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The characteristic of a field is the least positive integer $p$ with $p{1}_F=0$, or zero if none exists. Finiteness forces a positive characteristic, since the sequence of integer multiples of one must repeat. If this least integer factored into two smaller positive factors, their nonzero field multiples would multiply to zero, contradicting the absence of zero divisors. Hence it is prime. The field is a finite-dimensional vector space over its prime subfield ${\mathbb{F}}_p$, so choosing a basis gives $|F|=p^a$.

A nonzero polynomial of degree $n$ over a field has at most $n$ roots: each root supplies a linear factor by polynomial division, and induction applies to the quotient. In particular $X^n-1$ has at most $n$ roots. The elements of $F^{\times }$ whose orders divide $n$ are precisely those roots. The permitted finite-group criterion therefore proves $F^{\times }$ is cyclic.

For the final assertion take an odd Fermat prime $p=2^m+1$, $m\ge 1$. Primality forces $m$ to be a power of two: an odd factor $s>1$ in $m=2^{r}s$ would factor $(2^{2^r}{)}^s+1$ nontrivially. Since $2^m\equiv -1(\mathrm{mod}p)$, the order of two divides $2m$ but not $m$. As $2m$ is a power of two, its only divisor not dividing $m$ is $2m$ itself. ThusIt generates the full group exactly when $2m=2^m$, which holds for $m=1,2$ and fails for $m\ge 3$. Therefore the possibilities are $p=3,5$, equivalently $p\le 5$ among odd Fermat primes. The oddness is implicit in referring to two as a unit; $p=2$ is not an admissible case for that claim.

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.

The characteristic of a field is the least positive integer $r$ with $r{1}_K=0$, or zero if none exists. In a finite field a positive such integer exists because the sequence $0,1_K,2\cdot 1_K,\dots$ must repeat.

The prime subfield is the intersection of all subfields of $K$, equivalently the subfield generated by $1_K$. In characteristic $p$ it is the copy of ${\mathbb{F}}_p$ formed by the multiples of $1_K$. Every field automorphism preserves $1_K$ and addition, so it fixes every element of this subfield.

Finally, once the order-$n$ Frobenius automorphism below is established, all its $n$ powers are distinct. Any automorphism fixes $F$ and is determined by its value at the generator $\lambda$. That value must be a root of the degree-$n$ minimal polynomial of $\lambda$, so there are at most $n$ automorphisms. Hence

Suppose all primes congruent to $3$ modulo eight are $p_1,\dots ,p_r$, and put $N=p_1\cdots p_r$, an odd integer. The integer $N^2+2$ is odd and congruent to three modulo eight. If a prime $q$ divides it, then $N^2\equiv -2(\mathrm{mod}q)$ and $q\nmid N$, since a common divisor of $N$ and $N^2+2$ would divide two. Thus $(-2/q)=1$, so part (v) forces $q\equiv 1$ or $3(\mathrm{mod}8)$.

If all prime factors congruent to three had even multiplicity, their product, including the factors congruent to one, would be one modulo eight. Since $N^2+2$ is three modulo eight, at least one prime factor is congruent to three and occurs to odd multiplicity. But it is none of the $p_i$, since those divide $N$. This contradiction proves infinitely many primes are congruent to three modulo eight.

For $n\ge 1$, the zero residue contributes zero. Write $S={\sum }_{x\in {\mathbb{F}}_p^{\times }}{x}^n$. Multiplication by any nonzero $\alpha$ permutes these residues, so $S={\alpha }^{n}S$ in ${\mathbb{F}}_p$. If $(p-1)\nmid n$, choose a primitive root $g$; then $g^n\ne 1$, and $(1-g^n)S=0$ forces $S=0$.

If $(p-1)\mid n$, every nonzero $x$ has $x^n=1$, so $S=p-1\equiv -1(\mathrm{mod}p)$. ThereforeThe source also lists $n=0$, although its premise assumes $n\ge 1$. For that separate case, the integer sum has $p$ terms each equal to one, so it is zero modulo $p$.

