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Write $\chi (a)=(a/p)$. If $p\mid a$, the claimed identity has both sides zero. Otherwise let $ϵ=\chi (a)\in \{1,-1\}$. Applying the Euler criterion givesBoth sides are in $\{1,-1\}$, which are distinct residue classes because $p$ is odd. Therefore the modular congruence is an equality:

Take $\boxed{K=F[x]/(f)}$ and let $\alpha$ be the residue class of $x$. Division with remainder represents every class by a polynomial of degree less than $n$, so every element is a polynomial in $\alpha$ and $f(\alpha )=0$. A nonzero class represented by $u$ has $\gcd (u,f)=1$ because $f$ is irreducible. Bézout's identity gives $au+bf=1$, making the class of $a$ its inverse. Thus $K$ is a field, containing $F$ through its constant classes.

A splitting field of a nonconstant polynomial is an extension in which it factors into linear factors and which is generated over the original field by its roots. To construct one, choose an irreducible factor and use the preceding quotient construction to adjoin a root. Divide off a linear factor in that extension and repeat on the remaining polynomial. Each stage lowers the remaining degree, so finitely many stages suffice. The field generated by all roots in the resulting field is the required splitting field. This induction works with repeated roots and does not assume separability. The two roman clauses in the source specify properties of this one construction; they are not separate questions.

Euler's totient function $\varphi (n)$ counts the integers $1\le a\le n$ coprime to $n$, equivalently the invertible residue classes modulo $n$. Among $1,\dots ,p^e$, exactly the $p^{e-1}$ multiples of $p$ are not coprime to the prime power. HenceFactor $n={\prod }_p{p}^{e_p}$. Its distinct prime-power factors are pairwise coprime, so the permitted multiplicativity givesFor $n=1$, the empty product is one and agrees with $\varphi (1)=1$.

Since $n>1$, its factorial is even, so $(n!{)}^2+1$ is odd and its prime divisor $p$ is odd. Also $p$ cannot divide $n!$, since it would then divide1. Modulo $p$,Raise this equation to the integer power $(p-1)/2$. Fermat's little theorem givesAs $p$ is odd, $1$ and $-1$ are distinct residues. Thus the exponent is even, and

Milne-Thomson's circle theorem inserts an impermeable circular boundary into a two-dimensional irrotational incompressible flow. If $f(z)$ is analytic in a neighborhood of the closed disc and its flow singularities lie outside, the new complex potential outside $|z|=a$ isReflection and coefficient conjugation make the second term analytic outside the circle, with its image singularities inside. On $|z|=a$, $a^2/\bar{z}=z$, so the two terms sum to $2\mathrm{Re}f(z)$: the streamfunction is constant there. Thus no velocity crosses the circle. At infinity the image term tends to the constant $\overline{f(0)}$, which has no effect on velocity, and the far-field flow is retained. This proves the construction, with an additive potential constant immaterial.

Put $A=\mu e^{i\alpha }$, $B=\mu e^{-i\alpha }$. For the dipole the reflected potential is $B/(a^2/z-b)=Bz/(a^2-bz)$, givingIts image pole is $c=a^2/b$, inside the cylinder. The complex velocity isNear $c$, the second term is $C(z-c{)}^{-2}$ with $C=B{a}^2/b^2$, while $v_0(z)=-A/(z-b{)}^2$ is analytic there. The square of the pole term has no residue, nor does $v_0^2$. The cross term has residue $2C{v}_0^{\prime }(c)$, soTake the contour counterclockwise around the cylinder, as in the supplied Blasius force theorem. The residue theorem gives $\frac{1}{2}i\rho \oint (w^{\prime }{)}^{2}dz=-\pi \rho \mathrm{Res}(w^{\prime }{)}^2$. ThereforeThe force points toward the dipole. Its independence of dipole orientation follows from the product $AB={\mu }^2$ in the sole residue contribution.

Wilson's theorem states that for a prime $p$, $(p-1)!\equiv -1(\mathrm{mod}p)$. For odd $p$, pair each nonzero residue with its multiplicative inverse. Only the self-inverse residues remain unpaired; $x^2=1$ in a field gives $x=\pm 1$. Their product is $-1$, proving the result. For $p=2$ the assertion is immediate.

If composite $n\ge 6$ is not the square of a prime, choose a prime divisor $p$ such that $p$ and $n/p$ are distinct. Both occur among the factors of $(n-1)!$, so their product $n$ divides it. In the remaining case $n=p^2$, $p\ge 3$, and the two distinct factors $p,2p$ both occur in the factorial and have product divisible by $p^2$. Thus the factorial is zero modulo every such composite $n$.

