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For genuine degrees three and two, the resultant is the Sylvester matrix determinantEquivalently, if $r_1,r_2,r_3$ are the roots of $p$, counted with multiplicity, then $R(p,q)=a^2{\prod }_{i}q(r_i)$. Thus $R(p,q)\ne 0$ means that the polynomials have no common root, or equivalently are coprime over $\mathbb{C}$. One can also see this from the matrix: its singularity gives a nonzero relation $Ap+Bq=0$ with $\mathrm{deg}A<2$, $\mathrm{deg}B<3$. If $p,q$ were coprime, $p\mid B$ would force $B=0$, then $A=0$, a contradiction. Conversely, a common factor supplies such a relation after dividing $p,q$ by that factor. When a leading coefficient is zero, use the resultant for the actual polynomial degrees rather than assuming the displayed fixed-degree criterion unchanged.

An algebraic addition theorem for a meromorphic function $f$ means a nonzero polynomial $P(U,V,W)$ withidentically where the values are finite. A local relation extends meromorphically to the other regular arguments.

For a polynomial $f$ of positive degree $d$, let $z_i$ be the $d$ roots of $f(z)-U$ and $w_j$ the $d$ roots of $f(w)-V$. FormIts coefficients are separately symmetric polynomials in the two sets of roots. The Fundamental theorem of symmetric polynomials and the fixed nonzero leading coefficient of $f$ show they are polynomials in $U,V$. The expression is monic of degree $d^2$ in $W$, so it is nonzero. If $U=f(z)$, $V=f(w)$, one factor vanishes at $W=f(z+w)$. This proves the addition relation, including at exceptional multiple-root values by polynomial identity. A constant polynomial instead has the relation $W-f(0)=0$.

For the elliptic case, let $E=\mathbb{C}/\Lambda$. A nonconstant elliptic function $f$ defines a finite morphism $E\to {\mathbb{P}}^1$, so the function field of $E$ is a finite algebraic extension of $\mathbb{C}(f)$. The elliptic function-field decomposition expresses it as $\mathbb{C}(\wp ,{\wp }^{\prime })$, withThe Weierstrass addition formula makes translation algebraic:Differentiating and using the cubic differential equation also makes ${\wp }^{\prime }(z+w)$ rational in the four separate values. Therefore $f(z+w)$ lies in a finite algebraic extension of $\mathbb{C}(f(z),f(w))$. Its algebraic equation, after clearing denominators, is the required polynomial relation. This sketches why every elliptic function, not only $\wp$, has an algebraic addition theorem. Constants are already covered.

For the final assertion, consider $f(z)=e^{e^z}$, an entire function. If $c$ is a period, then $e^{e^z(e^c-1)}=1$ for every $z$. The continuous exponent must be a constant integer multiple of $2\pi i$; its derivative forces $e^c=1$. Hence its periods are exactly $2\pi i\mathbb{Z}$, making it a simply periodic function.

Suppose it had an algebraic addition theorem $P$. There are only finitely many values $V_0$ for which $P(U,V_0,W)$ is identically zero: a nonzero coefficient polynomial in $V$ already bounds this exceptional set. Choose a positive irrational $\alpha$ with $e^{\alpha }$ outside this set, and set $c=\log \alpha$. Specializing $w=c$ gives a nonzero polynomial $Q(U,W)=P(U,e^{\alpha },W)$ and, with $t=e^z$,All exponents with nonzero coefficients are distinct, by irrationality of $\alpha$. Divide by the exponential with largest real exponent and let $t\to +\infty$; its nonzero coefficient would have to tend to zero. This contradiction proves that simple periodicity does not imply an algebraic addition theorem. It is the simply periodic entire function without an algebraic addition theorem example.

The fundamental theorem of algebra says that every nonconstant polynomial with complex number coefficients has a root in the complex numbers. Repeated polynomial division by a factor of degree one therefore factors a degree-three polynomial with complex number coefficients into three factors of degree one, with algebraic multiplicities included.

