20170102, 15:18  #1 
Aug 2002
10000010011101_{2} Posts 
January 2017

20170215, 16:00  #2 
Aug 2002
3·11^{2}·23 Posts 

20170215, 16:32  #3 
"Robert Gerbicz"
Oct 2005
Hungary
2×3×11×23 Posts 
My sent solution with code(!):
x=(1+1/80)^80 y=(1+1/80)^81 the first and last ten digits of x: 2701484940, 6337890625 the first and last ten digits of y: 2735253502, 1142578125 For positive integer n value it is known that x=(1+1/n)^n and y=(1+1/n)^(n+1) gives a rational solution of the equation, because: x^y=((1+1/n)^n)^((1+1/n)^(n+1))=(1+1/n)^((n+1)^(n+1)/n^n) y^x=((1+1/n)^(n+1))^((1+1/n)^n)=(1+1/n)^((n+1)^(n+1)/n^n) hence x^y=y^x. If we want x, where we can see the last eight (or some) digits then n=2^a*5^b for some nonnegative a,b integers. Restricting the search for a,b<10 we have found 3 solutions, the simple PARIGP solution, using only integer arithmetic: for(a=0,9,for(b=0,9,n=2^a*5^b;e=max(a,b);M=10^8;\ r=Mod(n+1,M)^n*Mod(2,M)^(n*(ea))*Mod(5,M)^(n*(eb));\ r=lift(r);test=1;v=vector(10,i,0);for(i=1,8,\ res=1+(r%10);r\=10;v[res]+=1;if(v[res]>1,test=0));\ if(test,X=[a,b,n];print([a,b,n])))) what has found n=80;3125;1953125 as a solution, this gives for example that x=(1+1/80)^80 and y=(1+1/80)^81 a solution. A "cut" function used to get the first and last L digits of a positive rational number: cut(r,L)={tmp=r;d=10^L;while(type(tmp)!="t_INT",tmp*=10);\ t=tmp;while(tmp>=d,tmp\=10);return([tmp,t%d])} In the second email: Just for small correction obvioulsy x starts as 2.701484940 and y with 2.735253502 if you needed also the first (ten) digits with a decimal point. ps. use cut() only if r>0 has a finite representation in base 10. 
20170217, 01:20  #4 
Jan 2017
78_{16} Posts 
It's pretty easy to directly show starting from the question rationals of that form are answers and are the only rational answer:
x^y = y^x Let 'a' be such that x^a = y, and replace y by that in above x ^ x ^ a = x ^ (a*x) Take xbased logarithm x ^ a = a*x Since x^a = y, this means a = y/x, thus rational too Let 'b' be such that a = x^b x ^ x ^ b = x^b * x Take xbased logarithm x^b = b + 1 x^b equals a, thus b = a1 and rational Let x = p/q, b = n/m where p,q,n,m integers (p/q)^(n/m) = (n+m)/m If the value on the left is rational (as per above it equals a and must be rational for a valid solution), it must be n:th power of one. Thus the value on the right must be too. So (n+m) and m are n:th powers. But it's not possible for two integers n apart (m and n+m) to both be n:th powers unless n=1. Thus: (p/q)^(1/m) = (1+m)/m p/q = ((1+m)/m)^m And this gives a solution for any integer m. Last fiddled with by uau on 20170217 at 01:20 
20170222, 21:34  #5 
"Robert Gerbicz"
Oct 2005
Hungary
2×3×11×23 Posts 
A longer article on this topic: http://www.komal.hu/cikkek/loczy/pow...mpower.e.shtml in English, but the original was in Hungarian from 2000 in Kömal (http://db.komal.hu/KomalHU/showpdf.p...Cikk&id=200047).

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