For prime $n$, Wilson's congruence makes the numerator sine squared equal ${\sin }^2(\pi /n)$, so its ratio is one. For composite $n\ge 6$, the numerator vanishes. The only exceptional composite in the sum is four; its ratio is ${\sin }^2(3\pi /2)/{\sin }^2(\pi /4)=2$. Therefore for $x\ge 4$ the sum counts the primes plus two, and subtracting two gives exactly the required prime count. All statements concern the integers $2\le n\le \lfloor x\rfloor$.

The $d$ distinct powers $1,a,\dots ,a^{d-1}$ all solve $X^d=1$. By the allowed root bound they are all its solutions. A residue $b$ of order $d$ must therefore equal $a^j$; conversely $\mathrm{ord}(a^j)=d/\gcd (d,j)$, because $d\mid jt$ characterizes the exponents $t$ with $(a^j{)}^t=1$. This proves the equivalence with $\gcd (d,j)=1$ and gives exactly $\phi (d)$ elements of order $d$ whenever any such element exists.

To establish existence at the full order too, put $m=p-1$ and let $N_d$ count elements of order $d$. By Lagrange's theorem for finite groups, these orders divide $m$. The first result gives $N_d=0$ or $\phi (d)$. Thereforeusing the standard divisor-sum identity for Euler's function. Equality forces every deficit to vanish, so $N_m=\boxed{\phi (p-1)}$.

Finally let $h=g^{(p-1)/8}$, which has order eight. Its fourth power is the unique nonidentity square root of one, namely $-1$. Since $h^7=h^{-1}$,because $h^{-2}=h^6=-h^2$. This proves the stated residue squares to two.

Put $\kappa =Q/(2\pi )$, $a=1+i$ and $b=1/\bar{a}=a/2$. The circle theorem gives, up to an additive constant,Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem isIf $q=s/(z-z_0)+h(z)$ locally, the residue of $q^2$ is $2sh(z_0)$. The enclosed poles are zero and $b$, with residuesTheir sum is $2{\kappa }^2\{1/a+1/(b-a)\}=-{\kappa }^2(1-i)$. The residue theorem therefore yieldsThe force points towards the source. For the moment formula the double poles also contribute: ${\mathrm{Res}}_{z_0}(z{q}^2)=s^2+2z_{0}sh(z_0)$. HenceIt follows that $\oint z{q}^{2}dz=-4\pi i{\kappa }^2$ andThe zero torque agrees with pressure acting normally to a circle, hence radially through its centre.

The $d$ distinct powers $1,a,\dots ,a^{d-1}$ all solve $X^d=1$. By the allowed root bound they are all its solutions. A residue $b$ of order $d$ must therefore equal $a^j$; conversely $\mathrm{ord}(a^j)=d/\gcd (d,j)$, because $d\mid jt$ characterizes the exponents $t$ with $(a^j{)}^t=1$. This proves the equivalence with $\gcd (d,j)=1$ and gives exactly $\phi (d)$ elements of order $d$ whenever any such element exists.

To establish existence at the full order too, put $m=p-1$ and let $N_d$ count elements of order $d$. By Lagrange's theorem for finite groups, these orders divide $m$. The first result gives $N_d=0$ or $\phi (d)$. Thereforeusing the standard divisor-sum identity for Euler's function. Equality forces every deficit to vanish, so $N_m=\boxed{\phi (p-1)}$.

Finally let $h=g^{(p-1)/8}$, which has order eight. Its fourth power is the unique nonidentity square root of one, namely $-1$. Since $h^7=h^{-1}$,because $h^{-2}=h^6=-h^2$. This proves the stated residue squares to two.

Put $\kappa =Q/(2\pi )$, $a=1+i$ and $b=1/\bar{a}=a/2$. The circle theorem gives, up to an additive constant,Thus an image source lies at $b$ and an image sink at zero. With a counterclockwise contour around the cylinder, Blasius' force theorem isIf $q=s/(z-z_0)+h(z)$ locally, the residue of $q^2$ is $2sh(z_0)$. The enclosed poles are zero and $b$, with residuesTheir sum is $2{\kappa }^2\{1/a+1/(b-a)\}=-{\kappa }^2(1-i)$. The residue theorem therefore yieldsThe force points towards the source. For the moment formula the double poles also contribute: ${\mathrm{Res}}_{z_0}(z{q}^2)=s^2+2z_{0}sh(z_0)$. HenceIt follows that $\oint z{q}^{2}dz=-4\pi i{\kappa }^2$ andThe zero torque agrees with pressure acting normally to a circle, hence radially through its centre.

