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Suppose only finitely many primes are congruent to five, and list them as $p_1,\dots ,p_k$. Consider $N=6p_1\cdots p_k-1>1$. It is divisible by neither two nor three, so every prime divisor is congruent to one or five modulo six. If all its prime factors were congruent to one, their product would be congruent to one by part (i), whereas $N\equiv 5$. Hence some prime divisor $q$ has $q\equiv 5(\mathrm{mod}6)$.

But no listed $p_i$ divides $N$, since $N\equiv -1(\mathrm{mod}{p}_i)$. This gives a new such prime, a contradiction. Therefore infinitely many primes are congruent to five modulo six. The argument does not require the specially constructed integer itself to be prime.

If $\lambda =0$, only a linear function can solve the equation, and zero endpoint values make it zero. If $\lambda <0$, the corresponding hyperbolic-sine solution also cannot vanish at both endpoints unless it is zero. For $\lambda =k^2>0$, the first endpoint gives $y=C\sin kx$, and the second requires $k=n\pi$ for a positive integer $n$. Thus $\boxed{{\lambda }_{\min }={\pi }^2}$.

The admissible trial function $y=\sin \pi x$ has quotientTherefore the minimum, or infimum before its attainment is established, is $\boxed{\le {\pi }^2}$ as requested. The sharp Poincaré inequality supplies the reverse inequality on the finite-energy zero-trace class, so the exact minimum is also ${\pi }^2$ and is attained by this trial function. The one-sided assertion in the source does not require a full variational existence argument.

For hard-wall confinement, the stationary spatial modes and energies areTheir stationary time dependence is $e^{-i{E}_{n}t/\hslash }$. The correct triple-angle identity is $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$; the printed hint has its right side reversed in sign. Thus the specified initial state expands asThe coefficients have squared moduli summing to one. The first requested energy is $E_1$, so its probability is $\boxed{9/10}$. The other requested energy is $2E_1$, which is not $n^2{E}_1$ for any positive integer $n$, so its probability is $\boxed{0}$. The occupied excited energy is instead $E_3=9E_1$, with probability $1/10$.

Eisenstein criterion says that a primitive integer polynomial is irreducible over $\mathbb{Q}$ if some prime divides all coefficients except the leading one, does not divide the leading coefficient, and its square does not divide the constant coefficient. The passage from primitive integer factorization to rational factorization is Gauss lemma for polynomials.

Translate the variable:This is Eisenstein at two. Translation is an automorphism of $\mathbb{Q}[x]$, so $x^8+1$ itself is irreducible.

Set $\zeta =e^{i\pi /8}$ and $c=\cos (\pi /8)$. Since $\zeta$ is a root of that irreducible degree-eight polynomial, $[\mathbb{Q}(\zeta ):\mathbb{Q}]=8$. Also $c=(\zeta +{\zeta }^{-1})/2$, so $\mathbb{Q}(c)\subseteq \mathbb{Q}(\zeta )$ and $\zeta$ solves $T^2-2cT+1=0$. The extension degree is at most two. It is not one, since every element of $\mathbb{Q}(c)$ is real and $\zeta$ is not. The tower law for field extensions now gives

Let $A_n=\{x:f(x)\ge 1/n\}$. To avoid presupposing finite measure of $A_n$, first truncate it: $A_{n,k}=A_n\cap [-k,k]$ is measurable and has finite measure. Its indicator divided by $n$ is integrable and is bounded above by $f$. The allowed monotonicity property givesThus every $A_{n,k}$ is null. Their countable union $A_n$ is null by part (a), and so is $E={\bigcup }_n{A}_n$. Every positive value of $f$ is at least $1/n$ for some integer $n$, hence $\{f>0\}=E$. Consequently $f=0$ outside a null set.

Wilson's theorem states that an integer $p>1$ is a prime number if and only ifIn particular this holds for every prime, including $p=2$.

Fermat's little theorem says that for a prime number $p$ and any integer $a$,If $p\nmid a$, cancellation gives the equivalent statement $a^{p-1}\equiv 1(\mathrm{mod}p)$.

Because the primes are distinct, Fermat's little theorem gives $q^{p-1}\equiv 1(\mathrm{mod}p)$ and $p^{q-1}\equiv 1(\mathrm{mod}q)$. The other term in each modulus vanishes. ThereforeThe distinct primes are coprime, so their product divides this integer. Hence

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.

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

For a positive integer $n$, Euler's totient function counts the integers in a complete residue system that are coprime to $n$:For $n>1$ this is equally the count with $0<k<n$. The convention gives $\phi (1)=1$.

Map the integers into a field by $n\mapsto n1$. Its kernel is a prime ideal of $\mathbb{Z}$, since the image has no zero divisors. Thus the kernel is either zero or $(p)$ for a prime $p$. In the zero case, the map extends injectively to the rational numbers by taking quotients of nonzero integers. In the other case, its image is the field ${\mathbb{F}}_p$. This proves the prime-field assertion.

