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7. Algebraic-Geometric Synthesis: Minimal Polynomial of $\varepsilon_{21}$
7.1. Minimal Polynomial of $\varepsilon_{21}$ has Discriminant $21$
The minimal polynomial of $\varepsilon_{21}$ over $\mathbb{Q}$ is
$$m(x) = x^{2} - 5,x + 1.$$Proof. Since $\varepsilon_{21} = (5 + \sqrt{21})/2$ and its conjugate is $\bar{\varepsilon} = (5 - \sqrt{21})/2$, we have
$$\varepsilon_{21} + \bar{\varepsilon} = 5, \quad \varepsilon_{21} \cdot \bar{\varepsilon} = (25 - 21)/4 = 1.$$
Therefore $\varepsilon_{21}$ satisfies $x^{2} - 5x + 1 = 0$ by Vieta's formulas. $\blacksquare$
The discriminant of this polynomial is
$$\Delta(m) = b^{2} - 4 a c = 5^{2} - 4\cdot 1\cdot 1 = 25 - 4 = \mathbf{21}.$$
Discriminant identity. This is not a coincidence: by the relationship between minimal polynomial discriminant and field discriminant for quadratic fields $\mathbb{Q}(\sqrt{D})$ with $D$ squarefree (fundamental discriminant),
$$\Delta(m) = D \quad \text{if } D \equiv 1 \pmod 4,$$$$\Delta(m) = 4D \quad \text{if } D \equiv 2 \text{ or } 3 \pmod 4.$$
Since the fundamental discriminant of $\mathbb{Q}(\sqrt{21})$ is $d = 21 \equiv 1 \pmod 4$ (squarefree, and congruent to $1$ modulo $4$), the minimal polynomial discriminant equals $21$ identically.
Trace recurrence. The sequence $T_n = \varepsilon_{21}^{n} + \bar{\varepsilon}^{\,n}$ satisfies
$$T_n = (\varepsilon_{21} + \bar{\varepsilon}) T_{n-1} - \varepsilon_{21} \bar{\varepsilon} , T_{n-2} = 5,T_{n-1} - T_{n-2},$$
with $T_0 = 2$ and $T_1 = 5$. The first few values are
$$2,\ 5,\ 23,\ 110,\ 527,\ 2525,\ 12098,\ 57965,\ \ldots$$
This is OEIS A005247, the Pell-like sequence at $5$, with closed-form
$$T_n = \frac{(5+\sqrt{21})^{n} + (5-\sqrt{21})^{n}}{2^{n-1}}.$$
The value $T_3 = 110 = 2 \cdot 55$ recovers the fundamental Pell solution $(55, 12)$ of $x^{2} - 21 y^{2} = 1$ through $T_3 / 2 = 55$ and $\sqrt{T_3^{2} - 4}/(2\sqrt{21}) = \sqrt{12100 - 4}/(2\sqrt{21}) = \sqrt{12096}/(2\sqrt{21}) = \sqrt{576}/2 = 12$.
7.2. Grassmannian Dimensions Equal to $21$
The Plücker embedding realizes the Grassmannian$\mathrm{Gr}(k, n)$ as a subvariety of $\mathbb{P}^{\binom{n}{k}-1}$. The dimension of $\mathrm{Gr}(k, n)$ over any field is
$$\dim \mathrm{Gr}(k, n) = k (n - k).$$
The diophantine equation $k (n - k) = 21$ has positive integer solutions:
$(k, n)$
$\mathrm{Gr}(k, n)$
Geometric realization
$(1, 22)$
$\mathrm{Gr}(1, 22) = \mathbb{P}^{21}$
Lines in $22$-space
$(3, 10)$
$\mathrm{Gr}(3, 10)$
$3$-planes in $10$-space
$(7, 10)$
$\mathrm{Gr}(7, 10)$
$7$-planes in $10$-space (dual of Gr$(3, 10)$)
$(21, 22)$
$\mathrm{Gr}(21, 22) = \mathbb{P}^{21}$
$21$-planes in $22$-space
The Plücker embedding of $\mathrm{Gr}(3, 10)$ is the quadric hypersurface $\mathbb{P}^{119}$ defined by the Grassmann-Plücker relations on the $\binom{10}{3} = 120$ Plücker coordinates. The four Grassmannians of dimension $21$ therefore sit inside the same $119$-dimensional projective space. Their intersections reflect the arithmetic of $21$.
