Use the stated rational change of variables
$$
\widetilde x=-\frac6{v+w},
\qquad
\widetilde y=\frac{3(v-w)}{v+w}.
$$
Direct substitution turns the anomaly equation into
$$
\frac{216(v^3+w^3-1)}{(v+w)^3}=0,
\qquad\text{so}\qquad v^3+w^3=1.
$$
Writing $v=a/c$ and $w=b/c$ in lowest common denominator gives the integer equation $a^3+b^3=c^3$. The supplied special case of Fermat's Last Theorem says that one of $a,b$ must vanish. Thus $(v,w)=(1,0)$ or $(0,1)$, which gives
$$
\widetilde x=-6,
\qquad
\widetilde y=\pm3.
$$
The two signs merely exchange the names of the up- and down-type singlets. Choosing $q=1$ and the conventional sign $y=3$ gives the unique assignment up to overall scaling and that exchange:
$$
\boxed{l=-3,\qquad u=4,\qquad d=-2,\qquad x=-6}.
$$
These are six times the conventional Standard Model <hypercharges>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 305 / 3 / v / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-305/3/v/solution