<s>[INST] Solve $(a^2-1)(b^2-1)=\frac{1}4 ,a,b\in \mathbb Q$ Does the equation $(a^2-1)(b^2-1)=\dfrac{1}4$ have solutions $a,b\in \mathbb Q$?
I search $0<p<1000,0<q<1000$, where $a=\dfrac{p}q$, but  no solutions exist. I wonder  is this equation solvable?
 [/INST] If there are rational points on your curve than there are rational points on the curve:
$$ (x^2-4)(y^2-1)=1, $$
so $(y^2-1)=\frac{p}{q}$ and $(x^2-4)=\frac{q}{p}$, with $(p,q)=1$, must hold.
This implies that both $\frac{p+q}{q}$ and $\frac{4p+q}{p}$ are squares of rational numbers, so $p=u^2,q=v^2$ and both $u^2+v^2$ and $4u^2+v^2$ are squares, say $u^2+v^2=q_1^2$ and $(2u)^2+v^2=q_2^2$. By the parametrization of primitive pythagorean triples, the second identity implies that $u=ab$ and $v=a^2-b^2$, with $(a,b)=1$ and $a,b$ not both odd. By substituting into the first identity we get that
$$ (ab)^2 + (a^2-b^2)^2 = (a^2+b^2)^2-3(ab)^2 = \frac{1}{4}(a^2+b^2)^2+\frac{3}{4}(a^2-b^2)^2$$
must be a square. So we started with $q_2^2=4p+q$ that belonged to both the quadratic forms $x^2+y^2$ (since $4p+q=(2u)^2+v^2$) and $x^2+3y^2$ (since $4p+q=q_1^2+3u^2$) and get that $q_2$ belong to the intersection of the integers represented by the same quadratic forms: this leads to an infinite descent and the impossibility to find rational points on the curve $(x^2-4)(y^2-1)=1$.
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