Take $T_f$ to consist of $T$ together with the following sentences.

This includes $f(c)=c$ for each constant symbol $c$.

The first pair of axioms makes the interpretation $\vartheta$ a bijection. The remaining schemes say exactly that it preserves every operation and relation, so $(M,\vartheta )\models T_f$ exactly when $M\models T$ and $\vartheta \in \mathrm{Aut}(M)$.

Finally, $T$ is consistent and hence has a model $M$ by part (b). Expanding $M$ by interpreting $f$ as the identity automorphism gives a model of $T_f$, so $T_f$ is consistent.

The compactness theorem states that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model.

The forward implication follows by using the same model for every finite subset. Conversely, suppose every finite subset of $T$ has a model. If $T$ were inconsistent, a formal derivation of $\perp$ from $T$ would use only finitely many assumptions, say those in $T_0\subseteq T$. Then $T_0$ would be inconsistent. By part (b), $T_0$ would have no model, contrary to the hypothesis. Thus $T$ is consistent, and part (b) gives a model of $T$.

Suppose first that $M\models T$. By the soundness theorem for first-order logic, every theorem of $T$ is true in $M$. A contradiction cannot be true in a structure, so $T$ is consistent.

Conversely, the Godel completeness theorem says that every semantically valid consequence is derivable. Its equivalent model-existence form says directly that every consistent first-order theory has a model. Hence

Since $r$ is even,The first factor is not divisible by $N$, since that would contradict minimality of $r$, and the second is not divisible by $N$ by hypothesis. Nevertheless their product is divisible by $N$. Thereforeand Euclid's algorithm computes a nontrivial factor. One can also compute $\gcd (a^{r/2}+1,N)$; together the two gcds expose factors lying on opposite sides of the congruence $a^{r/2}\equiv \pm 1$ modulo the prime-power divisors of $N$.

Defineusing the inverse of $a$ for negative $x$. Since $a^r\equiv 1$, $f(x+r)=f(x)$. Conversely, if $f(x)=f(y)$ then $a^{x-y}\equiv 1$, so the definition of order gives $r\mid x-y$. Two integers in the same block of $r$ consecutive integers cannot differ by a nonzero multiple of $r$. Hence $f$ is one-to-one within each period.

The order of $a$ modulo $N$ is the least positive integer $r$ such thatIt exists because $a$ is a unit in the finite group $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$.

Seek $z=v{e}^{i\Omega t}$ and put $\mu =m/M$. The generalized characteristic equation isorAs a quadratic in $X={\Omega }^2$, this issoA corresponding displacement vector isThus the four normal modes are $v_{+}{e}^{\pm i{\Omega }_{+}t}$ and $v_{-}{e}^{\pm i{\Omega }_{-}t}$. Both squared frequencies are positive for every $\mu >0$, proving linear stability.

For $\mu \ll 1$,Therefore, labeling the larger and smaller positive frequencies by $\omega$ and ${\omega }^{\prime }$,so $\alpha ={\omega }_0$.

Put $\Delta ={\theta }_1-{\theta }_2$ and ${\omega }_0^2=g/l$. The Euler--Lagrange equations, after division by the appropriate powers of $l$, areTime-translation invariance conserves the total energy

Setting both angles and velocities to zero satisfies the equations, so the downward configuration is an equilibrium. Writing ${\theta }_i=z_i$ and retaining linear terms givesEquivalently, the mass and stiffness matrices are

With the pivot as origin and upward vertical coordinate,ThereforeThe potential energy isThus $L=T-V$ is

Take $n=1$ in part (c). The original equation $w_{xx}+w=0$ has the solution $\sin x=\sqrt{1-z^2}$. The solution $w_2$ is normalized by $w_2/(1-z{)}^{1/2}\to 1$ as $z\to 1$, soComparing this with the formula for $w_2$ and using $u=(1-z)/2$ givesThis also agrees with the general identity $F(A,C;C;u)=(1-u{)}^{-A}$.

SetThis Möbius change sends $z=1,-1,\infty$ to $u=0,1,\infty$, respectively, without changing the corresponding exponents. The hypergeometric P-symbol has exponentsMatching it with $(0,1/2)$ at both finite singularities and $(-n,n)$ at infinity givesThe exponent-zero solution normalized to one at $u=0$ is therefore

The second standard local solution at $u=0$ isAfter absorbing the constant $2^{-1/2}$ into its normalization, this becomesTheir exponents at $z=1$ are $0$ and $1/2$, so they are linearly independent.