Triple-coincidence non-occurrence under single-piece extension

Counterexamples at every original magic board. In each construction below, exactly one new piece is adjoined. Its fixed move set is the disjoint union of several fairy-leaper move sets; the constituent leapers are not separately added to the alphabet. The source explicitly permits such compound pieces.

Let $L_{a,b}$ denote the leaper with all sign changes and coordinate exchanges of displacement $(a,b)$. Counting starting squares for each displacement gives, whenever $n\ge b$,

$$T_{L_{a,b}}(n)= \begin{cases} 4n(n-b),&a=0,\\ 4(n-a)^2,&a=b>0,\\ 8(n-a)(n-b),&0<a<b. \end{cases}$$

Different leaper families have disjoint displacement sets, so their totals add.

Board $n=6$. Adjoin the single compound piece

$$P_6=L_{1,1}\cup L_{1,2}\cup L_{1,3}.$$

Its exact polynomial, valid for every $n\ge4$, is

$$T_{P_6}(n)=4(n-1)^2+8(n-1)(n-2)+8(n-1)(n-3) =4(n-1)(5n-11).$$

The source's existing Centaur and Archbishop satisfy

$$T_{P_6}(6)=T_{\mathrm{Centaur}}(6) =T_{\mathrm{Archbishop}}(6)=380.$$

Board $n=8$. Independently adjoin the single compound piece

$$P_8=L_{0,1}\cup L_{0,2}\cup L_{1,2}\cup L_{2,2}.$$

Its exact polynomial is

$$T_{P_8}(n)=20n^2-52n+32,$$

and

$$T_{P_8}(8)=T_{\mathrm{Rook}}(8) =T_{\mathrm{Archbishop}}(8)=896.$$

Board $n=12$. Independently adjoin

$$P_{12}=L_{0,3}\cup L_{3,3}\cup L_{4,4}.$$

Its exact polynomial, again valid on every prescribed board $n\ge4$, is

$$T_{P_{12}}(n)=4n(n-3)+4(n-3)^2+4(n-4)^2 =12n^2-68n+100.$$

Using the source's Bishop convention,

$$T_{P_{12}}(12)=T_{\mathrm{King}}(12) =T_{\mathrm{Bishop}}(12)=1012.$$

Each additional piece has one fixed finite move set, is distinct from every original piece, and has a degree-two move-count polynomial. Since all strengths use the same denominator $n^2(n^2-1)$, equal move totals give equal strengths. Thus Conjecture 9.2 fails separately at all three original magic board sizes $6,8,12$, always with exactly one added piece.

For completeness, the adjacent Conjecture 9.1 also fails under its stated extension rules: adjoin just the explicitly permitted Wazir $W=L_{0,1}$. Then

$$T_W(n)=4n(n-1),\qquad T_{\mathrm{Rook}}(n)=2n^2(n-1) =\frac n2 T_W(n),$$

giving a second nonconstant linear proportionality, distinct from the Bishop--King proportionality.

Source: Feys, The Arithmetic of Chess Piece Strength on the $n\times n$ Board, Conjectures 9.1 and 9.2.