let
x+y=a
xy=b
since the equation is symetric in x,y
we can assume WLOG x>=y and thus a>=b>=0
then
a^2=(x+y)^2=x^2+2xy+y^2
b^2=(xy)^2=x^22xy+y^2
a^2b^2=4xy
xy=(a^2b^2)/4
so we can transform the given equation to
a*b^n=(a^2b^2)/4
4ab^n=a^2b^2
a^24ab^nb^2=0
solving for a we get
a=(4b^n+sqrt(16b^(2n)+4b^2))/2
a=(4b^n+2bsqrt(4b^(2n2)+1))/2
a=2b^n+bsqrt(4b^(2n2)+1)
if n=1 we get
a=2b+b*sqrt(4+1)
a=2b+b*sqrt(5)
the only integer solutions for a,b is a=b=0 which gives
x=y=0 which violates positivity
so n>1
then we have
a=2b^n+b*sqrt(4b^(2n2)+1)
so for a to be a positive integer we need
4b^(2n2)+1 to be a perfect square
4b^(2n2)+1=m^2
m^24b^(2n2)=1
m^2(2b^(n1))^2=1
(m+2b^(n1))(m2b^(n1))=1
only integer solutions are when
m+2b^(n1)=m2b^(n1)=1
thus b=0 m=1
which again gives a=0 and x=y=0
thus there are no positive integer solutions to this problem

Posted by Daniel
on 20161021 07:58:47 