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Near an isolated point $P\in A$, choose source and target coordinate charts, using continuity of $\alpha$ to keep a sufficiently small source neighbourhood in the target chart. The coordinate representation is holomorphic on the punctured neighbourhood and is bounded near $P$ by continuity. The Riemann removable singularity theorem extends it holomorphically across $P$, with the given continuous value. Repeating at all points of $A$ proves that $\alpha$ is analytic.

For a nonconstant holomorphic function $f$ on a connected Riemann surface, let $A$ be the set where its coordinate derivative vanishes. This condition is independent of the chart. The derivative cannot vanish identically on an open set, because the identity theorem would then make $f$ constant globally. Its zeros are isolated and closed locally, so $A$ is a closed discrete subset. At every other point the holomorphic inverse function theorem makes $f$ a local coordinate chart.

Now suppose $\alpha$ is a homeomorphism and $g=f\circ \alpha$ is analytic. Outside ${\alpha }^{-1}(A)$, use $f$ as a target coordinate to write locally $\alpha =f^{-1}\circ g$. This is analytic. The preimage ${\alpha }^{-1}(A)$ is discrete because $\alpha$ is a homeomorphism, and the first argument extends analyticity across it. An injective holomorphic map cannot have a zero derivative: a zero of order at least two would give a locally multiple-to-one map. Its local inverse is therefore holomorphic, proving that $\alpha$ is a conformal equivalence.

The plane $\mathbb{C}$ and the unit disk are homeomorphic by $z\mapsto z/(1+|z|)$, with inverse $w\mapsto w/(1-|w|)$. They are not conformally equivalent: a holomorphic equivalence from the plane to the disk would be a bounded nonconstant entire function, contradicting Liouville's theorem.

The real Stone-Weierstrass theorem says that a subalgebra of $C(X,\mathbb{R})$ on a compact metric space is uniformly dense if it contains the constants and separates points. The finite sums ${\sum }_i{f}_i(x)g_i(y)$ form such an algebra on $[0,1{]}^2$: multiplication preserves this form, the constant one belongs to it, and the coordinate functions distinguish every pair of different points. Thus they approximate the continuous kernel uniformly.

A finite-rank operator has finite-dimensional range. A compact operator maps the unit ball to a set with compact closure, equivalently sends every bounded sequence to a sequence with a convergent subsequence. The identity on $C[0,1]$ is bounded but not compact: the functions $x^n$ have supremum norm one, and every subsequence has the discontinuous pointwise limit which is zero on $[0,1)$ and one at one. It therefore has no uniformly convergent subsequence.

A bounded finite-rank operator is compact because bounded sets in its finite-dimensional range have compact closure. Suppose $T_n\to T$ in operator norm. For any $\epsilon >0$, choose $n$ with $\Vert T-T_n\Vert <\epsilon /3$ and a finite $\epsilon /3$-net for $T_n$ of the unit ball. The same centres give an $\epsilon$-net for $T$ of that ball. Its closure is totally bounded and complete in the Banach space $Y$, hence compact. Thus an operator-norm limit of compact operators is compact.

For $K_n(x,y)={\sum }_i{f}_i(x)g_i(y)$ uniformly approximating $K$, the integral operator $T_n$ has range in $\mathrm{span}\{f_i\}$ and is bounded. MoreoverConsequently $T_n\to T$ in operator norm, proving that the stated integral operator is compact.

Put $s=\sqrt{-17}$. The ring of integers of a number field is $\mathbb{Z}[s]$ and the field discriminant is $-68$. The Minkowski bound for ideal classes says that each ideal class has an integral ideal representative of norm at most $(2/\pi )\sqrt{68}<6$.

The prime two ramifies as $(2)={\mathfrak{p}}_2^2$, where ${\mathfrak{p}}_2=(2,1+s)$. Three splits as $(3)={\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, choosing ${\mathfrak{p}}_3=(3,1+s)$, whereas five is inert because $-17\equiv 3(\mathrm{mod}5)$ is not a square. The norm bound therefore leaves only the classes of $1$, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and ${\overline{\mathfrak{p}}}_3$; a norm-four ideal is ${\mathfrak{p}}_2^2$ and is principal.