For $n\ge 1$, the zero residue contributes zero. Write $S={\sum }_{x\in {\mathbb{F}}_p^{\times }}{x}^n$. Multiplication by any nonzero $\alpha$ permutes these residues, so $S={\alpha }^{n}S$ in ${\mathbb{F}}_p$. If $(p-1)\nmid n$, choose a primitive root $g$; then $g^n\ne 1$, and $(1-g^n)S=0$ forces $S=0$.

If $(p-1)\mid n$, every nonzero $x$ has $x^n=1$, so $S=p-1\equiv -1(\mathrm{mod}p)$. ThereforeThe source also lists $n=0$, although its premise assumes $n\ge 1$. For that separate case, the integer sum has $p$ terms each equal to one, so it is zero modulo $p$.

A primitive root modulo $p$ is a residue class generating the cyclic multiplicative group $(\mathbb{Z}/p\mathbb{Z}{)}^{\times }$, so its order is $p-1$. If $g$ is one generator, all generators are $g^j$ with $\gcd (j,p-1)=1$. Thus the number iswhere $\phi$ is Euler's totient function. This includes $p=2$, with $\phi (1)=1$.

The Milne-Thomson circle theorem says that if a complex potential $f(z)$ has no singularities on or inside $|z|=a$, then a circular impermeable boundary can be introduced by takingThe new singularities are the reflected ones inside the circle. On $|z|=a$, $a^2/z=\overline{z}$, so $W(z)=f(z)+\overline{f(z)}$ is real; the stream function is constant and there is no normal velocity. At infinity the added term tends to $f^{\ast }(0)$, a harmless potential constant, so it preserves the far-field flow. This proves the theorem with zero added net circulation about the cylinder. A logarithmic term may then specify a different cylinder circulation.

For $f(z)=-i\Gamma \log (z-b)/(2\pi )$, put $c=a^2/b$. The reflected term iswith locally consistent logarithm branches. The circle-theorem solution has zero cylinder circulation: its image vortex at $c$ and its central vortex cancel. To impose circulation $-\Gamma$, add $i\Gamma \log z/(2\pi )$. The result isIt consists of the physical vortex and an opposite image vortex inside the obstacle. On the circle, $|z-b|=(b/a)|z-c|$, so its stream function is indeed constant.

The complex velocity isInside the positively oriented obstacle contour, $q^2$ has only the double pole at $c$. Its residue isThe residue theorem and the supplied Blasius force theorem therefore giveConsequentlyThe force points toward the held vortex, independent of the sign of its circulation.

The residue calculation is the force on a circular cylinder from a held exterior vortex.

In the analytic continuation inside the circle, only the image location $c=a^2/b$ is singular; $b$ lies outside. Squaring the velocity givesThe last term has no residue, and the middle term has residuebecause $AB=m^2{a}^2/b^2$ and $c-b=-(b^2-a^2)/b$. The residue theorem and part (a) therefore yieldThus $F_y=0$ and $F_x>0$. The force points towards the planar potential doublet and is independent of its orientation, since the factors $e^{i\alpha }$ and $e^{-i\alpha }$ cancel in $AB$.

Orient $\Gamma$ and label its left side plus and its right side minus. Assume the density is Hölder continuity near the point in question, or has sufficient regularity for the boundary limits below. The source does not explicitly state this regularity assumption. Parametrize locally by oriented arclength $\gamma (s)$, with $\gamma (0)=t$. Define the Cauchy principal value by removing equal arclengths on the two sides:Subtract $f(t)$ from the numerator near $t$. The Hölder continuity remainder is locally bounded in absolute value by a constant times $|s{|}^{\nu -1}$ for some $\nu >0$, so its integral has the same limit from the two sides. The constant part contributes half the residue, with sign determined by the side of the small semicircle. This proves the Sokhotski–Plemelj theoremIn particular $f=w_{+}-w_{-}$ and the Cauchy principal value term with coefficient $1/(\pi i)$ equals $w_{+}+w_{-}$. Substituting these two identities into the singular integral equation gives

Choose a canonical homogeneous solution $\tilde{w}$ whose boundary values do not vanish, and account for any permitted zeros or poles when forming a quotient. The assumptions $a\pm b\ne 0$ alone do not ensure that an arbitrary homogeneous solution has this property. Its jump relation isDividing the inhomogeneous relation by $(a+b){\tilde{w}}_{+}$ therefore givesThe Cauchy integral solves this additive jump. More generally every solution in the chosen analytic class haswhere $H$ has no jump across $\Gamma$ and obeys the applicable growth, endpoint and divisor conditions. Setting $H=0$ gives exactly the intermediate formula printed in the source.