A finite field cannot contain the infinite rational field, so it has characteristic $p$. It is a finite-dimensional vector space over its prime field ${\mathbb{F}}_p$, say of dimension $n\ge 1$, and choosing a basis counts its elements asFor existence, put $q=p^n$ and take a splitting field of $x^q-x$ over ${\mathbb{F}}_p$. Its derivative is $-1$, so it has exactly $q$ distinct polynomial roots. The polynomial roots are closed under addition and multiplication because the Frobenius identity gives $(u+v{)}^q=u^q+v^q$ and $(uv{)}^q=u^q{v}^q$. They include $0,1$, are closed under additive inverses, and for a nonzero polynomial root $u$, $(u^{-1}{)}^q=(u^q{)}^{-1}=u^{-1}$. They therefore form a field of exactly $q$ elements. Indeed this field is the splitting field because it contains all polynomial roots and is generated by them.

In order 4, characteristic is 2. For $\omega \ne 0,1$, the multiplicative group has order 3, so ${\omega }^3=1$ and ${\omega }^2\ne 1,\omega$. Factorization of $x^3-1$ gives ${\omega }^2+\omega +1=0$. The quadratic $x^2+x+1$ has no polynomial root in ${\mathbb{F}}_2$ and is the minimal polynomial. Write $s={\omega }^2=\omega +1$. The complete tables areFor an element $\xi \ne 0,1$ in order 8, the tower law makes $[{\mathbb{F}}_2(\xi ):{\mathbb{F}}_2]$ divide3. It is not1, so its minimal polynomial is an irreducible monic cubic. Its constant term must be1. A cubic $x^3+a{x}^2+bx+1$ has no polynomial root at 1 exactly when $a+b=1$, and absence of a polynomial root is equivalent to cubic irreducibility. Consequently the two possibilities areBoth occur: together with $x$ and $x+1$ they factor $x^8-x$ over ${\mathbb{F}}_2$, whose distinct polynomial roots are exactly the eight elements of the field. Each cubic contributes its three polynomial roots.

Euler's totient function $\varphi (n)$ counts the integers $1\le a\le n$ coprime to $n$, equivalently the invertible residue classes modulo $n$. Among $1,\dots ,p^e$, exactly the $p^{e-1}$ multiples of $p$ are not coprime to the prime power. HenceFactor $n={\prod }_p{p}^{e_p}$. Its distinct prime-power factors are pairwise coprime, so the permitted multiplicativity givesFor $n=1$, the empty product is one and agrees with $\varphi (1)=1$.

If a prime divisor $p$ of $(n!{)}^2+1$ satisfied $p\le n$, it would divide $n!$ and therefore divide the difference $(n!{)}^2+1-(n!{)}^2=1$, impossible. Hence $p>n$.

Every integer $(n!{)}^2+1>1$ has a prime divisor, and (ii) makes each of its prime divisors congruent to 1 modulo 4. Taking arbitrarily large $n$ produces such primes beyond every bound. ThereforeEquivalently, if their list were finite, choosing $n$ at least its largest member would produce a new member larger than that supposed maximum.

Since $n>1$, its factorial is even, so $(n!{)}^2+1$ is odd and its prime divisor $p$ is odd. Also $p$ cannot divide $n!$, since it would then divide1. Modulo $p$,Raise this equation to the integer power $(p-1)/2$. Fermat's little theorem givesAs $p$ is odd, $1$ and $-1$ are distinct residues. Thus the exponent is even, and

Fermat's little theorem states that for a prime $p$ and an integer $a$ not divisible by $p$,Equivalently, $a^p\equiv a(\mathrm{mod}p)$ for every integer $a$, with divisibility by $p$ allowed in that latter formulation.

Set $u=r\psi$, choose $\alpha >0$ for a negative bound-state energy, and write $\kappa =2mK/{\hslash }^2$. Multiplying the radial equation by $r$ givesFor $u=e^{-\alpha r}f$, its second derivative is $e^{-\alpha r}(f^{\prime \prime }-2\alpha f^{\prime }+{\alpha }^{2}f)$. Substitution cancels the two ${\alpha }^2$ terms and givesFor the physical hydrogen problem, the angular eigenvalue has $\ell$ a nonnegative integer. Regularity at the origin selects the Frobenius exponent $u\sim r^{\ell +1}$, rather than the singular branch. PutThe equation for the series factor $v=f/r^{\ell +1}$ is $r{v}^{\prime \prime }+(2\ell +2-2\alpha r)v^{\prime }+(\kappa -2\alpha (\ell +1))v=0$, and coefficient comparison yieldsThe regular confluent-hypergeometric solution generically grows like $e^{2\alpha r}$ in this series factor; after the extracted $e^{-\alpha r}$ it is not normalizable. A bound state requires termination at some degree $q\ge 0$, namely $2\alpha (q+\ell +1)=\kappa$. Thus $n=q+\ell +1$ is an integer greater than $\ell$, andRegularity, normalizability and the integer angular index are the physical conditions selecting this spectrum; the differential equation without those conditions would admit many other solutions.

