Exercise 1

Posted by: Reetesh Mukul | March 10, 2010

Exercise 1

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This exercise is taken from Nicolas Bourbaki, Algebra-I, Chapter 1, §1, Ex. 12:

Exercise 1:

The only distinct natural numbers $\neq{\;\;0}$ which are permutable under the law $(x,\;y)\mapsto x^y$ are $2$ and $4$ .

Solution:

By no loss of generality we can assume that $y\;>\;x$ ; $y=zx$ and $z\;>\;1$ . Permutability requires,

$x^y=y^x$
$\Rightarrow x^y/x^x=y^x/x^x$
$\Rightarrow x^{y-x}=\;(y/x)^x=z^x$

This means $z$ must be a natural number as left side of above equation is always a natural number. Thus $x | y$ .

Next,

$x^{zx}=(zx)^x$
$\Rightarrow x^{(z-1)x}=z^x$

$\Rightarrow (1+(x-1))^{z-1}=z$

This we can expand using binomial theorem, to get,

$1\;+\;(z-1)(x-1)\;+\;{{z-1}\choose{2}}(x-1)^2\;+\;\ldots\;=\;z$

By rearranging,

$(z-1)(x-2)\;+\;{{z-1}\choose{2}}(x-1)^2\;+\;\ldots\;=\;0$

As $z\;>\;1$ and $x\;\geq \;2$ and each term on the side of addition in above equation is non-negative,the only way above equation will be satisfied when $x=2$ (so that first term, $(z-1)(x-2)$ becomes zero) and $z=2$ (so that every combination ${z-k}\choose{k+1}$ becomes zero).