Let $C\rho =(2\pi i{)}^{-1}\mathrm{PV}{\int }_{\Gamma }\rho (\zeta )/(\zeta -t)d\zeta$. Applying Sokhotski–Plemelj theorem to the displayed solution and subtracting boundary values yieldsUseThe normalized particular solution is consequentlyand the full solution isThus the printed expression is a particular solution; homogeneous terms need separate normalization. To see why “general” cannot hold without qualification, take the positively oriented unit circle, $a=(1+z)/2$, $b=(1-z)/2$, and $c=0$. Then $a+b=1$ and $a-b=z$ are nonzero on the contour. The analytic pair ${\tilde{w}}_{+}=1$, ${\tilde{w}}_{-}=1/z$ satisfies the homogeneous jump and decays on the exterior side. Its nonzero density $f=1-1/z$ has these very Cauchy boundary values and solves the original homogeneous equation, whereas the source formula with $c=0$ gives zero. This is homogeneous freedom in a scalar singular integral equation; endpoint and index conditions determine which jump-free additions are permitted.

Orient the smooth contour so the plus side is to its left. Assume the density is Hölder continuous near the observation point and the integrals converge at any other endpoints. The Cauchy principal value isSubtract $f(t)$ from the numerator. The resulting quotient is locally integrable, so its limits from the two sides coincide. For the remaining constant density, straightening the contour gives a half-circle contribution $\pm i\pi f(t)$. Division by $2\pi i$ yields the Sokhotski–Plemelj formulaFor the requested real integral assume $0<\alpha <1$ and $-1<x<1$, ensuring integrable endpoints and an interior principal-value pole. Put $W(t)=(1-t{)}^{\alpha -1}/(1+t{)}^{\alpha }$. Choose $w(z)=(z-1{)}^{\alpha -1}/(z+1{)}^{\alpha }$ analytic off $[-1,1]$, positive for $z>1$, and $w(z)\sim 1/z$ at infinity. Its two boundary values are $w_{\pm }(t)=-e^{\pm i\pi \alpha }W(t)$.

The jump is $-2i\sin (\pi \alpha )W(t)$. The Cauchy representation for this decaying analytic function is therefore $w(z)=-\sin (\pi \alpha ){\int }_{-1}^{1}W(t)/(t-z)dt/\pi$. Taking the mean of its two boundary values removes the half-residue terms and gives

Use the Fourier convention $\hat{u}(k,y)={\int }_{\mathbb{R}}u(x,y)e^{ikx}dx$. Transforming the differential equation givesDefine $k_{+}(k)=\sqrt{k}$ analytic above the downward cut from zero, and $k_{-}(k)=\sqrt{k-i}$ analytic below the upward cut from $i$. More explicitly take $\arg k\in (-\pi /2,3\pi /2)$ and $\arg (k-i)\in (-3\pi /2,\pi /2)$. In the strip $0<\mathrm{Im}k<1$, put $q=k_{+}{k}_{-}$. Then $q^2=k^2-ik$ and $\mathrm{Re}q>0$; at $k=i/2$, $q=1/2$, fixing the choice throughout the strip. The decaying $y$ mode is $e^{-qy}$.

The positive-half-line boundary value has transform ${\int }_0^{\infty }{e}^{ikx}dx=i/k$ for $\mathrm{Im}k>0$. Thus $\hat{u}(k,0)=\overline{f}+i/k$ and ${\hat{u}}_y(k,0)=\overline{g}$. The transformed boundary relation isLet $c=k_{-}(0)=e^{-i\pi /4}$. Splitting off the pole at zero givesThe left side is analytic in the lower domain up to $\mathrm{Im}k<1$: its apparent pole at zero is removable. The right side is analytic for $\mathrm{Im}k>0$. By the supplied boundedness of $k\overline{f}$ and $\overline{g}$, both sides tend to zero at infinity in their half-planes. The gluing and Liouville theorem argument therefore makes their common value zero. ConsequentlyChoose the inversion contour $\Gamma =\{s+i\eta :s\in \mathbb{R}\}$ oriented left to right, with any fixed $0<\eta <1$. It passes above the pole at zero and below the branch point at $i$. Fourier inversion gives exactlyThus the appropriate value is $\sqrt{-i}=e^{-i\pi /4}$. The other proposed value reverses the residue of the boundary step and violates its prescribed positive value.

