Question 3 - Page 2

mr fantastic · #11 $Old$ 08-30-2008, 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.
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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,
$4^2=2^4$ .
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When, $a<b$ without loss of generality we have $a=2,b=4$ .
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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@ · #12 $Old$ 10-23-2008, 10:19 PM

How did you obtain this?

mr fantastic · #13 $Old$ 10-24-2008, 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 · #14 $Old$ 10-24-2008, 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