For every integer $j\ge 0$, the $j$th derivative with respect to $t$ of $\cos (t\cos x)$ is $(\cos x{)}^j$ times a sine or cosine with a sign. Its absolute value is at most $1$ on the finite interval $[0,\pi /2]$. This is an integrable dominating function independent of $t$. Repeated differentiation under the integral sign is therefore justified for every order, giving $J_0\in C^{\infty }(\mathbb{R})$.

For an odd prime number $p$, a nonzero class $a(\mathrm{mod}p)$ is a quadratic residue if $a\equiv x^2(\mathrm{mod}p)$ for some integer $x$ not divisible by $p$. The Legendre symbol isWhen discussing solvability of $x^2\equiv a(\mathrm{mod}p)$ without a coprimality assumption, zero is also a square. This convention matters in the final local characterization of integer squares.

Every integer square remains a square modulo every prime number, including the zero class at its prime divisors. Conversely, let $a$ be an odd integer which is not an integer square. If $a$ is the negative of a square, part (i) supplies an odd $k$ with $(a/k)=-1$. In every other case, the prime factorization of $|a|$ has an odd exponent $t$ at some odd prime $q$. Write $a=\pm q^{t}b$ with $b>0$ odd and $\gcd (q,b)=1$, and apply part (ii).

In either case, factor the resulting $k$ into primes. Since its Jacobi symbol is a product of Legendre symbols and equals $-1$, at least one prime factor $p$ has $(a/p)=-1$. Thus $a$ is not a square modulo that prime. This proves an odd integer is an integer square exactly when it is a square modulo every prime. If the phrase quadratic residue is restricted to nonzero residue classes, the final statement must instead say “a square modulo every prime”; for example, $9$ reduces to zero modulo $3$.

Choose a nonzero quadratic nonresidue $c(\mathrm{mod}q)$. There is one because the squaring map on ${\mathbb{F}}_q^{\times }$ is two-to-one, so exactly half its elements are squares. The moduli $4,q,b$ are pairwise coprime, so the Chinese remainder theorem gives a positive integer $k$ satisfyingWhen $b=1$, the last condition is vacuous. In particular $k$ is odd and coprime to $qb$. Since $k\equiv 1(\mathrm{mod}4)$, Jacobi reciprocity givesAs $t$ is odd, multiplicativity of the Jacobi symbol now yields

Here $b$ is odd and nonzero. Choose $k=4|b|-1$. This is a positive odd integer, $k\equiv 3(\mathrm{mod}4)$, and $\gcd (k,b)=1$. By multiplicativity of the Jacobi symbol,The coprimality condition ensures that the squared Jacobi symbol is $1$ rather than $0$.

False. LetThis is a bounded measurable function, and the intervals are disjoint, soNevertheless $k(n)=1$ for every positive integer $n$, while $k(n+1/2)=0$. Hence it has no limit at infinity. Narrow spikes separate Lebesgue integrability from pointwise decay.

A constructible regular polygon can be translated, rotated, and rescaled by constructible operations to one inscribed in the unit circle with a vertex at $(1,0)$. Thus the real and imaginary parts of $\zeta =e^{2\pi i/p}$ are constructible. Part (c) puts both in a real field $K$ of degree $2^n$ over $\mathbb{Q}$. Consequently $\zeta \in K(i)$, and $[K(i):\mathbb{Q}]=2^{n+1}$ because $K\subset \mathbb{R}$.

For prime $p$, the supplied irreducibility of $1+x+\cdots +x^{p-1}$ shows that $[\mathbb{Q}(\zeta ):\mathbb{Q}]=p-1$. The intermediate-field tower law implies that $p-1$ divides $2^{n+1}$. Every positive divisor of a power of two is a power of two, soThis necessary condition includes $p=2$ with $r=0$. It does not assert sufficiency for arbitrary integers of this form.

