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The quadratic form factors as a sum of two squares:The change of variableshas determinant $-1$, so its matrix is a unimodular matrix and it maps ${\mathbb{Z}}^2$ bijectively to itself. The original equation is therefore equivalent toThe sum of two squares theorem also decides this systematically: $365=5\cdot 73$, and every prime factor congruent to $3$ modulo $4$ occurs to an even power, vacuously here. Thus $365$ is a sum of two integer squares. Explicitly,Inverting the change of variables gives, for example, $(x,y)=(32,-15)$. ConsequentlyThis is the unimodular reduction of a discriminant-minus-four form to two squares.

Applying the Euclidean algorithm,Hence $\gcd (2508,3211)=19$. Since $19$ does not divide $55$, the linear Diophantine equation solvability criterion gives

The linear Diophantine equation solvability criterion says that $ax+by=c$ has an integer solution exactly when $\gcd (a,b)$ divides $c$. The Euclidean algorithm givesThus $\gcd (1007,2314)=1$, which divides $37$. Therefore

Put $\alpha =\sqrt[4]{7}$. The roots of $T^4-7$ are $\alpha ,i\alpha ,-\alpha ,-i\alpha$, so its splitting field isThe polynomial is irreducible over $\mathbb{Q}$ by the Eisenstein criterion at $7$, and $\mathbb{Q}(\alpha )\subset \mathbb{R}$ does not contain $i$. The tower law for field extensions therefore gives $[L:\mathbb{Q}]=8$.

Define automorphismsThey satisfy $r^4=s^2=1$ and $srs=r^{-1}$. The eight maps $r^j$ and $r^{j}s$ are distinct, so they exhaust the Galois group. Hencethe dihedral group of order eight.

The Galois correspondence gives all intermediate fields. The full list, with a single generator for each field, isFor example, $\sqrt{7}+i$ has stabilizer $⟨r^2⟩$, and $(1+i)\alpha$ has stabilizer $⟨rs⟩$; the other entries follow similarly. Also $\alpha +i$ has trivial stabilizer, so its orbit has eight elements and the primitive element theorem conclusion $L=\mathbb{Q}(\alpha +i)$ holds explicitly.

Finally, the normal subextension criterion says that normal intermediate fields correspond exactly to normal subgroups. In $D_8$ these are $D_8$, $⟨r⟩$, the two index-two Klein four-groups, $⟨r^2⟩$, and $\{1\}$, which gives exactly the “yes” rows above. This is the subfields of the splitting field of x to the fourth minus seven lattice.

A finite extension $L/K$ is normal when every irreducible polynomial in $K[T]$ having one root in $L$ splits completely over $L$.

Suppose first that $L/K$ is normal. Choose ${\alpha }_1,\dots ,{\alpha }_m$ with $L=K({\alpha }_1,\dots ,{\alpha }_m)$, and let $f_j$ be the minimal polynomial of ${\alpha }_j$ over $K$. Every $f_j$ splits in $L$ by normality, and $L$ is generated by their roots because it is already generated by the particular roots ${\alpha }_j$. Thus $L$ is the splitting field of $f_1\cdots f_m$.

Conversely, suppose that $L$ is the splitting field of $f\in K[T]$. Every $K$-embedding $\sigma :L\to \overline{K}$ permutes the roots of $f$, so $\sigma (L)=L$. By the embedding characterization of normality, $L/K$ is normal. This proves the finite normal extension as a splitting field criterion.

Fix $k\ge 1$ and sample $G$ from the Erdős-Rényi model with $p=\frac{1}{2}{n}^{-2/3}$. Let $X$ count copies of the complete bipartite graph $K_{3,3}$. By the expected subgraph count in the Erdős-Rényi model,

Put $m=\lceil n/k\rceil$ and let $Z$ count independent set of size $m$. If the chromatic number satisfies $\chi (G)\le k$, some colour class has at least $m$ vertices, so $Z\ge 1$. Moreover,because the positive part of the exponent is $O(n)$ whereas $p\left(\genfrac{}{}{0ex}{}{m}{2}\right)$ has order $n^{4/3}$. The first moment method therefore gives $\mathbb{P}(\chi (G)\le k)\le \mathbb{E}Z=o(1)$.

Using Markov inequality for $X$ and the union bound,for all sufficiently large $n$. Hence at least one such $G$ contains no $K_{3,3}$ and has $\chi (G)>k$.

The compactness theorem says that a set $T$ of first-order sentences has a model if and only if every finite subset of $T$ has a model. One direction is immediate: a model of $T$ models every subset. Conversely, suppose every finite subset is satisfiable. If $T$ had no model, then $T\models \perp$. The Godel completeness theorem would give a formal proof $T⊢\perp$. Every proof is finite, so it would use only a finite subset $T_0\subseteq T$, making $T_0$ inconsistent and hence unsatisfiable. This contradiction proves compactness.

