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Compare constant terms in the polynomial identity proved above. The constant term of $X^{p-1}-1$ is $-1$, while that of the product is $(-1{)}^{p-1}(p-1)!$. For odd $p$, this gives $(p-1)!\equiv -1(\mathrm{mod}p)$. For $p=2$, the same conclusion holds because $1\equiv -1(\mathrm{mod}2)$. Thus Wilson's theorem follows:

Reduce the polynomial modulo $p$, obtaining a nonzero degree-$d$ polynomial over the field ${\mathbb{F}}_p$. If $a$ is a root, polynomial division gives $\bar{f}(X)=(X-a)h(X)$. Any different root $b$ is a root of $h$, because $b-a\ne 0$ is invertible. Induction on degree therefore proves there are at most $d$ distinct roots modulo $p$.

Every nonzero residue is a root of $X^{p-1}-1$ by Fermat's little theorem. The difference between this monic polynomial and the product over all its $p-1$ nonzero roots has degree at most $p-2$, yet has $p-1$ roots. The root bound forces it to vanish identically over ${\mathbb{F}}_p$. Thus all coefficients of the corresponding integer polynomial are divisible by $p$.

For the requested failure modulo $p^2$, take $p=7$: $6!+1=721=7\cdot 103$, and $103$ is not divisible by seven. Hence the prime-square strengthening fails, even though some special primes do satisfy it.

The squared Vandermonde product defining the polynomial discriminant is symmetric in all roots. By the fundamental theorem of symmetric polynomials, it is a polynomial in the elementary symmetric functions, which are the signed coefficients of the monic polynomial. Hence it is a polynomial function of those coefficients over the base field.

For $f=X^3+pX+q$, $\Delta (f)=-\mathrm{Res}(f,f^{\prime })$. Eliminating the two leading rows in the Sylvester determinant reduces the resultant toThusFor $X^3-3X+1$, the only possible rational roots are $\pm 1$, and neither is a root. The cubic is therefore irreducible, so its Galois group is a transitive subgroup of $S_3$. Its discriminant is $81=9^2$. The Vandermonde product changes sign under odd permutations, and its square being a rational square implies that the product itself is rational; hence all Galois permutations are even. The only transitive subgroup of $A_3$ is $A_3$ itself. Consequently, writing $L$ for its splitting field,

We consider finite-dimensional continuous complex representations of SU(2). Its maximal torus is $\mathrm{diag}(e^{i\theta },e^{-i\theta })$. The irreducible representations areThey may be realized as homogeneous polynomials of degree $n$ in two variables, with the induced linear substitution action. Their torus weights are $n,n-2,\dots ,-n$, each once. The centre $-I$ acts by $(-1{)}^n$, so precisely the even-$n$ representations descend to SO(3). In physics $n=2j$ labels integral or half-integral spin $j$.

Here is a classification argument. Average any positive definite Hermitian form over the compact group using normalized Haar measure. The averaged form is invariant, so the orthogonal complement of an invariant subspace is invariant. Repeatedly splitting gives complete reducibility. Differentiate an irreducible representation and complexify to the Lie algebra ${\mathfrak{sl}}_2$, with generators $H,E,F$ satisfying $[H,E]=2E$, $[H,F]=-2F$, and $[E,F]=H$. Torus periodicity gives integer $H$-weights. Choose a vector $v$ of maximal weight $n$; then $Ev=0$. Induction on $k$ using the commutators givesThe chain terminates: if $F^{p}v\ne 0$ and $F^{p+1}v=0$, the displayed formula forces $n=p\ge 0$. The vectors $v,Fv,\dots ,F^{n}v$ have distinct weights, span an invariant subrepresentation, and hence span the irreducible representation. Their generator action agrees with ${\mathrm{Sym}}^n{\mathbb{C}}^2$. Invariance under the Lie algebra is equivalent to invariance under the connected group, completing the classification. Conversely, a nonzero invariant subspace of a symmetric power contains a weight vector by projection onto torus weights. Raising it to the highest weight and then lowering generates every monomial, proving that the symmetric power is irreducible.

