Birationally transform over $\mathbb{Q}$ the curve $Y^2=\frac{1}{3}X^3+4X^2+1$ to a curve of the form $Y^2=X^3+AX+B$, where $A$ and $B$ are integers.
I know that over $\mathbb{Q}$, we can birationally transform a curve of the form $y^2=$ cubic in $x$ to a curve of the form $y^2=x^3+Ax+B$, for integers $A,B$, using only transformations of the form $(x,y)\mapsto (ax+b,cy)$. So the goal is to find a suitable map. For example, by looking at the coefficient of $X^3$, I think we could take $(x,y)\mapsto (3x,y)$ or something similar? But I can't figure out what the right mapping is. Any ideas?
With $(a,b,c,d)$ integer, let's plug $\begin{cases}X=\frac ad x+\frac bd\\Y=\frac cd y\end{cases}\quad $ in $\ Y^2-(\frac 13X^3+4X^2+1)=0$
After reduction to the same denominator we get
$$-a^3x^3+(-12a^2d-3a^2b)x^2+(-24adb-3ab^2)x+3dc^2y^2-(12db^2+b^3+3d^3) = 0$$
First we eliminate the term in $x^2$ by setting $b=-4d$
We get $\ 3dc^2y^2-a^3x^3+48d^2ax-131d^3 = 0\ $ which is normalized to
$$y^2 = \frac{a^3}{3dc^2}\,x^3 - \frac{16ad}{c^2}\,x + \frac{131d^2}{3c^2}$$
And we try to make integer coefficients.
Let see if we can find something for simple choice $c=1$
$a$ must be multiple of $3$ (forced by $x^3$ coefficient).
But then $\frac{a^3}{3d}=\frac{27\alpha^3}{3d}=\frac{9\alpha^3}{d}$ so $d$ multiple of $9$ works and that clears also the constant coefficient.
So we can just take $a=3,b=-4,c=1,d=9$