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Split the factorial at $h$, using $k=p-1-h$:By Wilson theorem, this is $-1$. Therefore $h!k!\equiv (-1{)}^{k+1}(\mathrm{mod}p)$. If the prime number $p$ is odd, then $h+k=p-1$ is even, so $(-1{)}^k=(-1{)}^h$ andFor $p=2$, the signs $1$ and $-1$ are identical modulo $2$, and the two possibilities $(h,k)=(0,1),(1,0)$ also give the same result. The empty product convention $0!=1$ includes the endpoint cases. This is the complementary factorial congruence.

Suppose $X^2\equiv -1(\mathrm{mod}p)$. Then $X$ is a unit modulo n for modulus $p$, and Fermat's little theorem impliesSince $p$ is an odd prime number, $1$ and $-1$ are distinct residue classes. Therefore $(p-1)/2$ is even, and $p\equiv 1(\mathrm{mod}4)$.

Conversely, put $q=(p-1)/2$. Pair each integer $j\in \{1,\dots ,q\}$ with $p-j$ in the factorial. Wilson theorem yieldsWhen $p\equiv 1(\mathrm{mod}4)$, $q$ is even, so $X=q!$ satisfies $X^2\equiv -1(\mathrm{mod}p)$. This proves the first supplementary law for quadratic reciprocity without assuming a primitive root. HenceThe two solutions are $q!$ and $-q!$ modulo $p$: their difference is nonzero, and factoring $X^2-Y^2$ shows that a degree-two polynomial over the field of residues modulo $p$ has no further roots.

The Fermat-Euler theorem states that for an integer $a$ and a positive integer $n$ with $\gcd (a,n)=1$,where the Euler totient function $\phi (n)$ counts the residue classes coprime to $n$. For $n\ge 2$, choose a reduced residue system $r_1,\dots ,r_{\phi (n)}$. Multiplication by the unit modulo n $a$ preserves coprimality and is injective on these residue classes: $a{r}_i\equiv a{r}_j(\mathrm{mod}n)$ implies $r_i\equiv r_j(\mathrm{mod}n)$, because $a$ has a multiplicative inverse modulo $n$. It therefore permutes the reduced residue system, givingThe product is also a unit modulo n, so it can be cancelled. This proves the Fermat-Euler theorem. For $n=1$ every modular congruence is automatic; one may use $\phi (1)=1$.

For a prime number $p$, the Euler totient function has value $\phi (p)=p-1$. Thus $a^{p-1}\equiv 1(\mathrm{mod}p)$ whenever $p\nmid a$. Multiplying by $a$, and treating $p\mid a$ separately, gives the all-integer version of Fermat's little theorem:Finally, Wilson theorem states that for every prime number $p$, $\boxed{(p-1)!\equiv -1(\mathrm{mod}p)}$. Its converse is also true: an integer $n>1$ is prime exactly when $(n-1)!\equiv -1(\mathrm{mod}n)$. For the converse, any proper divisor $1<d<n$ would divide both $(n-1)!$ and $n$, contradicting that congruence.

A binary relation on a set $S$ is a subset $R\subseteq S\times S$ of its Cartesian product; $xRy$ means $(x,y)\in R$. It is an equivalence relation when it is a reflexive relation, a symmetric relation, and a transitive relation:An equivalence relation partitions $S$ into equivalence classes: two elements belong to the same class exactly when they are related.

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.

Reduce $z$ using modular congruence: its residue class modulo $3$ is $0$, $1$, or $-1$. A square number therefore has residue class $0$ or $1$, and raising either residue to the positive integer power $n$ leaves it unchanged. ConsequentlyThe exponent in the original PDF is $2n$; the polynomial identities in the converted TeX are a transcription error. The argument uses $n\ge 1$; it does not assign $0^0$ a value.

For $0\le k<1$, the first burn reduces the total mass from $M+m_0$ to $M+(1-k)m_0$. The rocket equation therefore givesDetachment exerts negligible impulse, so the surviving second stage keeps velocity $v_1$. Its initial and final masses are $(1-k)(M+m_0)$ and $(1-k)M$, respectively. A second application of the rocket equation givesFor the single-stage comparison from rest, $v_{\mathrm{single}}=u\log (1+m_0/M)$. HenceEquality holds at $k=0$; for positive fuel and exhaust speed it is strict when $0<k<1$. This two-stage rocket with proportional fuel and body masses gains speed by discarding inert body mass before the last burn.

