Euler's theorem states that $a^{\varphi (m)}\equiv 1(\mathrm{mod}m)$ whenever $\gcd (a,m)=1$, where $\varphi (m)$ counts the invertible residue classes. For each prime-power divisor, the same theorem gives $a^{\varphi (p_i^{e_i})}\equiv 1(\mathrm{mod}{p}_i^{e_i})$. Since the exponent defined in the source is a multiple of every one of these exponents, raising that congruence to the corresponding integer power givesThe prime powers are pairwise coprime, so their product divides $a^{\lambda (m)}-1$, proving the congruence modulo $m$.

For the displayed composite integer,Thus every integer coprime to 561 satisfies $\boxed{a^{560}\equiv 1(\mathrm{mod}561)}$. This is a Carmichael number. The source's prime-power least common multiple need not be the minimal universal exponent for all moduli; for powers of two its definition can exceed the usual Carmichael function. No minimality claim is needed here.

The degree $[K:F]$ is the dimension of $K$ as a vector space over $F$. The tower theorem states that for intermediate fields $F\subseteq E\subseteq K$ in a finite extension,If the degree is prime, one factor must be one, forcing $E=K$ or $E=F$; no strictly intermediate field exists.

Let $K$ have 32 elements. Its prime subfield is ${\mathbb{F}}_2$ and $[K:{\mathbb{F}}_2]=5$. Every element outside that subfield has minimal polynomial of degree five, by the tower theorem. Also every member of $K$ satisfies $x^{32}=x$: nonzero elements form a group of order 31. Hence $x^{32}-x$ splits over $K$, with distinct roots because its derivative is $-1=1$ in characteristic two.

Every irreducible degree-five polynomial over ${\mathbb{F}}_2$ divides $x^{32}-x$. Indeed its quotient field has 32 elements, and its residue class of $x$ satisfies that equation. Thus every such polynomial splits into five distinct roots in $K$. Conversely each of the 30 elements outside ${\mathbb{F}}_2$ has one such minimal polynomial, and different minimal polynomials have disjoint root sets. Consequently the number isThere is no extra leading-coefficient factor, since every nonzero leading coefficient over ${\mathbb{F}}_2$ is one.

Burnside's lemma counts the orbits of a finite group action:Identify each face with its opposite vertex. Every symmetry induces the same permutation on these two four-element sets, so a permutation with $c$ cycles fixes $3^c{2}^c=6^c$ colorings. The rotational symmetry group is $A_4$, with one identity, eight three-cycles, and three products of two disjoint transpositions. Their cycle counts are four, two and two. Thus the rotation count isAllowing reflections gives the full tetrahedral group $S_4$. In addition there are six transpositions with three cycles and six four-cycles with one cycle. The count becomesEquivalently, 30 of the 141 rotational classes form 15 mirror pairs, while 111 rotational classes are their own mirror class.

The product of fixed-coloring counts is an instance of the orbit count for simultaneous colorings.

The form is symmetric and bilinear by the corresponding properties of finite real sums. Its squared norm is $⟨p,p⟩={\sum }_{i}p(x_i{)}^2\ge 0$. Equality means $p$ vanishes at all $n+1$ distinct nodes. A nonzero polynomial of degree at most $n$ has at most $n$ roots, so equality forces $p=0$. Thus this is an inner product on ${\Pi }_n$.

On ${\Pi }_{n+1}$ it is only positive semidefinite: the nonzero node polynomial $\omega (x)={\prod }_{i=0}^n(x-x_i)$ satisfies $⟨\omega ,\omega ⟩=0$. Hence it is not an inner product on ${\Pi }_{n+1}$. The finite-evaluation inner product on polynomials works exactly when the evaluation map is injective on the chosen space.

Finally, if $q\ne 0$, positivity applied to $p-tq$ and the choice $t=⟨p,q⟩/⟨q,q⟩$ giveRearranging proves Cauchy-Schwarz inequality in its required form:If $q=0$ both sides of the asserted inequality are zero. Equality holds exactly when $p,q$ are linearly dependent, since in the nonzero case the minimizing $p-tq$ must have zero norm.

The Lagrange interpolation polynomial constructs existence explicitly:Distinct nodes make every denominator nonzero, and ${\ell }_j(x_k)={\delta }_{jk}$, so $p(x_k)=f_k$ and $p$ has degree at most $n$. If two such polynomials existed, their difference would have degree at most $n$ but at least $n+1$ distinct roots. The root bound for a nonzero polynomial forces that difference to vanish. Thus the interpolating polynomial exists and is unique.

