this has been confusing me for some time.

can someone help me find each possible pair?
thank you in advance

you can start with x = 2013 and y = 2013.

then increase/desrease either x or y and decrease/increase the corresponding var

This post was edited by carteblanche on Nov 9 2013 10:02pm

First notice that : 2013 = 3 * 11 * 61

So : 2013^2013 = 2013^( 3*11*61 )

Since 2013 is not a power, your x must be 2013.

No you can use the rule : A^(B*C) = (A^B)^C

and see how you can reform the initial power : 2013^( 3*11*61 ) = ( 2013^3) ^ (11*61) = ( 2013^11) ^ ( 3*61) = ...

You should find 2^3 = 8 different solutions (with y a natural number).

replace the bolded with
"since 2013 is not a power, x has to be 2013 or a multiple thereof for x and y to be natural numbers"
just as clarification, otherwise the post is correct

This post was edited by brmv on Nov 10 2013 06:33am

or a power of 2013 !

Thank you for the clarification

so would the 8 be

2013^(3*11*61)
2013^3^(11*61)
2013^11^(3*61)
2013^61^(11*3)
what would the other 4 be?
2013^(3*11)^61?
2013^(3*61)^11?
2013^(11*61)^3?
2013^2013?

This post was edited by lolcatslol on Nov 10 2013 01:37pm

apart from the notation you are correct,
notice how 'feanur' wrote ( 2013^3) ^ (11*61) and not 2013^3^(11*61), so it is

2013^2013
[2013^3]^(11*61)
...
[2013^(11*61)]^3
[2013^2013]^1

and you have the pairs you are looking for

alright thank you.


thank you




thank you