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?
$\endgroup$ 11 Answer
$\begingroup$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.
$\endgroup$