For a matrix with complex number entries $A$ of size three, its characteristic polynomial is ${\chi }_A(t)=\det (tI-A)$, and its characteristic equation is ${\chi }_A(t)=0$. A complex number $\lambda$ is an eigenvalue exactly when $\lambda I-A$ is singular, equivalently when its null space contains a nonzero vector. Thus the three roots of the characteristic polynomial are the three eigenvalues counted with algebraic multiplicity.

For the specified matrix, expansion along the first row givesHence the eigenvalues are $1,1,-1$. Corresponding eigenvectors areThe matrix with these three columns has determinant $2i\ne 0$, so these eigenvectors are linearly independent. A repeated eigenvalue does not in general guarantee two independent eigenvectors; here they have been exhibited directly.

Put $\mu =\cos \vartheta$ and separate $\Phi =R(r)Y(\mu )$. Multiplying Laplace's equation by $r^2/(RY)$ givesFor the regular polynomial angular modes, $\lambda =n(n+1)$ and $Y=P_n(\mu )$ is a Legendre polynomial. The radial equation has branches $r^n$ and $r^{-n-1}$. Regularity at the origin keeps only the first inside, and decay at infinity keeps only the second outside. Continuity at $r=1$ identifies their coefficients, so a mode hasIts outward-minus-inward radial derivative is $-(2n+1)a_n{P}_n(\mu )$. The prescribed angular dependence decomposes as $V{\mu }^2=V{P}_0/3+2V{P}_2/3$, since $P_0=1$ and $P_2=(3{\mu }^2-1)/2$. Hence $a_0=-V/3$ and $a_2=-2V/15$, with no other required modes. The spherical harmonic matching with a derivative jump givesBoth branches are harmonic in their respective regions. They agree at the sphere, and their derivative jump is $V/3+(2V/3)P_2=V{\cos }^2\vartheta$. The interior is regular at zero and the exterior decays at infinity. An additional homogeneous matching mode would satisfy $-(2n+1)a_n=0$, so its coefficient vanishes.

After choosing a basis of $V$, a linear endomorphism is specified by an $n$ by $n$ matrix, with $n^2$ freely chosen coefficients. ThereforeThe $n^2+1$ endomorphisms $I,\alpha ,\dots ,{\alpha }^{n^2}$ are linearly dependent. Some nonzero coefficient list gives a nonzero polynomial $p$ satisfying $p(\alpha )=0$.

For nonzero $V$, the minimal polynomial $m_{\alpha }$ is the monic polynomial of least degree annihilating $\alpha$. It is unique: subtracting two monic least-degree choices would give a lower-degree annihilator. More generally division of any annihilator $p$ by $m_{\alpha }$ gives $p=q{m}_{\alpha }+r$, with $\mathrm{deg}r<\mathrm{deg}{m}_{\alpha }$, and evaluation implies $r(\alpha )=0$. Minimality forces $r=0$, so every annihilator is divisible by $m_{\alpha }$. If $V=\{0\}$, the convention $m_{\alpha }=1$ handles the degenerate case.

The arguments work over the underlying scalar field. In part (b), roots in that field correspond to eigenvalues there; to discuss other roots one extends scalars, for example by complexifying a real vector space.

Relative to the coordinate basis, the linear map has matrixExpanding $\det (tI-F)$ in its first column gives the monic characteristic polynomialIts eigenvalues are $1$ and $(1\pm i\sqrt{7})/2$, three distinct complex numbers. Eigenvectors corresponding to distinct eigenvalues are linearly independent, so they form a basis of ${\mathbb{C}}^3$ and $f$ is diagonalizable.

Every annihilating polynomial must vanish at all three eigenvalues: applying $p(f)$ to a corresponding eigenvector gives $p(\lambda )v$. Conversely, in the eigenbasis the product of the three distinct linear factors annihilates $f$. Hence the minimal polynomial isIf $af+b{f}^2=0$ with scalars $a,b$ not both zero, the nonzero polynomial $at+b{t}^2$ of degree at most two would annihilate $f$, contradicting this minimal degree of three. Thus $f$ and $f^2$ are linearly independent endomorphisms.