The Upward Lowenheim-Skolem theorem says that a first-order theory with an infinite model has models of arbitrarily large cardinality. Fix a cardinal $\kappa$ and expand the language by constants $c_{\alpha }$ for $\alpha <\kappa$. Add the sentencesto $T$. Any finite subset mentions only finitely many new constants, which can be interpreted as distinct elements of the given infinite model. Compactness therefore gives a model of the expanded theory in which the $c_{\alpha }$ are all distinct. Its reduct is a model of $T$ of cardinality at least $\kappa$. Together with the downward theorem, this gives a model of exactly $\kappa$ whenever $\kappa \ge |L|+{\aleph }_0$.

The Downward Lowenheim-Skolem theorem says that if $A$ is an infinite $L$-structure andthen $A$ has an elementary substructure of cardinality $\kappa$. To see why, first choose $\kappa$ elements. For every existential formula, add a Skolem function selecting a witness whenever one exists, and repeatedly close the chosen set under all language operations and these witness functions. There are at most $\kappa$ functions of finite arity, so the closure still has size $\kappa$. Every existential statement true in $A$ with parameters from the closure has a witness in the closure; the Tarski--Vaught test therefore says that the resulting substructure is elementary.

For $\beta =4$,The branch expanding from $a=0$ hasuntil it approaches $a^2=2$. With $x=a^2$,Solving this linear equation gives the critical closed radiation--cosmological-constant solution

For $H_{0}t\ll 1$,For $H_{0}t\gg 1$,soexponentially from below.

Put $x=a^2$. Apart from the positive factor $H_0^2/a^4$,For $\beta >4$, its discriminant $\beta (\beta -4)$ is positive and its two roots areBoth are positive. Moreover $F(0)>0$ and $F(2)=4-\beta <0$, soA universe expanding from $a=0$ follows the allowed region $0\le a^2\le x_{-}$ and reaches $a^2=x_{-}$ with $H=0$. There,so $H$ crosses from positive to negative: expansion reverses into contraction.

As $a\to 0$ on the contracting branch,Hence $ada\sim -H_0\sqrt{\beta }dt$, so zero scale factor is reached in finite time. The universe therefore undergoes the supercritical closed radiation--cosmological-constant recollapse to a Big Crunch.

Separate continuity givesusing the scalings for radiation in cosmology and cosmological constant energy. At $t=t_0$, where $a=1$, the Friedmann equation becomessoAt general $a$,which rearranges to

Part (a), applied to $\rho ={\rho }_r+{\rho }_{\Lambda }$ and $P={\rho }_r/3-{\rho }_{\Lambda }$, givesSubtracting the displayed formula for $H^2$ yieldsThese relations define the closed radiation--cosmological-constant turning polynomial.

Write $A=8\pi G/(3c^2)$ and $H=\dot{a}/a$. Differentiating the Friedmann equation givesThe cosmological perfect-fluid continuity equation says $\dot{\rho }=-3H(\rho +P)$, so, first where $H\ne 0$ and then everywhere by continuity,Using $K{c}^2/a^2=A\rho -H^2$ from the original equation givesSince $\dot{H}+H^2=\ddot{a}/a$ for the Hubble parameter, this is the Friedmann acceleration equation.

Write $\lambda =N$ with $N\in {\mathbb{Z}}_{\ge 0}$. ThenTake ${\gamma }_1$ to be a small positively oriented circle around $t=0$. For ${\gamma }_2$, start at $t=1$, travel to $-\infty$, and avoid the pole at zero by a fixed upper or lower detour. The endpoint factor vanishes at $t=1$, and the exponential controls $-\infty$.

The finite contour extracts the residue at zero:Only powers $z^0,\dots ,z^N$ occur, and the coefficient of $z^N$ is nonzero. Thus this solution is a polynomial of degree $N$ and, up to normalization, is the Laguerre polynomial $L_N(z)$:This also follows from the recurrence in part (i), because $a_{N+1}=0$ and every later coefficient vanishes.

Write $\lambda =-m$ with $m\in \mathbb{N}$. The amplitude is now the rational functionChoose ${\gamma }_1$ to be a small positively oriented circle around the pole $t=1$, and choose ${\gamma }_2$ from $t=0$ along the negative real axis to $-\infty$. At zero the endpoint factor behaves as $t^m$ and vanishes; at negative infinity the exponential decays.

By the residue theorem, the finite-contour solution isEvery derivative term contains $e^z$ times a power of $z$ of degree at most $m-1$, and the highest-degree term is nonzero. Henceup to a constant, where $P_{m-1}$ is a polynomial of degree $m-1$. This is the negative-integer case of the Integer-parameter Laguerre contour residues.