I'm having trouble understanding a part of the proof for the final theorem of this article https://www.cambridge.org/core/services/aop-cambridge-core/content/view/S0017089506003272.
The discriminant of $g(x)=x^4+px^2+qx+r$ is
$-27p^4-108p^3q-162p^2q^2-108pq^3-27q^4+256r^3$
We can rewrite $f(x)=x^4+ax^3+bx^2+cx+d = (x+\frac{a}{4})^4+(b-\frac{3}{8}a^2)x^2+(c-\frac{a^3}{16})x+(d-\frac{a^4}{256})$.
If we let
$p= b-\frac{3}{8}a^2$, $q=c-\frac{a^3}{16}$ and $r= d-\frac{a^4}{256}$ then the discriminant of $f$ is
$2^{24} \cdot(-27p^4-108p^3q-162p^2q^2-108pq^3-27q^4+256r^3)$
I don't understand why this is true since $f$ does not have the same shape as $g$ because of $(x+\frac{a}{4})^4$.
Also, where did the constant $2^{24}$ come from? Does it depend on the linear transformation $x \mapsto x+\frac{a}{4}$? Does it have to do with the fact that the discriminant is a homogeneous polynomial of degree $2n-2=2(4)-2=6$ in the coefficients of $f$?
The discriminant of $f$ is the same as the discriminant of $f(x-a/4)$. The discriminant is the product of differences of zeros of the polynomial, so doesn't change when these zeros are all translated by the same amount.
Dietmann does not assert that the formula with the $2^{24}$ factor is the discriminant. Rather that is a square integer and when you multiply it by the discriminant it cancels off all the powers of two in the denominators, yielding an expression (i) which is an integer if $a,\ldots,d$ are integers, and (ii) is a square iff the discriminant is a square.