For $a=2$, the least positive residues in Gauss lemma for quadratic residues are simply $2,4,\dots ,p-1$. The number exceeding $p/2$ isFor $p\equiv 1,3,5,7(\mathrm{mod}8)$, respectively, this count is even, odd, odd, even. Substituting these parities into the lemma gives

Gauss lemma for quadratic residues states: for $p\nmid a$, take the least positive residues of $a,2a,\dots ,((p-1)/2)a$ modulo $p$, and let $s$ be the number greater than $p/2$. Then

Euler criterion states thatWhen $p\nmid a$, the power is congruent to $+1$ or $-1$, according as $a$ is or is not a quadratic residue.

For an odd prime number $p$, the Legendre symbol $(a/p)$ is zero if $p\mid a$, is $+1$ if $a$ is a nonzero square modulo $p$, and is $-1$ otherwise.

Choose an inscribed regular tetrahedron from one of the regular compounds of five tetrahedra in the dodecahedron, and let $H$ be its rotational stabilizer. Its twelve tetrahedral rotations preserve the dodecahedron. Concretely, with $\varphi =(1+\sqrt{5})/2$, use dodecahedron vertices $(\pm 1,\pm 1,\pm 1)$ together with cyclic coordinate placements of $(0,\pm {\varphi }^{-1},\pm \varphi )$. The tetrahedron has the four cube vertices whose coordinate product is $+1$. Cyclic coordinate permutations and even numbers of coordinate sign changes preserve both vertex sets, have determinant $+1$, and give twelve distinct rotations. No other rotation can preserve this tetrahedron, because a regular tetrahedron has only twelve orientation-preserving symmetries.

Their action on its four vertices is faithful and consists exactly of the even permutations: the eight rotations through $\pm 120^{\circ }$ induce three-cycles, the three half-turns induce double transpositions, and there is the identity. Thus $H\cong A_4$, independently of part (ii).

Finally a nonidentity element of $H$ has order two or three. In the dodecahedron's full rotation group, all fifteen half-turns about opposite-edge axes form one conjugacy class, and all twenty third-turns about opposite-vertex axes form one conjugacy class. These geometric conjugacies follow by carrying the corresponding oriented axes to one another by rotations. A normal subgroup containing such an element must contain its whole class, which already has more than twelve elements. Hence no nonidentity normal subgroup of $G$ is contained in $H$, verifying the required extra property.

Let $G$ act by left multiplication on its five left cosets of $H$. This gives a group homomorphism $\phi :G\to S_5$. Its kernel is the core of a subgroup, ${\bigcap }_{g\in G}gH{g}^{-1}$: an element fixes every coset precisely when all its conjugates lie in $H$. It is a normal subgroup contained in $H$, so the hypothesis forces the kernel to be trivial. Thus $G$ embeds in $S_5$.

The image has order $60$ and index two, hence is the kernel of a surjection $S_5\to C_2$. All transpositions are conjugate and generate $S_5$, so under any such surjection they must all have the same nonidentity image. This determines the map uniquely as the sign homomorphism. Its kernel is $A_5$, proving $G\cong A_5$.

To prove simplicity directly, the nonidentity conjugacy classes in $A_5$ have sizes $15,20,12,12$. A double transposition has centralizer of order four in $A_5$; a three-cycle has centralizer of order three; a five-cycle has centralizer of order five, yielding the two classes of size twelve. The class sizes together with the identity account for all sixty elements. A normal subgroup must be a union of these classes containing the identity, and its order must divide sixty. None of the proper nontrivial sums of these sizes with one is a divisor of sixty. Therefore $A_5$, and hence $G$, is simple.

A group action assigns to each $g\in G$ and $x\in \Omega$ a point $g\cdot x$ so that $1\cdot x=x$ and $(gh)\cdot x=g\cdot (h\cdot x)$. The orbit of $x$ is $Gx=\{g\cdot x:g\in G\}$. The action is transitive if there is exactly one orbit, equivalently any point can be carried to any other by some group element.

The multiplicative group $K^{\times }$ has order $p^n-1$, so Lagrange's theorem for finite groups gives $x^{p^n-1}=1$ for $x\ne 0$. Including zero, $x^{p^n}=x$ for every $x\in K$. Therefore ${\vartheta }^n$ is the identity. No positive smaller power is the identity by part (iii), soFor $n=1$ this says that Frobenius itself is the identity, as it should be.

Here the iteration index is a positive integer: $1\le m<n$. If ${\vartheta }^m(\lambda )=\lambda$, then ${\vartheta }^m$ fixes $F$ and the generator, so it fixes all of $K$. Every element of $K$ would then solve $T^{p^m}-T=0$. A nonzero polynomial of degree $p^m$ over a field has at most $p^m$ roots, contradicting $|K|=p^n>p^m$. Thus ${\vartheta }^m(\lambda )\ne \lambda$. The printed inequality alone omits positivity of $m$; at $m=0$ the claimed inequality would be false.

Define $\vartheta (x)=x^p$ for every $x\in K$. In characteristic $p$, all intermediate binomial coefficients vanish, so $\vartheta (x+y)=\vartheta (x)+\vartheta (y)$; multiplicativity and preservation of one are immediate. If $x^p=y^p$, then $(x-y{)}^p=0$ and the field property gives $x=y$. Thus this field homomorphism is injective, and finiteness makes it surjective. It is therefore a field automorphism, sending $\lambda$ to ${\lambda }^p$ as required.

