Suppose an integer right triangle with square area exists, and choose one with the least positive hypotenuse $c$. It is primitive. Indeed, dividing all sides by their common divisor $d$ gives an integer right triangle whose area is the square of the rational number $s/d$, where $s^2$ was the original area. Its area is an integer, since one of its legs is even. A rational number whose square is an integer is itself an integer, as follows by writing it in lowest terms. Thus division by $d>1$ would give another square-area triangle with smaller hypotenuse.

By the primitive Pythagorean parametrization, after ordering the legs the triangle has sides $n^2-m^2,2nm,n^2+m^2$, with coprime $n>m>0$ of opposite parity. Its area isThese four positive factors are pairwise coprime: the pair $n-m,n+m$ has gcd dividing $2m$ and is odd, and the other pairs have gcd one by $\gcd (n,m)=1$. Since the product is a square, each factor is a square. WriteBoth $w,z$ are odd. Odd squares are $1(\mathrm{mod}8)$, so $2m=w^2-z^2$ is divisible by eight; hence $m$ is divisible by four and $v$ is even. Define positive integersThey satisfyThus $(r,s,u)$ is another positive integer right triangle with square area, but its hypotenuse $u=\sqrt{n}$ is strictly smaller than $n^2+m^2=c$. This contradicts minimality. No integer right triangle has a nonzero square area. This is the Fermat right triangle theorem, with the actual infinite descent construction exhibited.

Use both Kähler identities from the preceding parts, so that ${\partial }^{\ast }=i[\Lambda ,\bar{\partial }]$ and ${\bar{\partial }}^{\ast }=-i[\Lambda ,\partial ]$. For odd operators $a,b$, write $\{a,b\}=ab+ba$. Direct expansion givesSince $\partial \bar{\partial }+\bar{\partial }\partial =0$, we obtainTogether with part (b), this proves the full Kähler Laplacian identity

Now take the exact form $\eta$. Being $\bar{\partial }$-exact, it is orthogonal to the harmonic space, so $H\eta =0$, and it is also $\bar{\partial }$-closed. Choose $\alpha =G\eta$ for the Dolbeault Green operator of the Dolbeault Laplacian. Commutation gives $\bar{\partial }\alpha =G\bar{\partial }\eta =0$. ThereforeSuppose also $\partial \eta =0$, and put $v={\bar{\partial }}^{\ast }\alpha$, $u=\partial v$. The mixed identities and $({\bar{\partial }}^{\ast }{)}^2=0$ giveThus $u=\partial {\bar{\partial }}^{\ast }\alpha$ is harmonic for the Dolbeault Laplacian. By the Kähler Laplacian identity it is also $\partial$-harmonic and hence ${\partial }^{\ast }u=0$. But it is $\partial$-exact, soConsequently $\boxed{\partial {\bar{\partial }}^{\ast }\alpha =0}$: the form $v$ is $\partial$-closed.

For every harmonic form $h$ of the same bidegree as $v$, $⟨v,h⟩=⟨\alpha ,\bar{\partial }h⟩=0$. Harmonic spaces for the two Dolbeault Laplacians coincide, so $v$ has no $\partial$-harmonic component. Apply the $\partial$ version of the Dolbeault Hodge decomposition. Since $v$ is $\partial$-closed, it has no ${\partial }^{\ast }$-coexact component either. Thus $v=\partial \varphi$, with $\varphi \in A^{p-1,q-1}$. We conclude the requested ddbar lemma:For example $\varphi ={\partial }^{\ast }Gv$ is a choice, since the same Green operator serves ${\Delta }_{\partial }={\Delta }_{\bar{\partial }}$. If $p=0$ or $q=0$, the relevant negative-bidegree space is zero and the argument forces $\eta =0$. Compactness, the absence of boundary and the Kähler condition are essential to this global harmonic argument.

Complex conjugation preserves the real Kähler form and interchanges $\partial ,\bar{\partial }$ and their formal adjoints. Conjugating the permitted identity among the Kähler identities therefore yields $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$. The same nilpotence calculation as in part (a) gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$. Now expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$:Both mixed anticommutators vanish, proving $\boxed{{\Delta }_d={\Delta }_{\partial }+{\Delta }_{\bar{\partial }}}$.

