Find all solutions to $x^2+x^2y^2+x^2y^4=525$ and $x+xy+xy^2= 35$

Find all solutions of the pair of equations $$x^2+x^2y^2+x^2y^4 = 525 \quad\text{and}\quad x+xy+xy^2=35$$

I got $(x,y)=(5,2)$, and $(x,y) = (20,1/2)$.

Is this correct?

1 Answer

The first equation can be factored as $$ 525=x^2(y^4+y^2+1)=x^2((y^2+1)^2-y^2)=x^2(y^2+y+1)(y^2-y+1)=35x(y^2-y+1) $$ so that $$ x(y^2-y+1)=15\implies xy=10. $$ Then $x+10y=25$, which makes $x$ and $10y$ the solutions of the quadratic equation $0=z^2-25z+100=\frac14[(2z-25)^2-225]=(z-20)(z-5).$ This gives exactly your two solutions.