Two distinct simple Lie algebras carry the integer $21$:
$\dim \mathfrak{so}(7) = \frac{7 \cdot 6}{2} = 21$ (orthogonal group). The Lie algebra $\mathfrak{so}(7)$ is the tangent space at the identity of the spin group $\mathrm{Spin}(7)$. It is one of only two classical Lie algebras, together with $\mathfrak{so}(8)$, that admit triality. The $8$-dimensional spin representation of $\mathrm{Spin}(7)$ realizes the octonions$\mathbb{O}$ as the imaginary octonions.
$\lvert\Phi^{+}(A_{6})\rvert = \frac{6 \cdot 7}{2} = 21$ (root system). The number of positive roots of the $A_{6}$ root system equals $\binom{7}{2}$.
The identity $\binom{n}{2} = 21$ at $n = 7$ is the shared arithmetic: $T(6) = \binom{7}{2} = 21$ and $\lvert\Phi^{+}(A_{6})\rvert = \binom{7}{2}$. This binomial structure appears on both sides of the bridge between the central node $T(6)$ and the Lie-theoretic dimension $21$.
7.4. The Harshad Property of $21$
A Harshad number, also called a Niven number, is an integer divisible by the sum of its digits (Harshad number). For $n = 21$:
$$\sigma_1(21) = 2 + 1 = 3, \qquad 21 = 3 \cdot 7.$$
So $21$ is Harshad. The Harshad numbers below $100$ form the set
$${1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 18, 20, \mathbf{21}, 24, 27, 36, 40, 42, 45, 54, 60, 72, 81, 84, 90}.$$
Among the Fibonacci-triangular semiprimes ${1, 21, 55}$, the integer $21$ is the largest (and only non-trivial) Harshad semiprime of this class. $55$ is not Harshad because $5+5 = 10$ does not divide $55$.
7.5. Cubic Residues Modulo $21$
The multiplicative group $(\mathbb{Z}/21\mathbb{Z})^{\times}$ has order $\varphi(21) = 12$ and group structure $C_2 \times C_6$ (multiplicative group of integers modulo n and Euler's totient function). The cubing endomorphism $\varphi: x \mapsto x^{3}$ has image of order $12 / \gcd(12, 3) = 4$. Direct verification:
$$1^3 \equiv 1,\quad 2^3 \equiv 8,\quad 3^3 \equiv 6,\quad 4^3 \equiv 1 \pmod{21},$$$$5^3 \equiv 20,\quad 8^3 \equiv 8,\quad 10^3 \equiv 13 \pmod{21}.$$
The cubic residue set is
$${1, 8, 13, 20},$$
a cyclic subgroup of order $4$ with generator $8$, because $8^{1} = 8$, $8^{2} = 64 \equiv 1$, and $8^{3} \equiv 8$ modulo $21$. The structure of cubic residues modulo $21$ is the canonical obstruction to a cube root lift from $\mathbb{Z}/21\mathbb{Z}$ to $\mathbb{Z}$, and the fact that the cubes ${1, 8}$ alone are $\pm 1$ in $(\mathbb{Z}/7\mathbb{Z})^{\times}$ extends the residue system across the prime factor $7 \equiv 3 \pmod 4$.
8. The Fano Plane and Number-Theoretic Properties
8.1. The Fano Plane $\mathrm{PG}(2, 2)$ has $21$ Incidences
The Fano plane $\mathrm{PG}(2, 2)$ is the projective plane of order $2$, realized over the field $\mathbb{F}_{2}$ (Fano plane). Its incidence structure is governed by:
Points: $7$ (the non-zero vectors of $\mathbb{F}_{2}^{3}$ up to scalar).