For an integer $a$ and an odd prime $p$, the Legendre symbol isThe value one refers to a nonzero quadratic residue; divisibility by $p$ is recorded separately by zero.

Reducing modulo $23$ gives $189\equiv 5$. Reciprocity with five has positive sign, sosince the nonzero squares modulo five are one and four. Modulo $59$, $189\equiv 12=4\cdot 3$, and $59\equiv -1(\mathrm{mod}12)$. Thus part (b) givesAs $1357=23\cdot 59$, a solution modulo $1357$ would in particular solve the modular congruence modulo $23$, contradicting the first symbol. Therefore there is no integer solution modulo $1357$.

Let $D=[0,1]\times [-1,1]$ and $q=p\circ i:D\to S/{\mathcal{R}}^{\prime }$. This is continuous. It is onto: given $(x,y)\in S$, choose an integer shift bringing $x$ into $[0,1]$, with the corresponding possible sign change in $y$; the resulting representative lies in $D$.

Two representatives in $D$ are in the same orbit only if their first coordinates differ by an integer. An interior point can therefore be equivalent only to itself. Distinct boundary representatives have first coordinates zero and one, and the iterate $g$ identifies exactly $(0,y)$ with $(1,-y)$. Thus the fibres of $q$ are precisely the equivalence classes defining the Möbius band $M$.

The rectangle $D$ is compact, and the target is Hausdorff by part (i). Applying part (a)(v) to $q$ therefore yieldsThe compactness argument supplies continuity of the inverse without requiring a discontinuous choice of one representative per orbit.

For nonnegative integers $j,k$, repeated integration by parts gives the beta integral ${\int }_0^1{y}^j(1-y{)}^{k}dy=j!k!/(j+k+1)!$. Multiplying the stated density by $y$ and using this formula yieldsCancellation gives

For $k=0$, the target is reached at time zero, so $h_0=1$. For $k<0$, the preceding almost sure convergence gives $Y_n\to +\infty$ with probability one. A sample path with unit upward and downward steps must visit every intermediate integer when it crosses from a negative value to a positive value. It must therefore visit zero at a finite time. ThusThe unit-step property matters: positive long-run displacement alone would not force a walk with larger upward jumps to land exactly on zero.

Work on the single event of probability one where $Y_n/n\to \mu =2p-1>0$. On every sample path in that event, $Y_n\to +\infty$ and the unit-step property ensures that every positive integer $j$ is reached, so $T_j<\infty$. Starting from zero requires at least $j$ steps to reach $j$, hence $T_j\ge j$ and $T_j\to \infty$.

The convergent numerical sequence $Y_n/n$ on this fixed sample path also converges along its possibly random subsequence $n=T_j$. Since $Y_{T_j}=j$,Taking the reciprocal is valid because $\mu >0$. We obtainThis is a first-passage law of large numbers for an upward skip-free random walk. No independence between the hitting times and the path is needed: almost sure convergence already supplies convergence along every diverging subsequence on the same event.

Since $Y_n\to +\infty$ almost surely, only finitely many early values can attain the global minimum; it is a finite integer. Also $W\le Y_0=0$. For every nonnegative integer $m$, the unit-step property and translation invariance of the increments giveSubtracting consecutive tails yields the probability mass functionThus $-W$ has the geometric distribution counting failures before the first success, with success parameterThis support convention is essential: the minimum of an upward-biased random walk can equal its starting position, so $-W=0$ has positive probability.

Source: Oxford's released examination PDF, pages 164–167, dated 21 June 2021. Questions 1, 2 and 3 occupy bundle pages 165, 166 and 167 respectively, or pages 2, 3 and 4 of the individual paper. All three questions and every part are solved below, including all six revisions requested in 3(d). In 3(c), $c$ is understood to be an integer: the printed transition rule does not define a transition matrix for noninteger $c$.