The ideal ${\mathfrak{p}}_2$ is not principal, since an element of norm two would solve $a^2+17b^2=2$. On the other hand $N(1+s)=18$, and $1+s$ is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ but not ${\overline{\mathfrak{p}}}_3$. Unique prime ideal factorization and the norm give $(1+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Thus $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2]$, which has order two, and

For the Diophantine equation, $x$ must be positive. If $x$ were even, $y^2\equiv 15(\mathrm{mod}32)$, impossible for a square. Thus $x$ is odd and $y$ even. If $17\mid x$, then $17\mid y$, but $y^2+17$ would have 17-adic valuation one while $x^5$ has valuation divisible by five, also impossible.

The ideals $(y+s)$ and $(y-s)$ are consequently coprime. A common prime divides $2s$ and hence lies over two or seventeen, both excluded above. Their product is $(x{)}^5$, so unique prime ideal factorization gives $(y+s)={\mathfrak{a}}^5$. Since the ideal class group has order four, $[\mathfrak{a}{]}^5=1$ forces $[\mathfrak{a}]=1$. Write $\mathfrak{a}=(a+bs)$ with integers $a,b$. The only units are $\pm 1$ by their norm, and the sign can be absorbed into the fifth power. Hence $y+s=(a+bs{)}^5$.

Comparing the coefficient of $s$ givesFor $b=1$, the bracket would be one, but it is four modulo five. For $b=-1$, the bracket would be minus one, but it is one modulo three for either possible square residue of $a$. Both cases are impossible. Therefore there are no integer solutions.

Use the reduction algorithm for a positive definite binary quadratic form. A determinant-one shear $(x,y)\mapsto (x+ky,y)$ puts the middle coefficient in $[-a,a]$. If the resulting last coefficient is smaller than $a$, the determinant-one substitution $(x,y)\mapsto (-y,x)$ decreases the positive integer leading coefficient. Repeating terminates with a reduced positive definite binary quadratic form satisfying $|b|\le a\le c$. These substitutions preserve proper equivalence of binary quadratic forms and the discriminant of a binary quadratic form.

The enumeration of reduced binary quadratic forms givesSince $b^2\equiv -163\equiv 1(\mathrm{mod}4)$, $b$ is odd. For $|b|=1,3,5,7$, the possible values of $(b^2+163)/4$ are respectively $41,43,47,53$. Each is a prime number greater than seven. But $ac=(b^2+163)/4$, so its positive divisor $a\le 7$ must be one. Then $|b|\le a$ forces $b=\pm 1$ and $c=41$. The two signs are properly equivalent: substituting $x\mapsto x-y$ in $x^2+xy+41y^2$ gives $x^2-xy+41y^2$. Thus every form in question is properly equivalent to

A matching in a graph is a set of edges with no shared endpoints. A matching from $X$ to $Y$ here covers every vertex of $X$. Necessity of the Hall marriage theorem condition is immediate: the partners of vertices of $A\subseteq X$ are distinct members of its neighbour set $\Gamma (A)$.

For sufficiency, take a matching of maximum size. If it leaves $x\in X$ unmatched, follow alternating unmatched and matched edges from $x$. Let $A\subseteq X$ and $B\subseteq Y$ be the reachable vertices. No vertex of $B$ can be unmatched, because the resulting augmenting path in a matching would increase the matching. Every neighbour of $A$ is in $B$, and the matching pairs $B$ bijectively with $A\setminus \{x\}$. Thus $|\Gamma (A)|=|B|=|A|-1$, contradicting Hall's condition. The maximum matching therefore covers $X$.

For the deficiency bound, adjoin $d$ new vertices to $Y$, each adjacent to all of $X$. Every nonempty $A$ then has $|\Gamma (A)|+d\ge |A|$ neighbours. Apply the Hall marriage theorem and remove the at most $d$ edges using new vertices. This leaves at least $|X|-d$ independent edges. The lower bound is vacuous if $d>|X|$.

Write the adjacency matrix of a graph in bipartite block form $\left(\begin{array}{cc}0& C\\ C^T& 0\end{array}\right)$. If Hall's condition failed for $A\subseteq X$, the linear map from vectors supported on $A$ to their neighbour coordinates in $Y$ would have fewer output than input coordinates. It therefore has a nonzero kernel, giving a nonzero vector $v$ supported on $A$ with $C^{T}v=0$. The vector $(v,0)$ would be a zero eigenvector of the adjacency matrix. Consequently absence of the zero eigenvalue implies a matching covering $X$.

The converse is false, even with equal part sizes. The four-cycle $K_{2,2}$ has a perfect matching, but its matrix $C=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ is singular. Its adjacency matrix has eigenvalues $2,-2,0,0$.