Lines: $7$ (each line is a $2$-dimensional subspace of $\mathbb{F}_{2}^{3}$, hence has $3$ points).
Incidence: $21$ (each point lies on $3$ lines, each line contains $3$ points).
The incidence count is
$$\text{number of incidences} = 7 \cdot 3 = 21.$$
The Fano plane is self-dual, so the dual calculation gives the same value.
Connection to $\binom{7}{2}$. The Steiner triple system $S(2, 3, 7)$ on the Fano plane partitions the $\binom{7}{2} = 21$ unordered pairs of $7$ points into $7$ triples (Steiner system). Therefore $21$ counts both the pairs and the triples, and each triple is a line.
Automorphism group:
$$\mathrm{Aut}(\mathrm{PG}(2, 2)) \cong \mathrm{PGL}(3, 2) \cong \mathrm{PSL}(2, 7),$$
with order $168 = 2^{3} \cdot 3 \cdot 7$. This is the smallest non-abelian simple group beyond $A_{5}$, and the coincidence with the prime factorization $2^{3} \cdot 3 \cdot 7$ shares the prime factors $3$ and $7$ with $21 = 3 \cdot 7$.
8.2. The Smarandache Function $S(21) = 7$
The Smarandache function $S(n)$, also called the Kempner function, is the smallest positive integer $m$ such that $n \mid m!$ (Kempner function). For $n = 21 = 3 \cdot 7$, the smallest $m$ such that both $3 \mid m!$ and $7 \mid m!$ is $m = 7$ (since $3 \mid 6!$ and $7 \mid 7!$). Therefore
$$S(21) = 7.$$Verification: $7! = 5040$ and $5040 / 21 = 240$, which is an integer. By contrast, $6! = 720$ and $720 / 21$ is not an integer.
The Kempner function satisfies $S(p) = p$ for a prime $p$, and $S(pq) = \max(p, q)$ for distinct primes $p < q$. For $21 = 3 \cdot 7$, that rule gives $S(21) = 7$.
8.3. The Dedekind $\psi$ Function Equals $\sigma$ at $21$
The Dedekind psi function is defined by
$$\psi(n) = n \prod_{p \mid n}!\left(1 + \tfrac{1}{p}\right).$$
For squarefree $n = p_{1} p_{2} \cdots p_{r}$:
$$\sigma(n) = \prod_{i=1}^{r}(1 + p_{i}) = n \prod_{i=1}^{r}!\left(1 + \tfrac{1}{p_{i}}\right) = \psi(n).$$
For $n = 21 = 3 \cdot 7$:
$$\psi(21) = 21 \cdot \frac{4}{3} \cdot \frac{8}{7} = \frac{21 \cdot 32}{21} = 32 = \sigma(21).$$
The factorization $\psi(21) = 32 = 2^{5}$ is a pure power of $2$, reflecting the relations $1 + 3 = 4 = 2^{2}$ and $1 + 7 = 8 = 2^{3}$.
By Jacobi's four-square theorem, the number of ordered representations of $n$ as a sum of four squares is
$$r_{4}(n) = 8 \sum_{d \mid n,, 4 \nmid d} d.$$
For $n = 21$, divisors are $1, 3, 7, 21$, all coprime to $4$, so
$$\sum_{d \mid 21,, 4 \nmid d} d = 1 + 3 + 7 + 21 = 32.$$
Therefore
$$r_{4}(21) = 8 \cdot 32 = 256 = 4^{4}.$$
The integer $21$ has exactly $256$ ordered representations as a sum of four squares. Symmetry note: by Jacobi's formula $r_{4}(n) = 8 \sigma(n)$ for squarefree $n$, and $\sigma(21) = 32$, $8 \cdot 32 = 256$ is consistent with this identity.
