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A cubic equation is a polynomial equation of degree of a polynomial three, $a{x}^3+b{x}^2+cx+d=0$, with $a\ne 0$. Its three complex roots of a polynomial, counted with multiplicity of a root, need not all have physical significance in an application; a model often selects a branch by continuity from a known limit.

A nonzero polynomial of degree $d$ over a field has at most $d$ distinct roots. Indeed, each root supplies a linear factor by polynomial division, and induction on $d$ gives the bound.

The leading coefficient of a nonzero polynomial $a_n{x}^n+\cdots +a_0$, with $a_n\ne 0$, is $a_n$; its degree of a polynomial is $n$. A monic polynomial has leading coefficient one. For polynomials over a field, a product has leading coefficient equal to the product of the two leading coefficients, which prevents cancellation of their highest-degree terms.

Here is an elementary polynomial pencil with four square members argument. Suppose first that $u,v$ are linearly dependent. Their coprimality of polynomials then forces both to be constant. Otherwise write the four distinct members as $L_j={\alpha }_{j}u+{\beta }_{j}v=s_j^2$. They are nonzero and pairwise coprime polynomials: a common nonconstant factor of two members would divide both $u$ and $v$.

Put $d=\max (\mathrm{deg}u,\mathrm{deg}v)>0$. At most one member of the pencil has degree of a polynomial smaller than $d$, since cancellation of its leading coefficient determines a unique projective pair. Choose a member $L_k$ of minimal degree $e$ and any independent member $L_l$ of degree $d$. The polynomialis nonzero: otherwise the rational function $L_k/L_l$ would have zero derivative, hence would be constant in characteristic zero. Its degree is at most $d+e-1$; if both members have degree zero the original $d>0$ assumption has already failed.

Replacing this pair by any other independent pair changes $W$ only by a nonzero scalar. Since $L_j=s_j^2$, each $s_j$ divides $W$. The pairwise coprimality of polynomials therefore gives ${\prod }_j{s}_j\mid W$. But three members have degree $d$, soa contradiction. Consequently $u$ and $v$ are constant. The possibility that a member is zero was already covered by linear dependence.

To apply this to an elliptic curve, complete the square in its Weierstrass equation of an elliptic curve and work over $\mathbb{C}$. Nonsingularity gives three distinct roots $e_1,e_2,e_3$, so the equation becomes $y^2={\prod }_{j=1}^3(x-e_j)$. A nonconstant rational parametrization of an algebraic curve would have $x=u/v$ with coprime polynomials. Clearing denominators givesThe four factors are pairwise coprime polynomials. A rational function whose square is a polynomial is itself a polynomial, by comparing numerator and denominator in lowest terms. Unique factorization, and the fact that every nonzero complex constant has a square root, make each of these four factors a square in $\mathbb{C}[t]$. They correspond to four distinct projective pairs. The result just proved forces $u,v$ to be constant, and the equation then forces $y$ to be constant as well. This proves the nonparametrizability of an elliptic curve.

A sequence of real orthogonal polynomials consists of real polynomials $p_n$ with degree of a polynomial $n$ andOrthogonality is relative to the specified inner product, rather than to the ordinary coefficients. A monic polynomial has leading coefficient one, so a monic orthogonal polynomial has the form $p_n(x)=x^n+\text{terms of lower degree}$. In particular, the monic degree-zero polynomial is $p_0=1$.

Since $x{p}_n-p_{n+1}$ has degree of a polynomial at most $n$, expand it in the basis $p_0,\dots ,p_n$. Multiplication by $x$ is self-adjoint for this weighted inner product: $⟨xp,q⟩=⟨p,xq⟩$. For $k\le n-2$, the polynomial $x{p}_k$ has degree at most $n-1$, soOnly the $p_n$ and $p_{n-1}$ coefficients remain. Taking their inner products gives the three-term recurrenceFor the second formula, $x{p}_{n-1}=p_n+$ a polynomial of lower degree, hence $⟨x{p}_n,p_{n-1}⟩=⟨p_n,x{p}_{n-1}⟩=⟨p_n,p_n⟩$. At $n=0$, take $p_{-1}=0$, ${\beta }_0=0$ and the same formula for ${\alpha }_0$. Positivity of the weight ensures that all the squared norms in the denominators are nonzero.