Assume each sampled response is truthful, card selection is independent of the student's status, and responses follow the usual independent sampling model. In randomized response, a yes answer occurs either for a user receiving the first question or for a nonuser receiving the complementary question. Its probability isConsequently $Y\sim \mathrm{Bin}(n,q)$, and the likelihood isFor $x\ne 1/2$, first maximize the binomial likelihood in the unrestricted probability $q$: its maximum occurs at $Y/n$. Inverting the affine relation givesFor an interior value $0<Y<n$, the same result follows from the scoreThe endpoint cases follow by monotonicity. This estimator has expected value $p$ and varianceThese identities follow immediately from $\mathbb{E}Y=nq$ and $\mathrm{Var}(Y)=nq(1-q)$.

The printed claim needs a parameter-space qualification. The admissible yes probabilities lie in $[\min (x,1-x),\max (x,1-x)]$, which may exclude $Y/n$. The constrained maximum-likelihood estimator isThe displayed untrimmed estimator is the maximum-likelihood estimator only when it already lies in $[0,1]$. Its unbiased estimator property does not carry over to the clipped estimator in general. For example, if $n=1$, $x=3/4$ and $p=0$, the clipped estimate equals 1 exactly when the answer is yes, so its expectation is $1/4$, not zero. At $x=1/2$, $q=1/2$ for every $p$: the parameter is unidentifiable and the printed quotient is undefined. Thus the unbiased formula requires $x\ne 1/2$, and the constrained likelihood assertion requires the stated clipping.

After pooling, there are eight categories: $0,1,\dots ,6,\ge 7$. Their frequencies have one constraint because they sum to $N$, and one model parameter was fitted. Thus the reference chi-squared distribution hasdegrees of freedom. The reported statistic satisfies $25.914>22.46$, the stated upper $0.001$ cutoff for six degrees of freedom. Therefore reject the Poisson model at the 0.1% significance level under the independence and common-mean assumptions. The discrepancy is especially pronounced in the zero-count class, but the test assesses the whole fitted distribution. The conclusion uses the approximate Pearson chi-squared goodness-of-fit test, not an exact finite-sample tail probability.

Convergence in probability means that for every $\epsilon >0$, $\mathbb{P}(|X_n-X|\ge \epsilon )\to 0$. The use of $>$ instead of $\ge$ gives the equivalent definition by replacing $\epsilon$ with a smaller positive number.

For $Y\ge 0$, the pointwise inequality $Y\ge a{1}_{\{Y\ge a\}}$ gives, on taking expected values, Markov's inequalityThis also holds if the expectation is infinite, though the resulting bound is then uninformative. Apply it to $Y=(X-\mu {)}^2$ with threshold ${\epsilon }^2$ to obtain Chebyshev's inequality:Set $S_n=n^{-1}{\sum }_{i=1}^n(Z_i-\mathbb{E}{Z}_i)$. Its mean is zero. Independence eliminates cross covariances, and boundedness gives $\mathrm{Var}(Z_i)\le \mathbb{E}{Z}_i^2\le 1$. HenceThus $\boxed{S_n\to 0\text{ in probability}}$. Identical distributions were never needed.

Start from the uniform-flow complex potential of a two-dimensional potential flow $Uz$. Milne-Thomson's circle theorem adds the image doublet $U{a}^2/z$, yielding constant stream function on $|z|=a$. Add a line vortex with positive counterclockwise circulation $\Gamma$:Indeed, on $z=a{e}^{i\theta }$ the uniform-flow and doublet terms sum to the real number $2Ua\cos \theta$, while the vortex has imaginary part $-\Gamma \log a/(2\pi )$, also constant. Thus the cylinder is a streamline and satisfies the impermeability condition. The vortex velocity is $\Gamma /(2\pi r)$ in the counterclockwise direction. Although the potential is multivalued when $\Gamma \ne 0$, its velocity is single-valued.

Differentiate and locate the coefficient of $z^{-1}$ in the square:For the counterclockwise contour in the printed Blasius force theorem, the residue theorem gives ${\oint }_C(w^{\prime }{)}^{2}dz=2U\Gamma$. ConsequentlyThese are force components per unit cylinder length; with the stated circulation convention the lift is downward for positive $U\Gamma$.