Description: Given the following parametric quartic polynomial 


$y^4 - 28 z y^3 - 14 (656 - 328 z + 83 z^2) y^2 +  4 z (-20464 + 10232 z + 3409 z^2) y +  91 (62208 - 62208 z + 41504 z^2 - 12976 z^3 + 675 z^4)$ 


how do I find all possible rational values of $z$ such that this polynomial is reducible.


I am performing Galois group classification on a family of polynomials of which this example is one case. Rational searching has identified the following values of $z$ that results in factorization: $z \in \left\{{-44,-2,1,\frac{44}{23},\frac{67}{19}, 4, \frac{134}{29}}\right\}$.


I do not know if there is an infinite or finite number of such points (I suspect that there is a finite number). Reposing the question: what is the procedure by which I can generate all know points $z$ in a systematic manner.



Find all possible rational values of a parametric quartic such that it is reducible
I essentially follow Igor Rivin's approach here. There are two possibilities for
how the polynomial can factor: (degree 1)$\times$(degree 3) or
(degree 2)$\times$(degree 2). Writing
$$ (y + a)(y^3 + by^2 + cy + d) = f $$ or
$$ (y^2 + ay + b)(y^2 + cy + d) = f, $$
where $f$ is your polynomial, and equating coefficents, we obtain two curves
$C\_1$ and $C\_2$
in the 5-dimensional affine space with coordinates $a,b,c,d,z$, and what you
want is the $z$-coordinates of their rational points.


**EDIT:** Actually, there is a somewhat simpler approach. $C\_1$ classifies
linear factors of $f$, which is equivalent to roots of $f$, so $C\_1$
is simply the affine plane curve $f(y,z) = 0$. Its projective closure
is a smooth quartic curve, so the genus of $C\_1$ is 3. Alternatively,
since the discriminant of $f$ as a polynomial in $y$ is a degree 12
polynomial in $z$ without multiple roots, it follows from Riemann-Hurwitz
that the genus of $C\_1$ is 3.


One finds 16 rational points on it; their $z$-coordinates
are those you also found (three of them ($z = -2, 1, 4$) give a factorization
into four linear factors, so they each give rise to four points on $C\_1$,
the other four give only one linear factor).


**EDIT:**
$C\_2$ classifies unordered pairs of (distinct) roots, so it can be obtained
as a degree 6 cover of the affine $z$-line by expressing the coefficients
of the polynomial whose roots are the products (or sums, or any other
symmetric expression) of pairs of roots in terms of those of the original
polynomial. This ramifies in 24 points (two above each branch point
of $C\_1 \to \mathbb P^1$), so by Riemann-Hurwitz its genus is 7.


One finds 18 rational points on it; they all come from $z = -2, 1, 4$
by writing the product of four linear factors as a product of two quadratics
(which can be done in six different ways).


Since both curves have genus $\ge 2$, they have only finitely many rational
points by Faltings's Theorem. So there are only finitely many $z \in \mathbb Q$
such that your polynomial factors. It is also very likely that there are no
further rational points on the curves than those mentioned above (points on such
curves are usually fairly small, so one is likely to find them all by search),
but this is also very likely very hard to prove. But for all practical
purposes, you can safely assume that the values you found constitute the
complete list.


