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An odd cycle is a cycle in a graph with an odd number of edges.

Take $N$ to be an even number for a perfectly bipartite periodic chain and make the bipartite spin rotation by $\pi$ about the $x$ axis on alternating sites. In the local frame the Néel state has all spins up, while a bond becomesThe linear spin-wave approximation keeps ${\tilde{S}}^z=S-a^{†}a$ and ${\tilde{S}}^{\pm }\simeq \sqrt{2S}(a,a^{†})$. ThereforeEvery site has two neighbours. Fourier transform gives ${\sum }_{⟨ij⟩}(n_i+n_j)=2{\sum }_k{n}_k$ and pairing coefficient ${\gamma }_k=\cos k$, soThe bosonic Nambu normal-ordering shift follows from $a_{-k}{a}_{-k}^{†}=1+a_{-k}^{†}{a}_{-k}$. It adds $NJS$ inside the matrix expression, which must be subtracted in its constant. HenceA periodic chain with an odd number of sites is frustrated at the boundary and lacks this exact two-sublattice reference; the bulk thermodynamic calculation uses the even-chain sequence.

For a trajectory connecting the degenerate minima in infinite Euclidean time, the first integral gives $m{\dot{q}}^2/2=V(q)$. The quartic double-well instanton and its reverse areTheir action isThe instanton fluctuation prefactor $K$ has dimensions of inverse time. It incorporates nonzero Gaussian fluctuation eigenvalues and the translation zero-mode Jacobian; a conventional determinant expression iswith common endpoint regularization and the prime removing the translation mode. Only its positive quantum tunnelling rate $\kappa =K{e}^{-S_0/\hslash }$ is needed here.

In the dilute instanton gas, well-separated crossings alternate in direction. Integrating $n$ ordered centers gives ${\tau }^n/n!$. A path returning to the same well has $n$ an even number, and one connecting opposite wells has $n$ an odd number. The common leading harmonic endpoint factor follows from the supplied harmonic oscillator transition kernel:Expanding these functions gives the sums over even numbers and odd numbers, with the position-kernel normalization ${\mathcal{A}}_0$ that its abbreviated formula suppresses. Validity requires $S_0/\hslash \gg 1$, $\omega \tau \gg 1$ and $\kappa \ll \omega$, so typical crossing separation greatly exceeds an instanton width. Local anharmonic corrections replace the harmonic well energy by its perturbatively corrected value; the leading formula does not claim uniformly negligible relative error at arbitrarily large $\tau$.

The two exponents identify the double-well tunneling splitting:The superposition with an even spatial wavefunction is the lower state. Coherent quantum tunnelling removes the classical degeneracy, and the exponentially small splitting sets the long quantum tunnelling timescale.

For a Pythagorean triple, each square number is $0$ or $1$ modulo $3$. If neither $x$ nor $y$ were divisible by $3$, modular congruence would give $z^2\equiv 1+1\equiv 2(\mathrm{mod}3)$, impossible for a square number. Thus at least one of $x,y$ is divisible by $3$; both can be divisible, as in $(9,12,15)$.

An odd number has square number congruent to $1$ modulo $4$, whereas an even number has square number congruent to $0$. If both $x$ and $y$ were odd numbers, their sum of square numbers would again be $2$ modulo $4$, which no square number can be. Hence $x$ and $y$ cannot both be odd. They can both be even; the conclusion does not require the Pythagorean triple to be primitive.

A reduced positive definite binary quadratic form $f=[a,b,c]=a{x}^2+bxy+c{y}^2$ satisfieswith $b\ge 0$ when $|b|=a$ or $a=c$. Its discriminant of a binary quadratic form is $d=b^2-4ac<0$. The first displayed inequality already gives $|b|\le a$, while $c\ge a$ givesTaking the positive square root yieldsMoreover, $d=b^2-4ac\equiv b^2\equiv b(\mathrm{mod}2)$, so $b\equiv d(\mathrm{mod}2)$. For fixed negative $d$, the displayed bounds leave only finitely many possible integers $a$ and $b$, and then $c=(b^2-d)/(4a)$ is uniquely determined. Hence only finitely many reduced forms have discriminant $d$.

For $d=-15$, the bound gives $a\le \sqrt{5}$, so $a\in \{1,2\}$. The modular congruence condition makes $b$ odd. If $a=1$, then $b=\pm 1$ and $c=4$; the boundary convention $|b|=a\Rightarrow b\ge 0$ retains $[1,1,4]$. If $a=2$, then $b=\pm 1$ and $c=2$; the boundary convention $a=c\Rightarrow b\ge 0$ retains $[2,1,2]$. Thus the complete list is

The binomial theorem statesThe required sum is the coefficient of $x^n$ inThere is no $x^n$ term when $n$ is odd. When $n$ is even, the relevant term has exponent $2(n/2)=n$ and coefficient $(-1{)}^{n/2}(\genfrac{}{}{0ex}{}{n}{n/2})$. Hence