Modulo $p$, the term $p^{q-1}$ is zero and Fermat's little theorem gives $q^{p-1}\equiv 1$, since $q\ne p$. Modulo $q$, the roles reverse. Hence the sum minus one is divisible by both distinct primes. Their coprimality gives

Combine the preceding congruences: $a^p\equiv a$ and $(p-1)!\equiv -1(\mathrm{mod}p)$. Therefore $a^p+(p-1)!a\equiv a-a=0$, and $(p-1)!a^p+a\equiv -a+a=0$. This proves both stated divisibilities for every integer $a$, with no assumption that $a$ is coprime to $p$.

Wilson's theorem states that an integer $p>1$ is prime if and only ifIn particular the congruence holds for every prime $p$, including two.

Fermat's little theorem states that for a prime $p$ and any integer $a$,If $p\nmid a$, the equivalent multiplicative form is $a^{p-1}\equiv 1(\mathrm{mod}p)$.

Replacing $x$ by $-x$ in the irreducible polynomial from part (i) gives $x^6-x^3+1$. This substitution is an invertible automorphism of $\mathbb{Q}[x]$, so it preserves irreducibility. The roots are the primitive eighteenth roots, and the splitting field is $\mathbb{Q}({\zeta }_{18})$. They are all powers of one such root, so its irreducible degree-six polynomial givesIn fact this is the same field as in part (i): ${\zeta }_9={\zeta }_{18}^2$ and ${\zeta }_{18}=-{\zeta }_9^5$.

The splitting field in $\mathbb{C}$ is the smallest subfield containing $\mathbb{Q}$ and every root of the polynomial. The sixth roots of unity are powers of ${\zeta }_6=e^{i\pi /3}$, so the splitting field of $x^6-1$ is $\mathbb{Q}({\zeta }_6)=\mathbb{Q}(\sqrt{-3})$.

For $x^6+x^3+1=(x^9-1)/(x^3-1)$, the roots are exactly the primitive ninth roots of unity, all powers of ${\zeta }_9=e^{2\pi i/9}$. Thus its splitting field is $\mathbb{Q}({\zeta }_9)$. The supplied irreducibility makes the displayed degree-six polynomial the minimal polynomial of ${\zeta }_9$, giving

The identity columns in the final tableau correspond to basic variables $x_2,x_3$. Their original basis matrix isalso visible directly in the two slack columns. Multiply the original constraint tableau and right side by $B^{-1}$ to obtain the missing entries. Subtract four times the first row and three times the second from the initial objective row, using the basic costs. The completed simplex tableau isThe last row uses the supplied convention of reduced costs $c_j-c_B^T{B}^{-1}{A}_j$, so its final right side is minus the objective. Setting the nonbasic variables to zero givesThe dual feasible vector $y=(1,1)$ has $A^{T}y=(3,4,3{)}^T\ge (1,4,3{)}^T$ and objective $4+6=10$. Part (a) independently certifies the optimum and confirms the objective-row sign.

The dual linear program isFor feasible $x,y$, nonnegativity and the two constraint inequalities imply weak duality:If a feasible pair has equal objectives, every primal feasible objective is at most that common value and every dual feasible objective at least it. Thus both feasible solutions are optimal. Strong duality is not needed for this certificate.

For the shifted normal proposal, $w(x)=e^{-\mu x+{\mu }^2/2}$ and the estimator isLet $\bar{\Phi }=1-\Phi$. Completing the square gives its unaveraged second momentThe variance is $[I(\mu )-\bar{\Phi }(z{)}^2]/N$, so only $I$ needs minimization. It tends to infinity at both ends of the real $\mu$ axis and has a finite minimizer. Differentiation at a minimizer yieldsFor a standard normal variable, integration of $x\phi (x)=-{\phi }^{\prime }(x)$ shows $E_p[X\mid X>a]=\phi (a)/\bar{\Phi }(a)$. ConsequentlyThe shift is chosen to reduce the second moment of the weighted rare-event indicator, rather than merely to increase the proposal's tail probability.

The importance weight for the two normal densities isUse ${\hat{\mu }}_N=N^{-1}{\sum }_i{w}_{\sigma }(X_i)f(X_i)$ with $X_i\sim N(0,1)$. Its second moment before averaging isFor the source's general assumption $E_p{f}^2<\infty$, a guaranteed range isIndeed $w_{\sigma }\le 1/\sigma$ there, so $E_q(w_{\sigma }^2{f}^2)=E_p(w_{\sigma }{f}^2)\le {\sigma }^{-1}{E}_p{f}^2<\infty$.