With the usual exterior-boundary interpretation that the harmonic function remains bounded as $r\to \infty$, positive radial powers and $\log r$ are excluded. HenceThe series must converge locally for $r>1$, with any required finite boundary values at $r=1$. These are the interior harmonic Fourier expansions in planar concentric domains after the inversion $r\mapsto 1/r$.

If “finite” means only finite at each finite point, without a condition at infinity, growing solutions such as $r\cos \theta$ and $\log r$ are allowed. Under that literal reading the general exterior expansion iswith local convergence. Boundedness at infinity is the condition selecting the decaying exterior expansion.

Set $F(z)=\exp (ig(z{)}^2)$. Since $\mathrm{Im}(g^2)=2uv$, its modulus is $e^{-2uv}\le 1$. The Liouville theorem again makes $F$ constant, and $F^{\prime }=2ig{g}^{\prime }F=0$ implies $(g^2{)}^{\prime }=0$. Hence $g^2$ is a constant $c$. If $c=0$, then $g=0$. If $c\ne 0$, then $g$ never vanishes and $g{g}^{\prime }=0$ implies $g^{\prime }=0$. In either case $g$ is constant. This is also an application of the one-sided product bound for an entire function after rotating $g$ by $i$.

The fixed-point property also fails on the real line. The continuous function $\boxed{f(t)=t+1}$ maps $\mathbb{R}$ into itself but has no fixed point, since $t+1=t$ is impossible.

In a homogeneous Ohmic conductor, a steady electric current density obeys $\nabla \cdot \mathbf{j}=0$ and $\nabla \times \mathbf{j}=0$. Uniform axial current in a straight cylinder follows if its side is insulated and its flat end faces are equipotentials. The electric potential then solves Laplace's equation with those boundary conditions, and uniqueness makes it linear in the axial coordinate. Steadiness alone does not impose these contact conditions: nonuniform end potentials can produce nonuniform current even with an insulating side. Bending also generally redistributes current; in an annular conductor the azimuthal current density is proportional to the inverse radius.

For any measurable real path $x$, put $A_n(t)=n{\int }_0^t{1}_{\{|x(s)|<1/n\}}ds$. If $n_k\le n\le n_{k+1}$, nested windows give $(n/n_{k+1})A_{n_{k+1}}(t)\le A_n(t)\le (n/n_k)A_{n_k}(t)$. Consequently, when $n_{k+1}/n_k\to 1$ and the subsequence $A_{n_k}$ converges uniformly on bounded time intervals, the full sequence does also, to the same limit. This deterministic interpolation turns summable-subsequence First Borel-Cantelli lemma estimates into almost-sure convergence without incorrectly claiming that convergence in $L^2$ alone implies full-sequence almost-sure convergence.

Over an algebraically closed field of characteristic different from two, a nondegenerate quadratic form of rank at least three is irreducible, its cone has only the vertex singular, and its hypersurface ring is Cohen-Macaulay. Thus $(R_1)$ and $(S_2)$ in the Serre criterion for normality hold. In characteristic two, the reduced zero variety of ${\sum }_i{t}_i^2$ is instead the hyperplane ${\sum }_i{t}_i=0$, which is normal. The unreduced double-hyperplane scheme is not normal.

A singular point on a normal two-dimensional algebraic variety has an integrally closed local domain which is not a regular local ring. A rational double point is an important isolated example with a crepant minimal resolution.

If a transcendental meromorphic function had only finitely many simple preimages over each of five distinct finite values, their simple counting functions would be $O(\log r)$. The displayed multiplicity inequality and the Nevanlinna first main theorem would bound the truncated sum by $5T_f/2+O(\log r)$. The Nevanlinna second main theorem requires at least $3T_f$ up to its $o(T_f)$ error, a contradiction. Thus at least one value has infinitely many simple preimages; four other values can remain totally ramified.

