Determine all possible pair(s) (X, Y) of positive integers that satisfy this equation.

                                        X^XX = Y^Y

Note: The order of calculation in X^XX is as given in this article.

(In reply to improved program by Charlie)

Rather than incrementing y by 1 to get to ever higher values needed for y, it's better to calculate the approximate needed y for a given x, and try the integer below and the integer above. That way, the following program goes up to x=800, needing values of y with over 2000 digits:

 10   point 16
 20   for X=2 to 800
 30    Lhs=(X^X)*log(X)
 40    Ly=100:PrevY=2:Y=2
 50    loop
 60      Y=abs(Lhs/log(Y))
 70      if abs(Y-PrevY)<0.0000000000000000000001 then goto *FoundOne
 80      PrevY=abs(Y)
 90    endloop
100   *FoundOne
110    Y1=int(Y):Y2=-int(-Y)
115    if abs(Lhs-Y1*log(Y1))<0.000000001 then print X,Y,Lhs,Y1*log(Y1)
116    if abs(Lhs-Y2*log(Y2))<0.000000001 then print X,Y,Lhs,Y2*log(Y2)
200   next

It still does not find any pairs beyond (1,1).

Posted by Charlie on 2009-03-18 15:50:28