This range cannot be enlarged uniformly over all such $f$. If $\sigma >1$, choose $f^2=e^{a{x}^2}$ withThen $E_p{f}^2<\infty$ but the displayed weighted second-moment integral diverges. Thus no variance range depending only on $\sigma$ can characterize every individual integrand beyond the guaranteed range.

If an additional boundedness assumption on $f$ is intended, the familiar sufficient range becomes $0<\sigma <\sqrt{2}$. For $f=1$ this condition is also necessary, but a compactly supported $f$ can have finite weighted variance for every $\sigma$. Distinguishing these assumptions prevents imposing an unstated restriction on the source integrand.

For unknown normalization constants, use self-normalized importance sampling with $\tilde{w}=\tilde{p}/\tilde{q}$:Under the same support condition, $E_q\tilde{w}=Z_p/Z_q>0$ and $E_q[\tilde{w}f]=(Z_p/Z_q)\mu$. Moreover $E_q|\tilde{w}f|=(Z_p/Z_q)E_p|f|<\infty$, by Cauchy-Schwarz inequality and the supplied $E_p{f}^2<\infty$. The strong law of large numbers applies to numerator and denominator sample averages. Their ratio tends almost surely, hence in probability, to $\mu$. If a finite sample happens to give zero denominator, it can be assigned an arbitrary value; this does not affect the eventual almost sure limit. The ratio estimator is generally biased at finite $N$.

Assume the necessary support condition for importance sampling, $p\ll q$; the source's density assumptions alone do not ensure it. Put $\mu =E_{p}f$ and $w=p/q=(Z_q/Z_p)\tilde{p}/\tilde{q}$. The importance sampling estimator isIntegration under $q$ gives $E_q[wf]=\int pf=\mu$, proving unbiasedness. Independence givesThis variance is finite precisely when the first integral is finite. The assumption $E_p{f}^2<\infty$ ensures target square integrability, but need not ensure square integrability after importance reweighting; the variance can be infinite.

In a Prüfer code, a vertex label occurs exactly its degree minus one times. A rooted binary tree with $2m+1$ vertices has one root of degree two, $m-1$ other interior vertices of degree three, and $m+1$ leaves. Its code of length $2m-1$ therefore has one label occurring once, $m-1$ labels occurring twice, and all other labels absent.

Conversely, these multiplicities force exactly those degrees in the tree decoded from the sequence. The unique once-occurring label is its root. This is the complete characterization.

If there are $i$ nonroot interior vertices and $\ell$ leaves, then $1+i+\ell =7$ and the tree degree sum is $2+3i+\ell =12$. Hence $i=2$ and $\ell =4$. Distribute these two nonroot interior vertices among the root's two descendant branches.

Either each branch contains one interior vertex, producing the balanced tree with two cherries, or both are in one branch while the other branch is a leaf. The latter gives a single unbalanced shape, because left and right children are not ordered. Thus there are two unlabelled rooted binary trees. A plane ordering of children would be additional structure not present in the question.

Take root one, its children two and three, the children of two to be four and five, and those of three to be six and seven. Its edge set is $\{12,13,24,25,36,37\}$. The root has degree two, vertices two and three have degree three, and the remaining vertices have degree one.

Removing the smallest available leaves in order gives $4,5,2,1,6$, with recorded neighbors $2,2,1,3,3$. Therefore its Prüfer code is

Decode the Prüfer code by repeatedly attaching the smallest label absent from the remaining code to its first entry. This produces the successive edges $1-2$, $3-6$, $4-2$, $5-2$, followed by the final edge $2-6$. Thus

Under the recessive alternative, CT and TT have the same status as non-CC individuals, so their separate labels do not distinguish the proposed risk groups. Pooling them makes the intended comparison CC versus non-CC. The table becomesThe independence test on this two-by-two table has one degree of freedom. Since $8.41>3.84$, reject no association at the 5% level; its chi-squared upper-tail probability is approximately $0.0037$. CC is more frequent among cases, consistent with the stated directional alternative. The supplied Pearson statistic is conventionally compared with its one-degree-of-freedom chi-squared reference.

The null reference is ${\chi }_2^2$, because the alternative has two additional parameters. The supplied statistic satisfies $8.42>5.99$, so reject no association at the 5% level. For two degrees of freedom the upper-tail probability is $e^{-8.42/2}\approx 0.0148$.

The result is evidence that genotype proportions differ between cases and controls. It is a statistical association test; it does not by itself identify a causal mechanism.

Under the null, the two groups share one vector of three category probabilities. The sum-to-one constraint leaves two free parameters. Under the unrestricted alternative, each group has its own such probability vector, leaving four free parameters. The fixed case and control totals introduce no further parameters; the difference is two, giving the chi-squared reference degrees of freedom.