For a basis $v_1,\dots ,v_d$ of the real vector space $V$, the Gram-Schmidt process constructs an orthonormal basis successively:The inner product makes $w_j$ orthogonal to all earlier $e_i$. Moreover $w_j\ne 0$, since otherwise $v_j$ would be a linear combination of its predecessors, contrary to linear independence. Induction gives $\mathrm{span}(e_1,\dots ,e_j)=\mathrm{span}(v_1,\dots ,v_j)$. Starting with a basis of a linear subspace $U$ and extending it to a basis of $V$ gives the same construction with the first $r=\dim U$ vectors spanning $U$.

The orthogonal complement is $U^{\perp }=\{v\in V:⟨v,u⟩=0\text{ for every }u\in U\}$. For every $v\in V$, set $u={\sum }_{i=1}^r⟨v,e_i⟩e_i$. Then $u\in U$ and $v-u\in U^{\perp }$. If $w\in U\cap U^{\perp }$, then $⟨w,w⟩=0$, so $w=0$ by definiteness of the inner product. Thus the direct sum isThis also covers $U=0$ and $U=V$.

For the polynomial space, the form is immediately a symmetric bilinear form, and $(f,f)=f(1{)}^2+f(2{)}^2+f(3{)}^2\ge 0$. A nonzero polynomial of degree of a polynomial at most two cannot have three distinct roots of a polynomial, so this form is an inner product for $n=1,2$. For $n\ge 3$, the nonzero polynomial $(x-1)(x-2)(x-3)$ has $(f,f)=0$. Therefore, among the stipulated positive integers,

The original PDF contains this subpart; the local TeX has omitted it. Write $a_n$ for the leading coefficient of $P_n$. The Legendre polynomial recurrence relation and the starting polynomials give $a_0=a_1=1$, and induction givesIndeed, $x{P}_n$ has degree of a polynomial $n+1$, while $P_{n-1}$ has smaller degree of a polynomial and cannot cancel its leading term. HenceAt $x=1$, the same linear recurrence relation gives $(n+1)P_{n+1}(1)=(2n+1)P_n(1)-n{P}_{n-1}(1)$. Starting from $P_0(1)=P_1(1)=1$, induction yields

Suppose $X=f_1\cdots f_s$ were a product of irreducible elements of $R$. Since degree of a polynomial is additive under products, exactly one factor has positive degree and all the others are constant. By the preceding classification, the positive-degree factor has constant term $\pm 1$, while every constant irreducible is $\pm p$ for a prime $p$. Their product therefore has a nonzero constant term, contradicting $X(0)=0$.

Thus $X$ is not a product of irreducibles, so this integral domain is not an atomic domain and in particular is not a unique factorization domain.

Let the smooth projective curve be the projective completion of $f(x,y)=0$, of degree $d$, and suppose after interchanging $x$ and $y$ if necessary that $f_y=\partial f/\partial y$ is nonzero. A basis of its holomorphic differential forms isEquivalently one may use $-x^i{y}^{j}dy/f_x$. These expressions agree on their common domain because differentiating $f(x,y)=0$ gives $f_{x}dx+f_{y}dy=0$. The allowed monomials are precisely the homogeneous polynomials of degree at most $d-3$, in accordance with the Adjunction formula $K_C\cong {\mathcal{O}}_C(d-3)$.

Lagrange theorem for polynomial congruences says that a nonzero polynomial of degree $d$ over the finite field ${\mathbb{F}}_p$ has at most $d$ roots. To prove it, use mathematical induction on $d$. A constant nonzero polynomial has no roots. If $f(a)=0$, polynomial division in the field ${\mathbb{F}}_p$ gives $f(X)=(X-a)q(X)$ with $\mathrm{deg}q=d-1$. Every root other than $a$ is a root of $q$, so the induction hypothesis gives at most $1+(d-1)=d$ roots.

Every nonzero residue class modulo $p$ has a unique multiplicative inverse. In the product $(p-1)!$, pair each class with its inverse. The unpaired classes satisfy $x=x^{-1}$, hence $x^2-1=0$. By the theorem these are exactly $1$ and $-1$, and thereforeThis is Wilson theorem.

Now write $p-1=kq$. By Fermat's little theorem, every nonzero class is a root ofThe second factor has degree $p-1-k$ and hence at most $p-1-k$ roots. Since their union contains all $p-1$ nonzero classes, $X^k-1$ has at least $k$ roots; Lagrange's theorem gives at most $k$. Thus $x^k\equiv 1(\mathrm{mod}p)$ has exactly $k$ solutions.

