原方程可化为(4x-2y+1)^2+(2y-3)^2=10.因仅有整数（±1)^2+(±3)^2=10.故由此可得8个方程组：{4x-2y+1=3,2y-3=1},{4x-2y+1=3,2y-3=-1.}...{4x-2y+1=-1,2y-3=3}.(注：若令4x-2y+1=m,2y-3=n.8个方程组为{m=3,n=1,},{m=...