[求助] 一道难题，求解

x^2+2ay=5 => 3x^2+6ay=15
将6a=y-x代入，3x^2+y(y-x)=15 => 3x^2-xy+y^2=15 => (y-x/2)^2+11/4x^2=15
=>(y-x/2)^2=15-11/4x^2>=0 x是正整数，因此1<=x<=2
(1)x=1 15-11/4x^2=49/4 此时y=4 a=1/2
(2)x=2 15-11/4x^2=4 此时y=3 a=1/6.