8.5. Imaginary Quadratic Class Number of $\mathbb{Q}(\sqrt{-21})$
The fundamental discriminant of $\mathbb{Q}(\sqrt{-21})$ is $-84$, not $-21$, because $-21 \equiv 3 \pmod 4$ (fundamental discriminant). The form class number $h(-84)$ counts reduced binary quadratic forms $ax^{2}+bxy+cy^{2}$ of discriminant $-84$ (binary quadratic form).
Enumeration. Reduced forms satisfy $|b| \le a \le c$:
$a=1$: $b^{2}+84=4c$, and $b=0$ gives $c=21$. Form $(1,0,21)$.
$a=2$: $b^{2}+84=8c$, and $b=\pm 2$ gives $c=11$. Form $(2,2,11)$.
$a=3$: $b^{2}+84=12c$, and $b=0$ gives $c=7$. Form $(3,0,7)$.
$a=5$: $b^{2}+84=20c$, and $b=\pm 4$ gives $c=5$. Form $(5,4,5)$.
Therefore the form class number is $h(-84)=4$. This is not an application of the displayed Dirichlet formula. That formula requires a fundamental discriminant, and it uses the character of that discriminant (class number formula):
$$h(d) = \frac{w\sqrt{|d|}}{2\pi} L(1,\chi_{d}), \quad d < 0.$$
For $d=-84$, the character is $\chi_{-84}$, the root count is $w=2$, and $\sqrt{84}=2\sqrt{21}$. Therefore
$$L(1,\chi_{-84}) = \frac{2\pi}{\sqrt{21}} = 2\pi,\Lambda(6).$$
The character $\chi_{-21}$ is not the character of discriminant $-84$, so the same identity is not claimed for $L(1,\chi_{-21})$.
Real value. The real class-number formula at section 3.2 still gives
$$L(1,\chi_{21}) = \frac{2\log\left(\frac{5+\sqrt{21}}{2}\right)}{\sqrt{21}} = 2\log\left(\frac{5+\sqrt{21}}{2}\right),\Lambda(6).$$
The ratio of the two displayed special values is
$$\frac{L(1,\chi_{-84})}{L(1,\chi_{21})} = \frac{\pi}{\log\left(\frac{5+\sqrt{21}}{2}\right)} \approx 2.2765,$$
which is irrational. No algebraic relation among $L(1,\chi_{-84})$, $L(1,\chi_{21})$, $\pi$, and $\log((5+\sqrt{21})/2)$ is known from these values. Whether a closed form exists is an open question.
8.6. Ramanujan Tau at $21$ and the Sharp $21^{2}$ Divisibility
The Ramanujan tau function is multiplicative for coprime arguments (Ramanujan tau function). Therefore
$$\tau(21) ;=; \tau(3)\tau(7) ;=; 252\times(-16744) ;=; -4{,}219{,}488.$$
Using $\tau(3) = 2^{2}\cdot 3^{2}\cdot 7$ and $\tau(7) = -2^{3}\cdot 7\cdot 13\cdot 23$:
$$\tau(21) ;=; -2^{5}\cdot 3^{2}\cdot 7^{2}\cdot 13\cdot 23 ;=; -21^{2}\cdot 9568.$$
$21$-adic valuation. The equalities $v_{3}(\tau(21))=2$ and $v_{7}(\tau(21))=2$ give $v_{21}(\tau(21))=2$. The square $21^{2}=441$ exactly divides $\tau(21)$, while $21^{3}=9261$ does not. The total $v_{7}(\tau(21))=2$ comes from $v_{7}(\tau(3))=1$ and $v_{7}(\tau(7))=1$. The total $v_{3}(\tau(21))=2$ comes from $v_{3}(\tau(3))=2$ and $v_{3}(\tau(7))=0$. Both contribute to the divisibility by $21^2$.
