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?
 A: Here are some observations which are too extensive really for a comment, and which reduce the search space somewhat, and which may assist in establishing a contradiction.
You can rewrite your equation as $$(ab+1)^2=\frac 14+(a+b)^2$$
Since this is a rational equation, clearing denominators gives a primitive pythagorean triple $$p^2=q^2+r^2$$ with $ab+1=\cfrac p{2q}$ and $a+b=\cfrac r{2q}$ whence $a$ and $b$ are roots of the quadratic $$2qx^2-rx+p-2q=0$$
This has rational solutions only if its discriminant $r^2-8q(p-2q)$ is a square. Now $r^2=p^2-q^2$ so this becomes $$p^2-8qp+15q^2=(p-4q)^2-q^2=s^2$$ This gives us a second pythagorean triple, and also $r^2-s^2=8q(p-2q)$.
Note that $p-2q$ or $p-4q$ may be negative. 
To summarise - a solution implies two related pythagorean triples, the first of which can be taken primitive
$$p^2=q^2+r^2$$$$(p-4q)^2=q^2+s^2 \text{ or }(4q-p)^2=q^2+s^2$$

[continued]
The second of these is then also primitive, because $(p-4q,q)=(p,q)=1$
We know that a primitive pythagorean triple is of the form $m^2+n^2, 2mn, m^2-n^2$ with $m,n$ coprime and one even and one odd.
If $p=m^2+n^2$ and $4q-p=v^2+w^2$ then $4q=m^2+n^2+v^2+w^2$, but the right-hand side is the sum of two even squares and two odd squares, and is congruent to $2$ mod $4$. So this case is impossible.
So we must have $p=m^2+n^2$ and $p-4q=v^2+w^2\gt0$, and in passing we note that $p\gt 5q$ since $(p-4q)^2\gt q^2$
A: 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 $(x^2-4)(y^2-1)=1$.