If the minimal polynomial of $\gamma$ has degree $d$, then
${\mathbb{F}}_p(\gamma )$ has $p^d$ elements and is an intermediate field of
$L/{\mathbb{F}}_p$. The tower law for field extensions givesso $d\mid n$.

The multiplicative group of a finite field is cyclic. Choose a generator
$\gamma$ of $L^{\times }$, of order $p^n-1$. If $\gamma$ lay in a proper intermediate field of degree $d<n$, its order would divide $p^d-1<p^n-1$, a contradiction. Thus
${\mathbb{F}}_p(\gamma )=L$ and its minimal polynomial has degree $n$.

For arbitrary $r\ge 1$, let $E$ be the splitting field of
$X^{p^r}-X$ over ${\mathbb{F}}_p$. Its roots form a field: the Frobenius endomorphism shows they are closed under addition, subtraction, multiplication, and inversion. The derivative is $-1$, so there are exactly $p^r$ distinct roots. Thus this root field has order $p^r$. Applying the preceding generator argument supplies an element whose minimal polynomial over ${\mathbb{F}}_p$ has degree $r$, proving that an irreducible polynomial of every positive degree exists.

The Lagrange root bound over a field says that a nonzero polynomial of degree $r$ over a field has at most $r$ roots. In particular, a polynomial modular congruence of degree $r$ modulo a prime number has at most $r$ incongruent solutions unless all its coefficients vanish modulo that prime.

Suppose that $d$ is good and that the positive integer divisor $e$ divides $d$. The $d$ roots of $X^d-1$ form a finite multiplicative subgroup $H$ of ${\mathbb{F}}_p^{\times }$. By the fact that every finite multiplicative subgroup of a field is cyclic, $H=⟨g⟩$ is a cyclic group of order $d$. The solutions in $H$ of $x^e=1$ areso there are at least $e$ of them. The Lagrange root bound over a field gives at most $e$ roots in all of ${\mathbb{F}}_p$, hence exactly $e$. Thus every divisor of a good number is good.

Now put $n=pq$. By the Chinese remainder theorem for unit groups, a base $b$ is a Fermat-pseudoprime base precisely when its two components satisfyThe power roots in a finite field andshow that each component has ten choices. Therefore there areFermat-pseudoprime bases.

To impose the strong pseudoprime condition, write $n-1=2^{s}m$ with $m$ odd. Since $10\mid n-1$, we have $5\mid m$. In each cyclic group of ten Fermat components, raising to the $m$th power sends five elements to $1$ and five elements to $-1$. A pair of components passes the strong test exactly when their signs agree: the pair $(1,1)$ satisfies the first alternative, and $(-1,-1)$ satisfies the second at $j=0$. Components of opposite sign become $(1,1)$ after squaring and can never jointly equal $-1$. Hence the number of strong-pseudoprime bases is

Write each polynomial of degree at most $n$ as $T\mathbf{a}$ and defineThis is a continuous function, and the reverse triangle inequality together with part (a) givesConsequently $F$ is a coercive function. Choose $R$ so large that $F(\mathbf{a})>F(0)$ outside the closed Euclidean ball of radius $R$. That ball is compact, so the extreme value theorem supplies a minimizer ${\mathbf{a}}_{\ast }$ there. Then $P=T{\mathbf{a}}_{\ast }$ is a best uniform approximation to $f$ among polynomials of degree at most $n$.

The Chebyshev alternation theorem says that a polynomial $P$ of degree at most $n$ is a best uniform approximation to $f\in C([0,1])$ if and only if there are $n+2$ pointsand a sign $\epsilon \in \{1,-1\}$ for whichThis is the equiripple criterion.

To prove sufficiency, put $E=\Vert f-P{\Vert }_{\infty }$ and suppose that another polynomial $Q$ of degree at most $n$ satisfies $\Vert f-Q{\Vert }_{\infty }<E$. At every $t_j$, the difference $R=Q-P$ must have the same sign as $f-P$; hence its signs alternate at the $n+2$ ordered points. The intermediate value theorem then gives at least one distinct root of $R$ in every interval $(t_j,t_{j+1})$, for at least $n+1$ roots in all. The Lagrange root bound over a field forces the nonzero polynomial $R$ to have degree at least $n+1$, contradicting $\mathrm{deg}R\le n$. Therefore no $Q$ gives a smaller error, and $P$ is best.