The character on the torus iswith the endpoints interpreted by continuity. Every representation has a unique decomposition $V\cong {⨁}_{n\ge 0}{m}_n{V}_n$. Its multiplicities can be computed from character inner products,The sine orthogonality directly verifies orthonormality of the irreducible characters. Alternatively, if $w_k$ is the dimension of the weight-$k$ subspace, the weight lists give $w_n=m_n+m_{n+2}+\cdots$, so $m_n=w_n-w_{n+2}$ for $n\ge 0$. This gives a constructive decomposition from torus weights. In particular, multiplying weight characters yields the Clebsch-Gordan decomposition $V_m\otimes V_n\cong V_{m+n}\oplus V_{m+n-2}\oplus \cdots \oplus V_{|m-n|}$. Arbitrary unitary Hilbert-space representations of this compact group similarly decompose into Hilbert direct sums of these finite-dimensional irreducibles; the finite-dimensional statement above is the standard representation-theory setting.

Since $\mathcal{O}=\mathbb{Z}[\theta ]$, reduction gives $\mathcal{O}/p\mathcal{O}\cong {\mathbb{F}}_p[X]/(\bar{g})$. The maximal ideals of this quotient are precisely those generated by the distinct irreducible factors ${\bar{g}}_i$. Their inverse images are thereforethe distinct prime ideals above $p$. In fact $\mathcal{O}/{\mathfrak{p}}_i\cong {\mathbb{F}}_p[X]/({\bar{g}}_i)$ is a field of size $p^{\mathrm{deg}{\bar{g}}_i}$. Every prime containing $p$ arises this way; independence of the choice of lift $g_i$ follows because another lift differs by a multiple of $p$.

To prove the ideal factorization, set $J={\prod }_i{\mathfrak{p}}_i^{e_i}$. A generator of the product either contains a factor $p$, and hence lies in $p\mathcal{O}$, or is the product ${\prod }_i{g}_i(\theta {)}^{e_i}$. The latter also lies in $p\mathcal{O}$, since its polynomial reduces to $\bar{g}$ and $g(\theta )=0$. Thus $J\subset p\mathcal{O}$. The ideal norm is multiplicative for nonzero ideals in the ring of integers; henceEqual finite indices together with the containment imply equality. ThereforeThis proves the Dedekind factorization theorem in the stated monogenic case by containment and norms, rather than assuming the desired factorization.

A binary linear-feedback shift register stores $L$ bits, shifts them at each clock step, and inserts a fixed linear combination over ${\mathbb{F}}_2$ of the stored bits. Its output obeys a recurrence $s_n+c_1{s}_{n-1}+\cdots +c_L{s}_{n-L}=0$. Write its connection polynomial as $C(D)=1+c_{1}D+\cdots +c_L{D}^L$; the corresponding monic feedback polynomial is $z^{L}C(z^{-1})$.

The Berlekamp-Massey algorithm maintains the shortest recurrence fitting the observed prefix. Initialize $C=B=1$, length $L=0$, displacement $m=1$, and previous nonzero discrepancy $b=1$. At symbol $n$, calculate $d=s_n+{\sum }_{j=1}^L{c}_j{s}_{n-j}$. If $d=0$, increase $m$. If $d\ne 0$, save $T=C$ and replace $C$ by $C-(d/b)D^{m}B$. If $2L\le n$, replace $L$ by $n+1-L$, $B$ by $T$, $b$ by $d$, and $m$ by $1$; otherwise only increase $m$. In the binary field subtraction equals addition. The shifted previous discrepancy polynomial cancels the current discrepancy while preserving earlier equations. A failed recurrence of length $L$ at position $n$ requires a corrected length at least $n+1-L$ when this exceeds $L$, explaining the length update. Thus the algorithm recovers the shortest compatible register; a known bound $L$ and at least $2L$ output symbols suffice to identify its minimal recurrence.

For the printed twelve-bit prefix, the nonzero discrepancies occur at positions $n=0,1,2,5,6,7$ (indexing from zero). The successive updated pairs $(L,C)$ areAll remaining discrepancies vanish. ThereforeThe first four observed bits are the fill. Length four is necessary: a length-three recurrence would need to hold already at $n=3$ and no such recurrence fits the prefix. Indeed the equations at $n=3,4,5$ force $c_2=1$, $c_1+c_3=0$, $c_1+c_3=1$, a contradiction. Finite observations cannot rule out longer nonminimal registers.

For a countable set $A$, the polynomials of polynomial degree at most $d$ with coefficients in $A$ are described by coefficient tuples in $A^{d+1}$. The finite Cartesian power of a countable set result makes this a countable set. Taking the union over $d\ge 0$ shows there are only countably many such polynomials altogether; restricting to the nonzero polynomials preserves countability.