As $k\to 1^{-}$ the expression tends to $2u\log (1+m_0/M)$, but the surviving second-stage mass also tends to zero. At exactly $k=1$ there is no second stage, so its final speed is undefined and that limiting formula cannot be assigned to it. The first stage alone reaches the single-stage burnout speed before detachment. If $m_0=0$, neither design gains speed.

Write the total remaining rocket mass as $\mathcal{M}=M+m(t)$, with $\dot{m}=-\alpha$. Over an interval $dt$, an exhaust mass $d{m}_e=\alpha dt$ is expelled backwards with inertial velocity $v-u$ to first order. Applying momentum conservation to the rocket and this exhaust givesCancellation and division by $dt$ give the rocket equationWith $M>0$, $m(t)=m_0-\alpha t$, and $0\le t\le m_0/\alpha$, integration givesAt burnout,The logarithm depends on the mass ratio, not on the burn rate. This follows from momentum conservation of the complete rocket–exhaust system; applying a constant-mass equation directly to the rocket alone would omit the momentum carried away.

The rotating-frame derivative formula is $(d\mathbf{A}/dt{)}_S=(d\mathbf{A}/dt{)}_{S^{\prime }}+\mathbf{\omega }\times \mathbf{A}$. Applying it twice to the position vector, with constant angular velocity, givesSolving for ${\mathbf{a}}_{S^{\prime }}$ identifies the displayed terms: its left side is the acceleration relative to the rotating reference frame, the first right-hand term is the inertial acceleration, $-2\mathbf{\omega }\times {\mathbf{v}}_{S^{\prime }}$ is the Coriolis acceleration, and $-\mathbf{\omega }\times (\mathbf{\omega }\times \mathbf{r})$ is the outward centrifugal acceleration.

For a particle of mass $m$, replace ${\mathbf{a}}_S$ by the actual force divided by $m$ using Newton's second law. The remaining terms then appear as apparent forces in $S^{\prime }$. There is no Euler acceleration because $\mathbf{\omega }$ is constant and no translational acceleration term because the common origin is fixed in the inertial frame.

A central force has no component in the angular direction. The angular component of the acceleration in polar coordinates therefore gives $2\dot{r}\dot{\theta }+r\ddot{\theta }=0$, orFor unit mass, $h=r^2\dot{\theta }$ is the signed angular momentum perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the torque $\mathbf{r}\times \mathbf{F}$ vanishes, which also fixes that plane.

The given potential energy is $k/r$, whose radial force is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The kinetic energy in polar coordinates is $({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$, so conservation of energy givesFor $h\ne 0$ and $k>0$, the effective potential is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa <0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa )$, and has its unique minimum atIt then approaches zero from below. The minimum corresponds to a stable circular orbit. If $h=0$, the centrifugal barrier disappears; the attractive effective potential is simply $-\kappa /r$ and has no minimum.

Use the usual setting of a continuously differentiable solenoidal vector field defined on all radial segments from the origin, including the origin. More generally an open star-shaped set containing zero suffices. Differentiation under the integral sign givesApply divergence and curl of a cross product with the second field $x$, using $\nabla \cdot x=3$ and $(D\cdot \nabla )x=D$:The scaling identity in part (a) now converts the last expression to a one-variable derivative:Continuity at the origin ensures the lower endpoint is zero. Hence the radial vector potential of a solenoidal vector field isThe order of the cross product matters: $x\times D$ would produce $-B$. This construction is not valid on every punctured or multiply connected domain. For example $B=x/|x{|}^3$ is solenoidal away from zero, but its radial integral diverges at zero and it has nonzero flux through a surrounding sphere. The stated formula therefore implicitly needs the regular radial-domain hypothesis, rather than just local zero divergence.

For a twice continuously differentiable vector potential $A$, write the curl with the Levi-Civita symbol:The second derivatives are symmetric in $i,j$, whereas ${ϵ}_{ijk}$ is antisymmetric, so the contraction is zero. Equivalently, writing out the three components shows that each mixed derivative cancels its reversed-order partner. ThusThis is divergence of a curl is zero. The regularity hypothesis is what licenses commutation of the mixed partial derivatives; it must not be omitted for a singular potential without specifying a weaker derivative interpretation.