The Legendre symbol is

Reduce the monic integer polynomial modulo the prime $p$. For every $x\in {\mathbb{F}}_p$, $x^p=x$, soIf the reduction were reducible, a proper monic factorization would, by part (2), force a root in ${\mathbb{F}}_p$, a contradiction. Its reduction is therefore irreducible of degree $p$. The monic reduction criterion, obtained from Gauss lemma for polynomials by reducing a hypothetical integer factorization, now proves

Interpret the displayed factorization as a proper factorization, with $1\le r\le p-1$; that restriction is necessary for the conclusion. In a splitting field, the monic degree-$r$ factor has $r$ of the distinct roots of the original polynomial. Comparing its coefficient of $X^{r-1}$ givesfor an $r$-element subset $J\subseteq {\mathbb{F}}_p$. Both $a_1$ and the final sum lie in $F$. Since $0<r<p$, the element $r$ is invertible in $F$, soPart (1) then gives complete splitting into distinct linear factors. Thus an Artin–Schreier polynomial of this form either is irreducible or splits completely; a trivial factorization with $r=0$ or $r=p$ would not imply a root in $F$.

In characteristic $p$, the binomial theorem gives $(\alpha +1{)}^p={\alpha }^p+1$. HenceIterating shows that all $\alpha +j$, with $j\in {\mathbb{F}}_p$, are roots. They are distinct, since the prime-field elements $j$ are distinct. There are $p$ of them, equal to the polynomial's degree. Thus if $\alpha \in F$,splits into distinct linear factors in $F[X]$. Equivalently, its derivative is $-1$, so no repeated roots are possible.

The transformation $N=k/p$ is strictly decreasing for $p>0$. Therefore an interval $[L,U]$ with $0<L<U\le 1$ transforms toThe event $L\le p\le U$ is exactly the event $k/U\le N\le k/L$, so the approximate coverage is inherited. If the lower endpoint for $p$ reaches zero, the transformed upper endpoint for $N$ is $+\infty$; if the upper endpoint reaches one, the lower endpoint is $k$. Finally intersect with integer values $N>k$, rounding the lower bound upward and the upper bound downward. This respects the original discrete population constraint.

For a fixed true parameter $0<p<1$, the usual asymptotic normality of maximum likelihood estimators givesEquivalently, for a large sample, $\hat{p}$ is approximately normally distributed with mean $p$ and variance $p^2(1-p)/(nr)$. Replacing the unknown variance by its consistent plug-in estimate gives the Wald confidence intervalHere $z_{1-\alpha /2}$ is a standard-normal quantile. Intersect with the physical parameter range $(0,1)$ if needed. This is an interior, large-sample approximation; the degenerate estimate $\hat{p}=1$ from $S=0$ does not justify a zero-width finite-sample interval.

Let $S={\sum }_i{x}_i$ and $\bar{x}=S/n$. Terms independent of $p$ can be dropped from the log-likelihood, leavingFor $S>0$, strict concavity gives the unique interior maximum likelihood estimatorIf $S=0$, the likelihood increases up to $p=1$: this is the boundary estimate on the closed parameter range, or a supremum if the model insists on $p<1$.

Using the given mean, the sample Fisher information isThe information per observation is $I_1(p)=r/[p^2(1-p)]$.

Let $u_n=\mathbb{P}(X_n=1\mid X_0=1)$. Conditioning on $X_n$ givesThe fixed value is $b/(a+b)$. Subtracting it and iterating the scalar recurrence yieldsThe formula includes $n=0$. If $a+b=1$, its value at $n=0$ is interpreted directly as $u_0=1$, while $u_n=b$ for all $n\ge 1$. Since $|1-a-b|<1$, the limiting probability is $b/(a+b)$.

For $n=0$, the conditional transition probabilities are ${\delta }_{ij}$, the entries of $P^0=I$; for $n=1$ the assertion is the definition of $P$. For the induction step, condition on the state at time $n$ and use the Markov property and time-homogeneity:The countable sum is justified by the law of total probability; all terms are nonnegative. Thus the formula follows for every $n$ by induction. It is the Chapman-Kolmogorov equation for repeated one-step transitions.

The preceding marginal calculation makes each $V_i$ uniformly distributed on $(0,1)$, with expectation $1/2$ and variance $1/12$. By independence,Apply part (a) to the centred random variable $Z=S_n/n-1/2$ to obtainThis is Chebyshev's inequality in this setting.

The inverse transformation is $x=uv$, $y=u(1-v)$, with support $u>0$, $0<v<1$. Its absolute Jacobian determinant isThe original joint probability density function, by independence, is ${\lambda }^2{e}^{-\lambda (x+y)}$ on the positive quadrant. The change of variables formula therefore givesIntegrating over $v$ givesthe gamma distribution of shape two and rate $\lambda$. Integrating over $u$ gives $f_V(v)=1$ on $(0,1)$. The factorization also shows that $U$ and $V$ are independent.

Pointwise,Taking expectations and dividing by $a^2>0$ gives the second-moment form of Markov's inequality:

Let $\mathbf{F}$ be the force on particle one, so the force on particle two is $-\mathbf{F}$, with $\mathbf{F}$ parallel to ${\mathbf{q}}_1-{\mathbf{q}}_2$. The combined torque isHence the combined angular momentum of particles one and two is conserved. Centrality of their interaction, in addition to equal and opposite forces, is what makes this torque sum zero.

The third particle is force-free, so its torque about the origin vanishes:Thus this angular momentum is conserved.

For the first particle, the derivative of its angular momentum about the origin is its torque:This need not vanish. For example take $f(r)=r^2/2$ and instantaneous positions ${\mathbf{q}}_1=(1,0,0)$, ${\mathbf{q}}_2=(0,1,0)$; the torque is $(0,0,1)$. Therefore it is not conserved in general, although it is in special configurations or for a constant potential.

Use constraints $h_1=x^2/4+y^2+z^2-1$ and $h_2=x+y-z$. Both constraints and the objective are continuously differentiable. The feasible set is a nonempty closed subset of a bounded ellipsoid, hence compact, so the extreme value theorem guarantees both extrema.

The gradients $(x/2,2y,2z)$ and $(1,1,-1)$ are independent everywhere feasible. If the first were $t$ times the second, then $x=2t$, $y=t/2$, $z=-t/2$. The plane constraint would give $3t=0$, forcing the origin, which is not on the ellipsoid. Thus the Lagrange multiplier theorem applies at every feasible point.

Its equations areHere $\lambda \ne 0$, since otherwise the first two equations require $\mu =1=2$. Subtract twice the first equation from the second, and use $\mu =2\lambda z$ and $z=x+y$; the result is $0=-3\lambda x$, so $x=0$. Then $z=y$ and the ellipsoid gives $y=\pm 1/\sqrt{2}$. The only candidates are therefore $(0,\pm 1/\sqrt{2},\pm 1/\sqrt{2})$, with matching signs. Consequently

There are real Lagrange multipliers ${\lambda }_1,{\lambda }_2$ such thatEquivalently $Dg(x_0)={\lambda }^{T}Df(x_0)$. The rank assumption says the two constraint gradients are linearly independent; it rules out an abnormal constraint point.