A nonzero polynomial of polynomial degree $d$ has at most $d$ distinct roots of a polynomial. To see the bound, a root $r$ gives the factorization $P(x)=(x-r)Q(x)$ by the factor theorem, and each other root is a root of $Q$, whose degree is $d-1$; induction starts with a nonzero constant, which has no roots. Thus $\varphi (A)$ is a countable union of countable sets, each of them finite, and is a countable set.

Starting with the given countable set $A_0$, mathematical induction now shows that every $A_n=\varphi (A_{n-1})$ is countable. A final application of the countable union of countable sets theorem givesThis countability argument does not require that the sequence of sets be increasing.

To obtain a countable closure under real polynomial roots, start with $B_0=\{1,\pi \}$ and explicitly retain previous elements at each stage:Part (b) shows that $\varphi$ takes countable sets to countable sets; consequently all $B_n$ and then $X$ are countable by the countable union of countable sets theorem. Also $1,\pi \in X$.

For a nonzero polynomial $P(t)=c_0+\cdots +c_d{t}^d$ with all $c_j\in X$, choose stages $n_j$ containing the finitely many coefficients and put $N={\max }_j{n}_j$. The inclusions $B_n\subseteq B_{n+1}$ ensure that every coefficient belongs to $B_N$. Every real root of a polynomial $P$ therefore belongs to $\varphi (B_N)\subseteq B_{N+1}\subseteq X$. This countable set $X$ has all the required properties. Retaining $B_n$ explicitly avoids assuming the generally false inclusion $A\subseteq \varphi (A)$: for example, $\varphi (\{1\})$ does not contain $1$, since a nonzero polynomial all of whose coefficients are one is positive at $1$.

Rouché's theorem states that if $f,g$ are holomorphic on a neighborhood of a closed disc with boundary $C$ and $|g|<|f|$ on $C$, then $f$ and $f+g$ have the same number of zeros in the disc, counted with multiplicity. More generally the conclusion holds for a simple closed contour whose interior lies in their domain. To prove it, set $f_t=f+tg$ for $0\le t\le 1$. The strict boundary inequality ensures $f_t\ne 0$ on $C$. By the argument principle,is the integer number of zeros inside. The integrand is continuous in $t$ uniformly on $C$, since its denominator is bounded away from zero on the compact product $C\times [0,1]$. Thus $N(t)$ is continuous and integer-valued, so constant. This proves the theorem.

On $|z|=1$, the term $8z^6$ dominates the remaining terms because $3+1+2+1=7<8$. Thus the polynomial has six zeros inside the unit circle. On $|z|=2$, $3z^9$ dominates because its modulus is $1536$, whereas the sum of the other bounds is $512+32+16+1=561$. There are therefore nine zeros inside radius two. The strict inequalities exclude boundary zeros, so

For the fixed-degree polynomial limit, choose $d+1$ distinct complex nodes ${\zeta }_0,\dots ,{\zeta }_d$. Write $p_n(z)={\sum }_{j=0}^d{a}_{j,n}{z}^j$, inserting zero coefficients if necessary. The vector of values at the nodes equals the fixed Vandermonde matrix times the coefficient vector. Its determinant is ${\prod }_{i<j}({\zeta }_j-{\zeta }_i)\ne 0$. Convergence of the values therefore implies coefficient convergence, say $a_{j,n}\to a_j$. For $p(z)={\sum }_{j=0}^d{a}_j{z}^j$ and a compact set with $|z|\le R$,This proves that the given limit is a polynomial of degree at most $d$ without invoking any theorem about uniform limits of holomorphic functions.

When the limit has $d$ distinct roots, choose pairwise disjoint small closed discs around them, with no limit root on their boundaries. On each boundary $|p|$ has a positive minimum. Uniform convergence makes $|p_n-p|<|p|$ there for all sufficiently large $n$, so Rouché's theorem puts exactly one root $z_{i,n}$ of $p_n$ in each disc. For every smaller disc about $z_i$ the same argument applies eventually. The unique root in the original disc must then lie in the smaller one, provingThe literal demand for a root at every initial index needs this finite-index qualification: $p_1=1$, $p_n=z$ for $n\ge 2$ satisfies the hypotheses with limit $z$, but $p_1$ has no root. If every initial polynomial has a root, arbitrary choices at the finitely many exceptional indices extend the convergent sequences; otherwise the sequences necessarily start at an eventual index.