Solving an equation for positive integral values of variables. Question > Solve > $$y^3+3y^2+3y=x^3+5x^2-19x+20$$ > for positive integers $x$ and $y$. My approach: > I factorised the equation as > $(y-x)(x^2+y^2+3x+3y+3)=2[(x-1)(x-10)]$ > and got two ordered pairs of $(x,y)$ i.e. $(1,1);(10,10)$ I can't go further. Thanks for your help. PS: Please consider this question without my approach also, because it may mislead you.

Write $$(y+1)^3= y^3+3y^2+3y+1=x^3+5x^2-19x+21$$ now, since a discriminant of $5x^2-19x+21$ is $19^2-20\cdot 21<0$ we have

$$x^3
Also $$x^3+5x^2-19x+21
since $x^2+31x-13>0$ for positive $x$. So $$x^3<(y+1)^3<(x+2)^3$$ which is only possible iff $y+1=x+1$ so $x=y$...