For a simple field extension, the powers $1,\lambda ,\dots ,{\lambda }^{d-1}$ form a basis when the minimal polynomial has degree $d$: reduction modulo that polynomial spans the extension, and a relation of smaller degree would contradict minimality. Hence $[F(\lambda ):F]=d$. The field is $K$, whose dimension is $n$, giving $d=n$.

Regard $K$ as a vector space over its prime subfield ${\mathbb{F}}_p$. Since $K$ is finite, a maximal linearly independent list terminates and is a finite basis, of some dimension $n\ge 1$. Each field element has a unique coordinate vector in ${\mathbb{F}}_p^n$, so

If the positive characteristic were composite, say $r=uv$ with $1<u,v<r$, then $(u{1}_K)(v{1}_K)=0$ while neither factor is zero, by minimality of $r$. This contradicts a field having no zero divisors. Thus the characteristic is a prime $p$.

Starting with $A_0=A$, choose real shifts ${\sigma }_j$, factor $A_j-{\sigma }_{j}I=Q_j{R}_j$, and setEach step is an orthogonal similarity, so the eigenvalues are preserved and all iterates are similar. If $A_j^T=A_j$, transposing $Q_j^T{A}_j{Q}_j$ proves symmetry of $A_{j+1}$, completing the induction.

The given matrix is the Householder reflectionIt has eigenvalue $-1$ along $1$ and $+1$ on its orthogonal complement, hence $A^{T}A=I$. In the positive-diagonal QR convention its unshifted factorization is $Q=A$, $R=I$, so the next iterate is again $A$. Thus unshifted QR is stationary, despite the matrix generally having nonzero off-diagonal entries. If arbitrary diagonal signs are allowed in $R$, a step can instead replace $A$ by $DAD$ for a diagonal sign matrix $D$; every off-diagonal magnitude remains unchanged. Such sign changes still make no progress towards diagonalization.

Only the first and third coordinates occur in the first two columns. Normalize the first column to get $q_1=(4,0,3,0{)}^T/5$. Removing its component from the second column leaves $(21,0,-28,0{)}^T/5$, of length seven, so $q_2=(3,0,-4,0{)}^T/5$. The remaining orthonormal columns can be $q_3=e_2$ and $q_4=e_4$. Taking their dot products with the columns of $A$ givesThe columns of $Q$ are orthonormal, and direct multiplication gives the source matrix. Exploiting its two-coordinate first block avoids unnecessary four-dimensional elimination.

A QR factorization is $A=QR$, where $Q^{T}Q=I$ and $R$ is upper triangular. If $A$ is orthogonal, so is $R=Q^{T}A$. Its inverse is upper triangular, but $R^{-1}=R^T$ is lower triangular, forcing $R$ to be diagonal. Orthogonality then givesWith the customary positive diagonal convention, $R=I$.

The residual is orthogonal to $q\in {\Pi }_k$, so Pythagoras givesTaking nonnegative square roots proves $\boxed{\Vert q\Vert \le \Vert f\Vert }$.

The singular weight requires care about the space of admissible polynomials. Since $n\ge 1$, $x^n$ has finite squared norm. A polynomial $q$ has finite error relative to $x^n$ near zero exactly when $q(0)=0$: a nonzero constant term makes the integrand asymptotic to $q(0{)}^2/x$, whose integral diverges. If $q(0)=0$, the residual is $O(x)$ and the integrand is $O(x)$.

Thus the full space ${\Pi }_{n-1}$ is not a subspace of $V$, so the projection theorem cannot be applied to it as written. Nevertheless a unique best finite-error approximation does exist inFor $n=1$, this is the zero space and the answer is $q=0$. For $n\ge 2$, $W$ is finite dimensional and its Gram matrix is positive definite, because a nonzero polynomial cannot vanish on an interval. Solving the normal equations therefore gives its unique orthogonal projection of $x^n$. All omitted polynomials have infinite error, so this also minimizes the extended error over the full polynomial set. The distinction is an invalid full-space projection premise, not nonexistence of every finite-error best approximation. As throughout weighted $L^2$, use equivalence classes for the underlying measure.

Write $q=A+Bx$ and $\ell =\log 3$. Orthogonality of $x^2-q$ to $1$ and $x$ givesThe Gram matrix is positive definite, with determinant $4(\ell -1)>0$, so the solution is unique. Solving yields

Use the usual weighted $L^2$ interpretation, identifying functions equal almost everywhere; on raw pointwise functions the displayed integral is only a seminorm. Assume $q\in {\Pi }_k\subset V$. If $f-q$ is orthogonal to ${\Pi }_k$, then for every $p\in {\Pi }_k$,Conversely, suppose $q$ minimizes the error. For $r\in {\Pi }_k$ and every real $t$, compare $q$ with $q+tr$:If the linear coefficient were nonzero, a sufficiently small $t$ with the appropriate sign would make the expression negative. Thus $⟨f-q,r⟩=0$ for every $r$, proving best approximation is equivalent to orthogonality of the residual.