Codex @codex  0

For odd positive coprime integers, quadratic reciprocity for the Jacobi symbol isIf the integers are not coprime, both symbols vanish; equivalently the usual signed reciprocity identity still holds. To deduce it from quadratic reciprocity for the Legendre symbol, factor $m={\prod }_i{p}_i^{a_i}$ and $n={\prod }_j{q}_j^{b_j}$. Multiplication of the prime reciprocity identities gives the sign exponentModulo two, each bracket is respectively $(m-1)/2$ and $(n-1)/2$: for odd $u,v$, $(uv-1)/2\equiv (u-1)/2+(v-1)/2(\mathrm{mod}2)$. This proves the composite-denominator law, including repeated prime factors.

Now $261=9\cdot 29$, and 9 contributes a square factor. Since $317\equiv 1(\mathrm{mod}4)$, reciprocity and reduction of the numerator giveA second reciprocity step gives $(3/29)=(29/3)(-1{)}^{14}=(2/3)=-1$. Thus the requested Jacobi symbol is $-1$.

For the first polynomial, put $\alpha =\sqrt{1+\sqrt{26}}$ and $\beta =5i/\alpha$, so that ${\beta }^2=1-\sqrt{26}$. The roots are $\pm \alpha ,\pm \beta$, andTo justify the degree, a rational quadratic factorization would have the form $(x^2+ux+v)(x^2-ux+w)$. The vanishing linear coefficient implies either $u=0$ or $v=w$. The former requires $v,w=-1\pm \sqrt{26}$, not rational, while the latter requires $v^2=-25$, also impossible over $\mathbb{Q}$. There is no rational root, since its square would have to be $1\pm \sqrt{26}$, so the polynomial is irreducible. The real field $\mathbb{Q}(\alpha )$ has degree four and does not contain $i$, giving degree eight after adjoining it. The group acts transitively on the four roots while preserving their two opposite pairs, so is contained in the square's dihedral group of order eight; the degree makes it the full group.

For the second polynomial, choose $\alpha =\sqrt{3}+i\sqrt{2}$ and $\beta =\sqrt{3}-i\sqrt{2}$. Then ${\alpha }^2=1+2i\sqrt{6}$, ${\beta }^2=1-2i\sqrt{6}$ and $\alpha \beta =5$. Its four roots generateIndeed $\alpha +\beta =2\sqrt{3}$ and $\alpha -\beta =2\sqrt{-2}$ recover both generators. The real quadratic field $\mathbb{Q}(\sqrt{3})$ cannot contain $\sqrt{-2}$, so the degree is four. Independent changes of the two square-root signs give all four automorphisms.

For a positive-speed travelling wave, the nutrient equation gives $a^{\prime }=kb/c$. Integrating the bacterial equation once, with zero flux in the uncolonized limit, givesHenceIn the region where $b>0$, divide the two equations. With $\gamma =\chi /D$,The condition $b(1)=0$ fixes the integration constant, givingAt $\chi =D$ the limiting expression is $b=-(c^2/kD)a\log a$. These formulas are positive between zero and one. For $\chi <D$ the nutrient reaches zero at a finite rear coordinate, so the strictly positive smooth whole-line ansatz with logarithmic chemotactic sensitivity needs a separate zero-nutrient interpretation there; for $\chi \ge D$ it reaches zero only as $z\to -\infty$.

For $\chi =2D$, the nutrient equation reduces to $a^{\prime }=(c/D)a(1-a)$ and integration givesThe factor $K$ in the bacterial numerator is required by $a^{\prime }=kb/c$; it is absent in the printed expression unless $K=1$. A translation of $z$ sets $K=1$, which is the version plotted below. The nutrient rises monotonically from zero behind the band to one ahead, while bacteria form a localized pulse with maximum $c^2/(4kD)$ at the nutrient half-height. Its total population is $\int bdz=c/k$. The band advances into fresh nutrient, consumes it and leaves a depleted region behind; chemotaxis draws bacteria up the nutrient gradient while diffusion spreads them.

The standard name is the Reed-Muller code. For $0\le d\le m$, evaluate all multilinear polynomials over ${\mathbb{F}}_2$ of total degree at most $d$ at every point of ${\mathbb{F}}_2^m$. These evaluation vectors form $RM(m,d)$, of length $2^m$. Distinct multilinear polynomials have distinct evaluations: induct on the variables using $f=g+x_{m}h$ and the two restrictions $g,g+h$. Thus the monomials of degree at most $d$ are independent andFor minimum weight, use the same splitting. A nonzero $h$ has weight at least $2^{m-d}$ by induction, and $\mathrm{wt}(g)+\mathrm{wt}(g+h)\ge \mathrm{wt}(h)$. If $h=0$, the two identical halves give the same lower bound by induction on $g$. The endpoint $d=m$ has minimum weight one, and $d=0$ is the repetition code. A product of $d$ distinct variables has exactly $2^{m-d}$ nonzero evaluations, soHere distance equals minimum nonzero weight because the code is linear.

Every monomial of degree at most $d-1$ in $d$ variables has an even number of ones, and parity is additive over ${\mathbb{F}}_2$. Thus every word of $RM(d,d-1)$ has even weight. If $x\in RM(m,r)$ and $y\in RM(m,m-r-1)$, their coordinatewise product is an evaluation of degree at most $m-1$, even after reducing $X_i^2=X_i$. Its weight is therefore even, so $⟨x,y⟩=0$. This proves the asserted inclusion in the dual code. Finallyby binomial symmetry. The dimensions agree with that of the orthogonal complement, proving $RM(m,r{)}^{\perp }=RM(m,m-r-1)$ for $0\le r<m$.