PR 4 # 12
 

Find unequal rational numbers, a, b, ( other than 2 and 4 ) such that [tex]a^b = b^a[/tex]

[spoiler] There are an infinite number of such pairs. Give the general solution. [/spoiler]

general solution
 

[QUOTE=Wacky]Find unequal rational numbers, a, b, ( other than 2 and 4 ) such that [tex]a^b = b^a[/tex]

[spoiler] There are an infinite number of such pairs. Give the general solution. [/spoiler][/QUOTE]
:razz: [QUOTE=Mally][spoiler] Hint: Euler proved that 2 and the 4 are the only integers satisfying the eqn. So now give the general solution for rationals[/spoiler][/QUOTE] :flex:
Mally :coffee:

There is one more set of integers satisfying the equation: [spoiler]{-2, -4}[/spoiler]

+ve and -ve integers.
 

[QUOTE=retina]There is one more set of integers satisfying the equation: [spoiler]{-2, -4}[/spoiler][/QUOTE]
:yucky:

I got this info from the book 'Numbers' by David Wells who seldom makes a mistake in the data he presents.

It clearly says ' Euler showed that the only integer solution to a^b = b^a is 4^2 =2^4 =16.'
Unless Wells takes the answer 1/16 with the negative no.s to be a non integer solution. Possibly thats the correct interpretation.

However when it comes to rational numbers the -ve sign cancels out I would say, as is the case with (-2,-4) but I' not sure.

Mally :coffee:

[spoiler]
a = (1 + 1/n)[sup]n[/sup]
b = (1 + 1/n)[sup]n+1[/sup]
n can be any integer different from zero.
[/spoiler]

[quote]n can be any integer different from zero[/quote]Actually, n can be any integer except zero or minus one.