Solving an equation for positive integral values of variables.

Question

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.

Answer 1

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

$$x^3<x^3+5x^2-19x+21$$

Also $$x^3+5x^2-19x+21<x^3+6x^3+12x+8 =(x+2)^3$$

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 $x=y$...