I am trying to parametrize $p$ such that $12(p+1/p)^2+18$ is a rational square. My attempt is as follows. Let $p+1/p=a/b$ where $a, b\in\mathbb{Z}$. Since $bp^2-ap+b=0$ should have rational roots, $a^2-4b^2$ must be a square. So we can parametrize $a=m+n, b=mn, m, n\in\mathbb{Z}$. Now, the problem becomes finding $(m, n)$ where $12(m+n)^2+18(mn)^2$ is an integer square. Can someone provide a parametrization or prove that there exists no such $(m,n)$? Thank you.
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COMMENT.-You do want to have $12(m+n)^2+18(mn)^2=z^2$ with integer $z\ne0$ which implies the diophantine $12X^2+18Y^2=Z^2$ and obviously also the equation of the type $ax^2+by^2+cz^2=0$, $$3X^2+2Y^2-Z^2=0\hspace 2cm(1)$$ in which $\boxed{-bc=-(2)(-1)=2}$ is not a quadratic residue modulo $a=3$. The consequence is that equation $(1)$ has not non-trivial integer solutions because of the following theorem (pag $46$ in Diophantine Equations, L.J. Mordell. Academic Press, 1969).
Theorem 3.- If $a,b,c$ are square-free and coprimes and do not have the same sign, then the equation $ax^2+by^2+cz^2=0$ has non trivial integer solutions if and only if $-bc$ is a quadratic residue of $a$ and so for every prime factor of $a$, i.e. $x^2+bc\equiv0\pmod a$ is solvable, and similarly for $-ca$ and $-ab$ and if $$ax^2+by^2+cz^2\equiv0\pmod 8$$ is solvable.
Ataulfo
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