There is also a far-field qualification in the source. The formula is the intended bounded mixed-boundary solution, but it does not tend to zero along every path to infinity. This can be checked explicitly: in parabolic coordinates $x=a^2-b^2$, $y=2ab$, with $a,b>0$, the mixed-boundary convection-diffusion solution in parabolic coordinates isIndeed $\Delta u=u_{bb}/[4(a^2+b^2)]$ and $u_x=-b{u}_b/[2(a^2+b^2)]$; the identity $u_{bb}=-2b{u}_b$ proves the differential equation. At $b=0$ the boundary value is one, and at $a=0$ its normal derivative is zero. Its traces are $f(x)=\mathrm{erfc}\sqrt{-x}$ for $x<0$ and $g(x)=-1/\sqrt{\pi x}$ for $x>0$, whose one-sided transforms are the expressions derived above.

At a fixed angle away from the positive $x$ axis, $b\to \infty$ and $u\to 0$. But at fixed $y>0$ and $x\to +\infty$, $b\sim y/(2\sqrt{x})$ and $u\to 1$. Hence the radial decay statement cannot hold uniformly as printed; the intended decay is away from the downstream boundary layer. The exact expression also identifies its thickness as $O(\sqrt{x})$.

The $d$ distinct powers $1,a,\dots ,a^{d-1}$ all satisfy $X^d=1$. The supplied root bound says they exhaust those roots. Thus every element of order $d$ has the form $a^n$. If $g=\gcd (n,d)$, the least positive $k$ with $a^{nk}=1$ is $d/g$, soThis proves both directions of the first assertion.

For existence of a primitive root one may apply the finite-field cyclicity criterion proved in the preceding section. There is also a direct counting proof: for each divisor $d$ of $p-1$, either no elements have order $d$, or the preceding argument shows there are exactly $\varphi (d)$. Lagrange's theorem partitions all $p-1$ nonzero residue classes by such orders, while ${\sum }_{d\mid p-1}\varphi (d)=p-1$. Every count is at most its corresponding $\varphi (d)$, so equality of the totals forces equality at every divisor. In particular there are $\boxed{\varphi (p-1)}$ primitive roots.

Put $t=g^{(p-1)/8}$. It has order eight, so $t^4=-1$ and $t^8=1$. The specified element is $k=t+t^7=t+t^{-1}$. Since $t^{-2}=-t^2$, squaring gives $\boxed{k^2=t^2+2+t^{-2}=2(\mathrm{mod}p)}$.

Take source strength $Q$ to mean total two-dimensional flux, so an isolated source has potential $Q\log (z-a)/(2\pi )$. Let $a=1+i$, $b=1/\bar{a}=(1+i)/2$, and put $c_Q=Q/(2\pi )$. The circle theorem gives, up to an additive constant,The derivative is single valued, which is what the force integrals require. With a positively oriented contour around the cylinder, the force form of Blasius' theorem isOnly the poles at zero and $b$ are enclosed; the physical source at $a$ lies outside. The residues of $q^2/c_Q^2$ are $2(1/a+1/b)$ at zero and $2[1/(b-a)-1/b]$ at $b$. Their sum is $2[1/a+1/(b-a)]=-2/a=-(1-i)$. ThereforeFor the second integral, the residue of $z{q}^2/c_Q^2$ at zero is one. At $b$ it is $1+2b[1/(b-a)-1/b]=-3$. Their sum is $-2$. Using precisely the supplied complex-moment convention givesThe vanishing physical moment $M$ is also consistent with pressure acting normally to a circle, hence radially about its centre. If strength is instead defined as the coefficient of $\log (z-a)$, replace $c_Q$ by that strength throughout; stating the flux convention fixes all factors of $2\pi$.

For an odd prime $p$, the Legendre symbol is $\boxed{(a/p)=0}$ if $p\mid a$, $\boxed{(a/p)=1}$ if $a$ is a nonzero square modulo $p$, and $\boxed{(a/p)=-1}$ otherwise. It depends only on the residue class of $a$.