Substitution in diophantine equation

Question

If I have an equation $$3(1+x+x^2)(1+y+y^2)+1=4x^2y^2 $$ and I am interested in non negative integer solutions. I let $x$ be the smaller of the positive integer solution so I substitute $x+\lambda=y$ to get $$3(1+x+x^2)(1+x+\lambda+(x+\lambda)^2)+1=4x^2(x+\lambda)^2$$ I do this to apply Descartes Rule of signs to see how many non negative integer solutions exist only in terms of x. So I expand the equations and combine like terms and I know $\lambda>0$ so Descartes Rule of signs tells me only one non negative integer solution exists because theres only on change of signs. Am i right or could a change in $\lambda$ change my solution $x$ ?

Answer 1

Since $(4,64)$ is one solution then $4$ is one positive solution in terms of $x$. Check can $x$ have another solution with the only condition being that if we rewrite $y$ in terms of $x+ \lambda$ with $ \lambda \in \mathbb N$. If this equation had 2 non negative solutions for x with $\lambda \in \mathbb N$ then looking at the expanded equation: $$-\lambda^2 x^2 + 3 \lambda^2 x + 3 \lambda^2 - 2 \lambda x^3 + 9 \lambda x^2 + 9 \lambda x + 3 \lambda - x^4 + 6 x^3 + 9 x^2 + 6 x + 4 $$ or $$-x^4+(-2\lambda +6)x^3+\lambda(9-\lambda)x^2+(9\lambda+6+3\lambda^2)x+(4+3\lambda+3\lambda^2) $$ so it is easy to see that there is only one sign change so only one positive solution exists for x with $\lambda \in \mathbb N$

Answer 2

Here is a picture of the curve in the first quadrant. There are horizontal and vertical asymptotes at $\frac{3+ \sqrt{21}}{2} \approx 3.79.$ The point where $x=y$ is at about $7.346.$ Therefore, the smaller integer coordinate can be only $4,5,6,7$ and these can be simply checked. Apparently $(4,64)$ is on the graph