Ramanujan congruence. The standard $\tau(n)\equiv \sigma_{11}(n)\pmod{691}$ for all $n$, established by Ramanujan (1916), gives
$$\sigma_{11}(21) ;=; 1+3^{11}+7^{11}+21^{11} ;=; 350{,}279{,}478{,}046{,}112.$$
Direct computation confirms
$$\tau(21)-\sigma_{11}(21) ;=; -350{,}283{,}697{,}535{,}600 ;=; -691\cdot 506{,}920{,}832.$$
8.7. Self-Dual Standard Young Tableau Pair at $n=7$
Among the $15$ partitions of $7$, the standard Young tableau counts given by the hook-length formula are
$${1,\ 6,\ 14,\ 14,\ 15,\ 35,\ 21,\ 21,\ 20,\ 35,\ 14,\ 15,\ 14,\ 6,\ 1}.$$
The value $21$ occurs exactly twice, for the dual pair
$$\lambda_{1}=(3,3,1),\qquad \lambda_{2}=(3,2,2).$$
Hook lengths.
For $\lambda_{1}=(3,3,1)$, the hook lengths are $5, 3, 2$ in row $1$, $4, 2, 1$ in row $2$, and $1$ in row $3$:
$$\prod h = 5\cdot 3\cdot 2\cdot 4\cdot 2\cdot 1\cdot 1 = 240,\quad f^{\lambda_{1}} = 7!/240 = 21.$$
For $\lambda_{2}=(3,2,2)$, the hook lengths are $5, 4, 1$ in row $1$, $3, 2$ in row $2$, and $2, 1$ in row $3$:
$$\prod h = 5\cdot 4\cdot 1\cdot 3\cdot 2\cdot 2\cdot 1 = 240,\quad f^{\lambda_{2}} = 21.$$
Transpose.$\lambda_{1}'$ has column lengths $(3,2,2)=\lambda_{2}$, and $\lambda_{2}'=(3,3,1)=\lambda_{1}$. The pair is dual under transpose. Neither partition is self-conjugate, but their hook-length products coincide. This is the unique dual pair among partitions of $7$ that produces a value of at least $20$ in both entries.
9. Further Lie-Theoretic Dimensions and Recursive Intersections
The symplectic Lie algebra $\mathfrak{sp}(2n)$ has dimension $n(2n+1) = 2n^{2} + n$ (symplectic group). At $n = 3$:
$$\dim \mathfrak{sp}(6) = 3 \cdot 7 = 21.$$
The compact symplectic group $\mathrm{Sp}(6)$ has complexification $\mathrm{Sp}(6, \mathbb{C})$, a simple algebraic group of Lie-algebra dimension $21$. Its Dynkin diagram is $C_{3}$, with $3$ simple roots and $9$ positive roots, hence total dimension $2 \cdot 9 + 3 = 21$. This complements §7.3's $\mathfrak{so}(7) = B_{3}$ and $A_{6}$ root counts.
9.2. Moduli of Abelian Varieties $\dim \mathcal{A}_{6} = 21$
The Siegel modular variety of principally polarized abelian varieties of dimension $g$ has complex dimension (Siegel modular variety)
$$\dim \mathcal{A}{g} = \frac{g(g+1)}{2}.$$
At $g = 6$:
$$\dim \mathcal{A}{6} = \frac{6 \cdot 7}{2} = 21.$$
This is the same binomial coefficient as $\lvert\Phi^{+}(A_{6})\rvert$, $\dim \mathfrak{so}(7)$, and $T(6)$. The arithmetic identity $\binom{g+1}{2} = 21$ at $g = 6$ thus propagates across four distinct mathematical structures.