For distinct interpolation nodes $x_0,\dots ,x_n$, let $p_n$ be the unique polynomial of degree at most $n$ satisfying $p_n(x_i)=f(x_i)$. Existence follows from the Lagrange polynomial expressionuniqueness follows because the difference of two interpolants has $n+1$ distinct zeros but degree at most $n$.

Define the divided difference $f[x_0,\dots ,x_n]$ to be the coefficient of $x^n$ in $p_n$, even when that coefficient is zero. For $n=0$ this gives $f[x_0]=f(x_0)$. Permuting the nodes does not change the interpolation conditions, so it does not change their unique interpolant or this coefficient. The divided difference is symmetric in all the nodes. Equivalently its symmetric expression is

Let $p(x)={\prod }_{i=0}^n(x-x_i)$ be the monic node polynomial. Suppose first that the quadrature is exact through degree $2n+1$. For any polynomial $r$ of degree at most $n$, $pr$ has degree at most $2n+1$, while every sampled value is zero. Exactness givesThus $p$ is an orthogonal polynomial of degree $n+1$, proving necessity in the orthogonal-node criterion for Gaussian quadrature.

Conversely, suppose the nodes are the zeros of such a polynomial. Rescale it to be monic without changing its zeros or orthogonality. For any $f$ of degree at most $2n+1$, polynomial division givesOrthogonality makes $\int pq=0$; the quadrature of $pq$ is zero because $p(x_i)=0$. The assumed degree-$n$ exactness applies to $r$, soHence degree-$2n+1$ exactness is equivalent to orthogonal nodes. The given distinct-zero property ensures the $n+1$ node values are independent interpolation data; the degree-$n$ exactness fixes the corresponding interpolatory weights. This is the construction of Gaussian quadrature.

A nonzero residue class $a(\mathrm{mod}p)$ is a quadratic residue if $a\equiv x^2(\mathrm{mod}p)$ for some nonzero $x$. If it is a square, Fermat's little theorem gives $a^{(p-1)/2}=x^{p-1}\equiv 1$. Conversely, the map $x\mapsto x^2$ on the nonzero classes identifies exactly the pairs $\{x,-x\}$, giving $(p-1)/2$ distinct squares. They are all roots of $X^{(p-1)/2}-1$ over the field ${\mathbb{F}}_p$; a polynomial of this degree has at most that many roots. Its roots are therefore precisely the squares. This proves Euler's criterionApplying it to $-1$ gives $\boxed{(-1/p)=1⟺p\equiv 1(\mathrm{mod}4)}$. The exclusion of $a=0$ is the usual nonzero-residue convention needed for this equivalence.

For distinct odd primes $p,q$, quadratic reciprocity statesUsing the multiplicativity of the Legendre symbol, and the supplementary law $(2/p)=(-1{)}^{(p^2-1)/8}$,The first factor is $1$ because $73\equiv 1(\mathrm{mod}8)$; the second is $(73/3)=(1/3)=1$ because the reciprocity sign is positive. Thus $\boxed{73\text{ is a quadratic residue modulo }127}$.

Let the finite poles be $a_1,\dots ,a_N$. Their Laurent series have finite principal parts $P_j(z)={\sum }_{k=1}^{m_j}{c}_{jk}(z-a_j{)}^{-k}$. Subtracting them removes all finite singularities: $h=f-{\sum }_j{P}_j$ is entire. Each $P_j$ tends to zero at infinity, so $h$ still has at most a finite-order pole at infinity.

The Laurent expansion of $h(1/w)$ near $w=0$ has only finitely many negative powers. Thus $h(z)=P(z)+c+O(z^{-1})$ for large $|z|$, with $P$ a polynomial. The entire function $h-P$ is bounded outside a large disk and, by continuity, on that disk. Liouville's theorem makes it constant. ConsequentlyCombining denominators gives a rational function $p/q$. This proves that meromorphic functions on the sphere are rational without assuming rationality at the start.

Write the nonconstant bijective map as $p/q$ with coprime polynomials and set $d=\max (\mathrm{deg}p,\mathrm{deg}q)$. For a finite value $w$, its finite preimages solve $p(z)-wq(z)=0$. Except for a possible leading-coefficient cancellation value, this polynomial has degree $d$.

