $-1 < t < 0$ the behaviour is not so straightforward.

Quote:

Originally Posted by chiph588@ $View Post$

How did you obtain this?

$x^y = y^x \Rightarrow \frac{\ln y}{y} = \frac{\ln x}{x}$ .... (1)

Try a parametric solution of the form

$x = (A + B)^C$ .... (2)

$y = (A + B)^D$ .... (3)

This form is chosen so that the ln terms in (1) cancel. A, B, C and D are unknown at this stage.

Substitute (2) and (3) into (1):

$\frac{C}{(A+B)^C} = \frac{D}{(A+B)^D} \Rightarrow \frac{C}{D} = (A+B)^{C-D}$

Choose C and D such that $C - D = 1 \Rightarrow C = D + 1$ .... (a)

$\Rightarrow \frac{D + 1}{D} = A + B$

$\Rightarrow \frac{1}{D} + 1 = A + B$ .... (4)

For equation (4) to be true choose

$\frac{1}{D} = A$ .... (b)

and

$B = 1$ .... (c)

Substitute (a), (b) and (c) into (2) and (3):

$x = \left(\frac{1}{D} + 1\right)^{D+1}$

$y = \left(\frac{1}{D} + 1\right)^{D}$

Replace D with t and there you have it. Non-rigorous but it works for me.

SimonM · #15 $Old$ October 24th, 2008, 04:38 AM

This is another way to prove that using different motivation.

Try $z = \frac{y}{z}$

Therefore $x^{zx} = (zx)^{x} \Leftrightarrow x^z = (zx) \Leftrightarrow x = z^{\frac{1}{z-1}} \, y = y^{\frac{z}{z-1}}$

Form which the answer follows.

The interesting thing is that mr_fantastic's parametrisation for integer D is the parametrisation for rational solutions