đặt
http://dientuvietnam...x.cgi?gcd(m,n,p)=1
Nếu
http://dientuvietnam...mimetex.cgi?m=n thì
http://dientuvietnam...mimetex.cgi?x=y thỏa mãn.Trường hợp x, y nguyên thì chỉ có nghiệm (4, 2). Ta xét
http://dientuvietnam...mimetex.cgi?x>y và x, y không nguyên. khi đó
http://dientuvietnam...mimetex.cgi?m>n và
http://dientuvietnam...mimetex.cgi?p>1 .
Ta có:
http://dientuvietnam...tex.cgi?x^y=y^x http://dientuvietnam.net/cgi-bin/mimetex.cgi?(\dfrac{m}{p})^{\dfrac{n}{p}}=(\dfrac{n}{p})^{\dfrac{m}{p}}
http://dientuvietnam.net/cgi-bin/mimetex.cgi?(\dfrac{m}{p})^n=(\dfrac{n}{p})^m
http://dientuvietnam.net/cgi-bin/mimetex.cgi?m^np^m=n^mp^n
Đặt http://dientuvietnam.net/cgi-bin/mimetex.cgi?gcd(p,d)=1
thay vào ta có:
http://dientuvietnam.net/cgi-bin/mimetex.cgi?(ad)^np^m=(bd)^mp^n
http://dientuvietnam.net/cgi-bin/mimetex.cgi?a^np^{m-n}=b^md^{m-n}
suy ra http://dientuvietnam.net/cgi-bin/mimetex.cgi?d^{m-n} mà http://dientuvietnam.net/cgi-bin/mimetex.cgi?d=1
suy ra http://dientuvietnam.net/cgi-bin/mimetex.cgi?m=a;n=b
ta có http://dientuvietnam.net/cgi-bin/mimetex.cgi?a^np^{m-n}=b^m
mà http://dientuvietnam.net/cgi-bin/mimetex.cgi?gcd(a,b)=1 suy ra http://dientuvietnam.net/cgi-bin/mimetex.cgi?p^{m-n} chia hết cho http://dientuvietnam.net/cgi-bin/mimetex.cgi?b^m suy ra p chia hết cho b. Đặt http://dientuvietnam.net/cgi-bin/mimetex.cgi?p=kb ta có:
http://dientuvietnam.net/cgi-bin/mimetex.cgi?a^n(kb)^{m-n}=b^m suy ra http://dientuvietnam.net/cgi-bin/mimetex.cgi?a^nk^m=b^n
hay http://dientuvietnam.net/cgi-bin/mimetex.cgi?m^nk^m=n^n mấu thuẫn với http://dientuvietnam.net/cgi-bin/mimetex.cgi?m>n