let
x+y=a
x-y=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=(x-y)^2=x^2-2xy+y^2
a^2-b^2=4xy
xy=(a^2-b^2)/4
so we can transform the given equation to
a*b^n=(a^2-b^2)/4
4ab^n=a^2-b^2
a^2-4ab^n-b^2=0
solving for a we get
a=(4b^n+-sqrt(16b^(2n)+4b^2))/2
a=(4b^n+-2bsqrt(4b^(2n-2)+1))/2
a=2b^n+-bsqrt(4b^(2n-2)+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^(2n-2)+1)
so for a to be a positive integer we need
4b^(2n-2)+1 to be a perfect square
4b^(2n-2)+1=m^2
m^2-4b^(2n-2)=1
m^2-(2b^(n-1))^2=1
(m+2b^(n-1))(m-2b^(n-1))=1
only integer solutions are when
m+2b^(n-1)=m-2b^(n-1)=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
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Posted by Daniel
on 2016-10-21 07:58:47 |
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