Can we say that a Diophantine equation can have rational coefficients?

Question

Because when we eliminate the denominators of these rational numbers through the multiplication of LCM of denominators on both the sides of the equality then we get an equivalent Diophantine equation with integer coefficients.

Answer 1

For convenience, let us call "rational equation" (resp. diophantine equation) a polynomial equation with rational (resp. integral) coefficients and unknowns. It is true, as you say, that multiplication by an adequate common denominator transforms a rational equation into a diophantine one. The converse operation is also possible in particular cases, e.g. when the diophantine equation is homogeneous. But the diophantine equation is generally more difficult than the rational one. The main methodological point is that the two types of problems do not belong to the same mathematical domain - very roughly speaking, according to the number of variables, this is the difference between algebraic geometry and arithmetic geometry, or between Galois theory and algebraic number theory.

As a concrete illustration, let us consider the Fermat-Pell equation $x^2 - dy^2 = 1$. As is well known, the solutions of this diophantine equation are given by Dirichlet's unit theorem. But the rational equation is much less deep: it is equivalent to the equation $N(z) = 1$, where $z\in Q(\sqrt d)$, whose solutions are directly given by Hilbert's theorem 90 : $N(z) = 1$ iff $z$ is of the form $z = s(t)/t$, where $s$ is the automorphism of $Q(\sqrt d)$ sending $\sqrt d$ to $-\sqrt d$. So in this particular example, it makes no sense to transform "rational" into "diophantine".

It can happen, as noticed above, that the diophantine and rational problems are equivalent, in which case it makes more sense to study the rational one. Consider for instance the determination of the Pythorean triples, which amounts to solving the diophantine equation $x^2 + y^2 = z^2$, or equivalently the rational equation $x^2 + y^2 = 1$: here Hilbert's 90 applied to $Q(i)/Q$ gives immediately all the solutions. But this approach is not a cure-all panacea. Just think of FLT !