I assumed we already knew about that too. It turns out that an irregular integer-degree bicyclic pentagon is also nontrivial; if we specify two integer-degree angles (which is sufficient to define the shape of the bicyclic pentagon) we generally do not get integer degrees at the other angles. If, for instance, the two equal adjacent angles are set to 90° instead of the 80° that actually works, we get this.

Curious where you get the pentagonal construction. I get a cubic equation for the inradius and it is difficult to see where it may become reducible.

We certainly have $f_3(3)\ge12$ with the regular icosahedron.

@KConrad Maybe if $p=2$ then $p=8$. Squares prime to $p$ for most $p$-adics are just quadratic residues $\bmod p$, but for $p=2$ you need a quadratic residue $\bmod8$ (which would be $1$).

@MarkFischler I am not entirely sure that $pp(3)$ has no solutions. If you allow negative arguments then $p(-1)+p(0)+p(1)=p(-2)$.

Also for $a=-1$, that being a unique case for no positive integers.

Make that $\mathbb{Q}[\sqrt{-30}]$.

In some cases tighter bounds on $\pi$ enter the reckoning. For example, $\pi^2>480/49$ enters into rendering the class group of $\mathbb{Q}[\sqrt{30}]$.