For $n=1$, necessarily $\ell =0$ and $q=0$. With $a_0={\hslash }^2/(mK)$,A normalized spatial ground state is ${\Psi }_1=(\pi a_0^3{)}^{-1/2}{e}^{-r/a_0}$. In spherical coordinates the radial probability has weight $r^{2}dr$. Constants and the angular integral cancel in the tail ratio. The supplied integral with $\lambda =2/a_0$ gives

A primitive root modulo $p$ is an integer whose residue generates the multiplicative group ${\mathbb{F}}_p^{\times }$, equivalently has order $p-1$.

The Chinese remainder theorem states that, for pairwise coprime positive integers $m_1,\dots ,m_r$, the system $x\equiv a_i(\mathrm{mod}{m}_i)$ has a unique solution modulo $M={\prod }_i{m}_i$. For a constructive proof put $M_i=M/m_i$. Bézout's identity gives an integer $u_i$ with $M_i{u}_i\equiv 1(\mathrm{mod}{m}_i)$. Then $x={\sum }_i{a}_i{M}_i{u}_i$ has all the required residues. If two solutions exist, their difference is divisible by every $m_i$; pairwise coprimality makes it divisible by their product, proving uniqueness modulo $M$.

Here $3,5,7$ divide $p-1$, and none is zero modulo $p$. Choose a primitive root $g$ and exponents $a,b,c$ withUse the theorem with moduli $3,5,7$ to choose $r$ such that $r\equiv a(\mathrm{mod}3)$, $r\equiv b(\mathrm{mod}5)$, $r\equiv c(\mathrm{mod}7)$. Define nonzero residuesThe exponents are integers; negative ones are interpreted using inverses in ${\mathbb{F}}_p^{\times }$. Taking integer representatives now givesThe theorem aligns the exponent modular congruences, which is why a primitive root converts the three power conditions into a linear modular congruence problem.

A possibly randomized statistical test has rejection probability $\phi (x)\in [0,1]$ after observing $x$. Its statistical power at parameter $\theta$ is ${\mathbb{E}}_{\theta }\phi$, and its test size is the supremum of that statistical power over the null parameter set.

For simple hypotheses with densities $f_0,f_1$, the Neyman-Pearson lemma states that a level-$\alpha$ test is most powerful if it rejects where $f_1>k{f}_0$, accepts where $f_1<k{f}_0$, and randomizes on equality to obtain null rejection probability $\alpha$. Here $k\ge 0$ is chosen to reach that test size. To prove optimality, let ${\phi }_{\ast }$ be this test and $\phi$ any test of level at most $\alpha$. Pointwise,Integrating and using $\int ({\phi }_{\ast }-\phi )f_0\ge 0$ gives ${\mathbb{E}}_1{\phi }_{\ast }-{\mathbb{E}}_1\phi \ge 0$. This proves maximal statistical power. For $k>0$, equality also forces any equally powerful test to use all its allowed test size and agree with the indicated decisions away from the likelihood-ratio tie set, up to null sets.

For the Poisson sample, $S={\sum }_i{X}_i\sim \mathrm{Poisson}(n\mu )$ and, for every fixed ${\mu }_1>{\mu }_0>0$, the likelihood ratio isIt increases strictly with $S$. Choose an integer $k$ so that ${\mathbb{P}}_{{\mu }_0}(S>k)\le \alpha \le {\mathbb{P}}_{{\mu }_0}(S\ge k)$, and reject for $S>k$, with boundary rejection probabilityThis has exact test size $\alpha$. Its rejection ordering and calibration do not depend on which ${\mu }_1>{\mu }_0$ is selected, so the lemma proves it uniformly most powerful for the whole one-sided alternative. Randomization is generally needed for an exact arbitrary test size in a discrete model.

For the approximate design, the central limit theorem under ${\mu }_0=1$ gives critical sum $n+1.645\sqrt{n}$. Under ${\mu }_1$, the sum is approximately normal with mean $n{\mu }_1$ and variance $n{\mu }_1$. Target statistical power $0.9$ therefore givesor $\sqrt{n}({\mu }_1-1)\approx 1.645+1.282\sqrt{{\mu }_1}$. Thus the requested design expression isRound upward to obtain an integer design. This is a normal approximation to the requested statistical power, rather than an exact discrete sample-size guarantee; the exact Poisson test above provides its exact calibration.

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.