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The Uniform boundedness principle states that a pointwise bounded family $\mathcal{T}$ of bounded linear maps from a Banach space $X$ to a normed space $Y$ is uniformly bounded in operator norm. Define closed sets $E_n=\{x:{\sup }_{T\in \mathcal{T}}\Vert Tx\Vert \le n\}$. Pointwise boundedness makes their union $X$. The Baire category theorem gives a ball $B(x_0,r)\subset E_n$ for some $n$. For $\Vert v\Vert <r$, both $x_0+v$ and $x_0$ lie in $E_n$, so $\Vert Tv\Vert \le 2n$ for every $T$. Scaling yields ${\sup }_T\Vert T\Vert \le 2n/r$, proving the principle.

For the separately continuous bilinear map, consider $T_x(y)=F(x,y)$ with $\Vert x\Vert \le 1$. Each $T_x$ is bounded and linear. For fixed $y$, continuity of the other linear map gives ${\sup }_{\Vert x\Vert \le 1}\Vert F(x,y)\Vert <\infty$. Apply the Uniform boundedness principle to this family to getBilinearity then gives $F(x,y)-F(x_0,y_0)=F(x-x_0,y)+F(x_0,y-y_0)$; the bound proves joint continuity.

For the trilinear map, fix $w$. The just-proved bilinear result bounds $G(\cdot ,\cdot ,w)$. Thus the bounded linear maps $T_{x,y}:w\mapsto G(x,y,w)$, for $\Vert x\Vert ,\Vert y\Vert \le 1$, are pointwise bounded. Apply the Uniform boundedness principle once more to obtain $\Vert G(x,y,w)\Vert \le C\Vert x\Vert \Vert y\Vert \Vert w\Vert$. Expansion of the difference in three slots proves joint continuity. Separate continuity suffices for these multilinear maps.

Without linearity it does not suffice. On ${\mathbb{R}}^2$ set $H(x,y)=xy/(x^2+y^2)$ away from the origin and $H(0,0)=0$. Each coordinate section is continuous, but $H(t,t)=1/2$ for $t\ne 0$, so $H$ is not continuous at the origin.

Suppose an odd continuous map $f:S^2\to S^1$ existed. Restrict to the equator, parametrized by $e^{i\theta }$. Choose a continuous real lift $F$ with $f(e^{i\theta })=e^{iF(\theta )}$. Oddness gives $F(\theta +\pi )-F(\theta )=\pi +2\pi k$ for a fixed integer $k$, since the integer-valued difference is continuous. Thus $F(\theta +2\pi )-F(\theta )=2\pi (1+2k)$, and the equatorial map has odd degree of a continuous mapping, hence nonzero degree. But it extends over a hemisphere, which is a disk, so the equatorial loop is null-homotopic and has degree zero. This contradiction proves the dimension-two Borsuk-Ulam theorem in the stated form.

If a continuous injection $h:S^2\to {\mathbb{R}}^2$ existed, then $h(x)-h(-x)$ would never vanish. Normalization would definea continuous odd map, already ruled out. In particular the sphere cannot be homeomorphic to any planar subset.

For an integral binary quadratic form $[a,b,c]=a{x}^2+bxy+c{y}^2$ with $a>0$ and $D=b^2-4ac<0$, the reduced positive definite binary quadratic form convention is $|b|\le a\le c$, with $b\ge 0$ on either boundary $|b|=a$ or $a=c$. Since $|D|=4ac-b^2\ge 3a^2$, the enumeration of reduced binary quadratic forms for $D=-23$ has $a\le \sqrt{23/3}<3$. The integer $b$ is odd. For $a=1$, only $b=1$ is retained and $c=6$; for $a=2$, both $b=\pm 1$ give $c=3$. Thus the complete list isA proper representation of a prime uses a primitive representing vector. Completing that vector to a matrix in ${\mathrm{SL}}_2(\mathbb{Z})$ transforms a representing form to $[p,b,c]$, so its discriminant of a binary quadratic form gives $b^2\equiv -23(\mathrm{mod}p)$. Conversely, choose an odd $b$ satisfying this congruence and put $c=(b^2+23)/(4p)$. This is integral, and $[p,b,c]$ is positive definite and represents $p$ at $(1,0)$. This proves the discriminant criterion for prime representation by a binary quadratic form here.

For $p\ne 23$, quadratic reciprocity gives $(-23/p)=(p/23)$. Including the ramified prime, the answer is thereforeFor example $[23,23,6]$ supplies a proper representation of $23$. Reduction transfers every such representation to one of the three displayed forms.

There are seven irreducible characters, one for each conjugacy class. Besides the trivial character and the four supplied rows, let the remaining degrees be $d,e$. The sum of squared degrees gives $d^2+e^2=360-(1+25+64+64+100)=106$, whose positive integer solutions are $\{d,e\}=\{5,9\}$. Complex conjugation permutes irreducible characters while preserving degree; the known rows are real and each unknown degree occurs just once among the remaining characters, so the two missing characters are real.