Multiple roots can occur only at zeros of $p^{\prime }q-p{q}^{\prime }$, since at a root $q\ne 0$ by coprimality. This derivative numerator is a nonzero polynomial for a nonconstant rational map. There are therefore only finitely many finite critical values to exclude. Choose $w$ outside them and outside the leading cancellation value. The fundamental theorem of algebra then gives exactly $d$ distinct finite preimages. Bijectivity forces $d=1$.

Hencewhich is a Möbius transformation. Conversely, the inverse is $(d_{0}w-b)/(a-cw)$, interpreted at its pole and at infinity, proving bijectivity on the Riemann sphere. The argument also establishes the generic-preimage interpretation of the degree of a rational map of the Riemann sphere.

The dual space $V^{\ast }$ consists of all real linear functionals on $V$. If $v_1,\dots ,v_d$ is a basis, its dual basis ${\epsilon }^1,\dots ,{\epsilon }^d$ is characterized by ${\epsilon }^i(v_j)={\delta }_{ij}$; explicitly ${\epsilon }^i$ extracts the $i$th coordinate. The natural map $J:V\to V^{\ast \ast }$ is $J(v)(\varphi )=\varphi (v)$. It is linear and injective, because the coordinate functionals separate nonzero vectors, and both spaces have dimension $d$. Hence it is an isomorphism. Its definition uses no choice of basis, which is the meaning of “naturally” here.

Evaluation $e_x(p)=p(x)$ is linear, so belongs to $V_n^{\ast }$. For distinct nodes $x_i$, define the Lagrange interpolation polynomialsThey lie in $V_n$ and obey $e_{x_j}({\ell }_i)={\delta }_{ij}$. Applying a putative relation $\sum a_i{e}_{x_i}=0$ to ${\ell }_j$ gives $a_j=0$, so the $n+1$ evaluations are independent and form a basis of $V_n^{\ast }$. The polynomials ${\ell }_i$ give their dual basis in $V_n$, using the canonical identification with $V_n^{\ast \ast }$.

Every $p\in V_n$ has the interpolation expansion $p(t)={\sum }_{i}p(x_i){\ell }_i(t)$: the difference has $n+1$ distinct roots but degree at most $n$. Integrate to get the numerical integration weights ${\lambda }_i={\int }_{-1}^1{\ell }_i(t)dt$. For the specified symmetric five nodes, symmetry gives weights $(a,b,c,b,a)$. Exactness on $1,t^2,t^4$ givesOdd moments vanish automatically. Solving yields Boole's rule

Write $a=|A|$ and $b=|B|$, so $p\ge a>b\ge 1$. To cover both the unsaturated and saturated cases with the same coefficient calculation, setChoose $B^{\prime }\subseteq B$ of size $b^{\prime }$. These parameters satisfy $1\le b^{\prime }<a\le p$ and $1\le r\le p$.

Assume, for a contradiction, that the restricted sumset $C$ has fewer than $r$ elements. Extend it to a subset $D\subseteq {\mathbb{F}}_p$ of size $r-1$, and formOn $A\times B^{\prime }$, this polynomial vanishes: when $x=y$ the first factor vanishes, and when $x\ne y$ the sum belongs to $C\subseteq D$. Its total degree is $r=(a-1)+(b^{\prime }-1)$. Only the leading homogeneous part $(x-y)(x+y{)}^{r-1}$ contributes to the coefficient of $x^{a-1}{y}^{b^{\prime }-1}$. That coefficient isThe equality includes $b^{\prime }=1$, with an out-of-range binomial coefficient interpreted as zero. All factorial arguments are between zero and $p-1$, and $0<a-b^{\prime }<p$, so this coefficient is nonzero in ${\mathbb{F}}_p$. The Combinatorial Nullstellensatz applied to $A\times B^{\prime }$ forces a nonzero value, contradicting the vanishing just proved. Hence the restricted sumset bound for unequal subsets of a prime field isThe strict size inequality is used precisely in the nonzero factor $a-b^{\prime }$. The argument includes $r=p$ without dividing by $p!$, so no separate saturation assumption is hidden.