For a nonconstant meromorphic function on the unit disc and $q\ge 3$ distinct spherical values, this version of the Nevanlinna second main theorem holds outside a set $E$ with ${\int }_{E}dr/(1-r)<\infty$, as $r\uparrow 1$. The truncated Nevanlinna counting function counts distinct preimages rather than multiplicity. The boundary-distance error is essential and need not be small relative to the characteristic. It follows from the disk Nevanlinna logarithmic derivative lemma and the growth lemma in the variable $\log (1/(1-r))$. The disk error estimate and weighted exceptional-set argument appear in mathweb.tifr.res.in/Documents/Publications/Lectures/tifr17.pdf .

For a nonnegative increasing continuously differentiable function not identically zero, choose $x_0$ with $\varphi (x_0)>0$. On $I=\{x\ge x_0:{\varphi }^{\prime }>{\varphi }^2\}$, $1<{\varphi }^{\prime }/{\varphi }^2$. Thus $|I|\le {\int }_{x_0}^{\infty }{\varphi }^{\prime }/{\varphi }^2\le 1/\varphi (x_0)$. Adding the finite initial interval controls all exceptions. For the zero function there are none. Taking logarithms turns derivative growth into logarithmic auxiliary-function growth in value-distribution estimates.

The truncated counting function counts each distinct $a$-point of a meromorphic function once, rather than with its local multiplicity. It uses the same logarithmic radial weights as the Nevanlinna integrated counting function, including an origin contribution $\log r$ when appropriate. The difference $N-\overline{N}$ measures integrated excess multiplicity.

The rational composition law for the Nevanlinna characteristic gives $T_f(r)=3r/\pi +O(1)$. Its zeros are precisely $2\pi i\mathbb{Z}$, all of multiplicity two, so $N(r;0)=2r/\pi +O(\log r)$ and $\overline{N}(r;0)=r/\pi +O(\log r)$. Hence both its Nevanlinna deficiency and Nevanlinna ramification index at zero equal one third.

For a degree-$d$ rational function $U$ of the Riemann sphere, choose relatively prime homogeneous polynomials $(P,Q)$ of degree $d$. Their joint norm on the unit three-sphere has positive lower and finite upper bounds. Thus $\log \Vert (P,Q)(v)\Vert -d\log \Vert v\Vert$ is bounded. Averaging over a circle using a nonvanishing entire homogeneous lift of a meromorphic function proves the displayed law for the spherical characteristic. Jensen's formula and $0\le \frac{1}{2}\log (1+x^2)-{\log }^{+}x\le \frac{1}{2}\log 2$ show that it also holds for the ordinary Nevanlinna characteristic.

The growth order of a meromorphic function is the upper power exponent of its Nevanlinna characteristic, with constants assigned order zero. It is different from the local signed order at a zero or pole. A Nevanlinna integrated counting function has an analogous growth order. The Nevanlinna first main theorem bounds every counting order by the function order.

With the standard string oscillator algebra and $G_0={\sum }_m{\alpha }_m\cdot {\psi }_{-m}$, pairing positive and negative modes gives $G_0^2={\alpha }_0^2/2+{\sum }_{m>0}({\alpha }_{-m}\cdot {\alpha }_m+m{\psi }_{-m}\cdot {\psi }_m)$. This is the vacuum-normal-ordered Ramond sector generator $L_0^R$, so $\{G_0,G_0\}=2L_0^R$. Rescaling the superconformal current rescales its square quadratically; its normalization cannot be ignored when comparing it with a fixed Virasoro generator.

The graph $z=x^2+y^2$ in Euclidean three-space has the displayed induced Riemannian metric and positive Gaussian curvature. It is smooth at its vertex; the polar-coordinate singularity is only a coordinate artifact. The metric dominates the Euclidean plane metric, so upward stability of Riemannian completeness proves completeness. It has infinite diameter. In dimension two Ricci curvature equals Gaussian curvature times the metric, so this is a counterexample to compactness under merely pointwise positive Ricci curvature. Its curvature tends to zero at infinity.

Here $\psi$ is a nontrivial additive character of the finite field. For nontrivial $\chi$, substituting $x=ty$ in the squared absolute value givesThe inner sum is $q-1$ at $t=1$ and $-1$ otherwise; multiplicative character orthogonality completes the identity.