Call their entries $x_j,y_j$, with degrees five and nine. Orthogonality with the regular character gives $5x_j+9y_j=-{\sum }_{\mathrm{known}}\chi (1)\chi (C_j)$ for $j>1$. Column norms give $x_j^2+y_j^2=360/|C_j|-{\sum }_{\mathrm{known}}\chi (C_j{)}^2$. These equations yield two possible rational pairs in some columns. For this integral character reconstruction from column orthogonality, a further constraint is essential: character values are algebraic integers, so rational candidates must be integers. For example at $C_6$, $5x+9y=-9$ and $x^2+y^2=1$ allow $(0,-1)$ and $(-45/53,-28/53)$; the second pair is not integral. The integral choices complete the table:Every nontrivial row has $\chi (g)\ne \chi (1)$ for $g\ne 1$. In a unitary realization, $\chi (g)=\chi (1)$ exactly when all eigenvalues of the representing matrix are one, so each nontrivial irreducible representation is faithful. If a nontrivial proper normal subgroup existed, a nontrivial irreducible representation of the nontrivial quotient would pull back to a nontrivial representation with that subgroup in its kernel. This contradicts faithfulness. The group is simple.

Here graphs are finite and simple. To prove Dirac theorem, take a longest path $v_1,\dots ,v_{\ell }$. All neighbours of either endpoint lie on the path. Among indices $1\le i<\ell$, let $S=\{i:v_1{v}_{i+1}\text{ is an edge}\}$ and $T=\{i:v_i{v}_{\ell }\text{ is an edge}\}$. Since $|S|+|T|=d(v_1)+d(v_{\ell })\ge n>\ell -1$, some $i$ lies in both. Following the path forward to $v_i$, jumping to $v_{\ell }$, and following it backwards to $v_{i+1}$ closes a cycle containing every vertex of the path. The degree condition makes the graph connected: two components would each have at least $\delta +1>n/2$ vertices. If $\ell <n$, connectedness gives an edge from an outside vertex to the cycle, and breaking the cycle there makes a longer path. This contradiction proves a Hamiltonian cycle.

For even $n=2m$, the complete bipartite graph $K_{m-1,m+1}$ has minimum degree $m-1$ and no Hamiltonian cycle. For odd $n=2m+1$, $K_{m,m+1}$ has minimum degree $m$ and likewise fails: a bipartite cycle uses equal numbers from its two parts. These examples include the smallest applicable $n$.

Adding one universal vertex gives $n+1$ vertices and minimum degree at least $(n+1)/2$ when the original minimum degree is at least $(n-1)/2$. Apply the proved theorem and remove the universal vertex from its cycle. The remaining sequence is a Hamiltonian path in the original graph.

A necessary condition for a Hamiltonian cycle is that deleting a nonempty vertex set $S$ leaves at most $|S|$ components, since deleting those vertices breaks a spanning cycle into at most that many path pieces. For any $k\ge 1$, take the connected star $G=K_{1,k+2}$. In $G_k$, deleting its center and the $k$ new vertices leaves $k+2$ components, more than $k+1$. Even two-connectivity does not suffice: take $G=K_{2,k+3}$, which is two-connected. Deleting its two-vertex part and the $k$ new vertices from $G_k$ leaves $k+3$ components, more than $k+2$. Both connectivity requirements admit counterexamples for every $k$.

Put $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c=m\dot{\mathbf{q}}$. From Hamilton's equations,For the stated time-independent vector potential, differentiating ${\pi }_i$ gives $m{\ddot{q}}_i=(e/c){\dot{q}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, orUse the Poisson bracket convention $[F,G]={\sum }_i(F_{q_i}{G}_{p_i}-F_{p_i}{G}_{q_i})$. Then direct differentiation givesFor $\mathbf{A}=(0,0,F(r))$ with $r=\sqrt{x_1^2+x_2^2}$, the Hamiltonian is independent of $x_3$, so ${\dot{p}}_3=0$. The radial force in the horizontal plane is ${\dot{p}}_r=e(p_3-eF/c)F^{\prime }/(mc)$. For circular motion the radial acceleration is $-r{\Omega }^2$, and $p_r$ is the mechanical radial momentum because the horizontal vector potential vanishes. ThusA circular orbit at that radius is possible when the right side is nonnegative; positive values give either sense of horizontal rotation, and the longitudinal velocity is the constant $(p_3-eF(r)/c)/m$ along that orbit.

In the Rabin cryptosystem, choose distinct large odd primes $p,q$, publish $N=pq$, and encrypt $m$ as $c=m^2\mathrm{mod}N$. Usually $p,q\equiv 3(\mathrm{mod}4)$, so a square root modulo each prime is obtained by exponentiating to $(p+1)/4$ or $(q+1)/4$. Combine the two independent choices of signs by the Chinese remainder theorem to obtain the four roots modulo $N$ for a message coprime to $N$. Redundancy or a specified encoding distinguishes the intended message. Factoring $N$ makes decryption easy; a general square-root oracle conversely factors $N$ by obtaining two roots not congruent up to sign and taking their greatest common divisor with $N$.

For the Repeated-message attack on the Rabin cryptosystem, first compute $d=\gcd (N,N^{\prime })$. If $d=1$, use the two ciphertexts to reconstruct $s\in [0,N{N}^{\prime })$ with $s\equiv m^2(\mathrm{mod}N)$ and $s\equiv m^2(\mathrm{mod}{N}^{\prime })$. Since $0<m<\min (N,N^{\prime })$, we have $m^2<N{N}^{\prime }$, so this reconstruction is the ordinary integer $m^2$. ThereforeIf distinct semiprime moduli have $d>1$, they share a prime; the gcd instead exposes their factors, allowing ordinary Rabin decryption with the usual encoding check. The two cases cover the stated protocol with distinct standard moduli.

In the coprime case, learning this message does not normally reveal the factors or a decryption key, so other messages encrypted only once remain protected by the square-root problem. Exceptional messages with $\gcd (m,N)>1$ do reveal a factor, and the shared-prime-modulus case compromises other messages too. Repetition under fresh coprime moduli is the vulnerability used here, rather than a universal attack on every ciphertext.