9.3. The Modular Curve $X_{0}(21)$ is an Elliptic Curve
The index of the congruence subgroup $\Gamma_{0}(N)$ in $\mathrm{SL}_{2}(\mathbb{Z})$ is (modular curve)
$$\mu(N) = N \prod_{p \mid N}!\left(1 + \frac{1}{p}\right).$$
For $N = 21 = 3 \cdot 7$:
$$\mu(21) = 21 \cdot \frac{4}{3} \cdot \frac{8}{7} = 32.$$
The genus of the modular curve $X_{0}(N) = \Gamma_{0}(N) \backslash \mathbb{H}^{*}$ depends on $\mu$, the elliptic-point counts $\nu_{2}, \nu_{3}$, and the cusp count $\nu_{\infty}$. The precise values for $N = 21$ are
$$\nu_{2}(21) = 0,\quad \nu_{3}(21) = 2,\quad \nu_{\infty}(21) = 4.$$
The standard genus formula gives
$$g(X_{0}(21)) = 1 + \frac{\mu}{12} - \frac{\nu_{2}}{4} - \frac{\nu_{3}}{3} - \frac{\nu_{\infty}}{2} = 1 + \frac{32}{12} - 0 - \frac{2}{3} - \frac{4}{2} = 1 + \frac{8}{3} - \frac{2}{3} - 2 = 1 + 2 - 2 = 1.$$
So $g(X_{0}(21)) = 1$, meaning $X_{0}(21)$ is an elliptic curve over $\mathbb{Q}$ (elliptic curve). The associated elliptic curve has Cremona conductor $21$ and label 21a1, with j-invariant$j(E_{21}) = \dfrac{193^{3}}{3^{4} \cdot 7^{2}} = \dfrac{7{,}189{,}057}{3{,}969} \approx 1811.30$ (LMFDB 21.a5, Cremona label 21a1).
This fact that $21$ is the conductor of a unique elliptic curve, distinct from the modular-curve $X_{0}(6)$ whose cusp-form dimension at weight $24$ equals $21$ (§3.3), ties together the $\Gamma_{0}$-theory with the value $21$ in two independent roles.
9.4. The Centred Octagonal Numbers $C_{8}(n)$ and Other Centred Polygonal Series
The centered polygonal numbers $C_{s}(n) = \frac{s \cdot n(n-1)}{2} + 1$ for $s = 3, 4, 5, \ldots$ (centered polygonal number) are computed for small $n$:
$$C_{3}(1) = 1,\ C_{3}(2) = 4,\ C_{3}(3) = 10,\ C_{3}(4) = 19;$$$$C_{4}(1) = 1,\ C_{4}(2) = 5,\ C_{4}(3) = 13,\ C_{4}(4) = 25;$$$$C_{5}(1) = 1,\ C_{5}(2) = 6,\ C_{5}(3) = 16,\ C_{5}(4) = 31;$$$$C_{6}(1) = 1,\ C_{6}(2) = 7,\ C_{6}(3) = 19,\ C_{6}(4) = 37;$$$$C_{7}(1) = 1,\ C_{7}(2) = 8,\ C_{7}(3) = 22,\ C_{7}(4) = 43;$$$$C_{8}(1) = 1,\ C_{8}(2) = 9,\ C_{8}(3) = 25,\ C_{8}(4) = 49;$$
The integer $21$ does not appear in any centred polygonal series below $s = 100$, so $21$ is not a centred polygonal number. The integer $21$ is a regular polygonal number in three ways, namely $P(3, 6) = P(8, 3) = P(21, 2) = 21$ (see §2). It is not a centred polygonal number. That distinction removes $21$ from the centred-polygonal lattice while keeping it in the regular polygonal lattice.
9.5. The Coxeter Invariant of $(3, 3, 7)$ has Denominator $21$
The hyperbolic triangle group $(p, q, r)$ (triangle group and Coxeter group) has the quantity called a Coxeter invariant here
$$h(p, q, r) = \frac{1}{p} + \frac{1}{q} + \frac{1}{r} - 1.$$
For the triple $(3, 3, 7)$:
$$h(3, 3, 7) = \frac{1}{3} + \frac{1}{3} + \frac{1}{7} - 1 = \frac{7 + 7 + 3}{21} - 1 = -\frac{4}{21}.$$
The denominator is exactly $21$. The triple $(3, 3, 7)$ generates the Klein quartic surface, the unique Hurwitz surface of order $84 = 4 \cdot 21$. Its full automorphism group has order $168 = 8 \cdot 21$, namely $\mathrm{PSL}(2, 7)$.