[proofplan] The theorem classifies the splitting of the rational prime $2$ in a quadratic field $L = \mathbb{Q}(\sqrt{d})$ purely in terms of $d \bmod 8$. The proof runs [Dedekind's Criterion](/theorems/1598) at $p = 2$: this reduces the factorization of $\langle 2 \rangle$ to the factorization of the minimal polynomial of a generator of $\mathcal{O}_L$ modulo $2$. Because Dedekind's criterion requires $p \nmid [\mathcal{O}_L : \mathbb{Z}[\alpha]]$, the choice of generator depends on $d \bmod 4$: use $\alpha = \frac{1+\sqrt{d}}{2}$ when $d \equiv 1 \pmod 4$ (index $1$) and $\alpha = \sqrt{d}$ when $d \equiv 2, 3 \pmod 4$ (index $1$). In each of the three arithmetic progressions $d \bmod 8$, the reduction of the minimal polynomial modulo $2$ takes one of only three shapes — two distinct linear factors (split), irreducible quadratic (inert), or a repeated linear factor (ramified). [/proofplan]

[step:Split into cases by $d \bmod 4$ and identify the correct generator of $\mathcal{O}_L$]By [Integers of Quadratic Fields](/theorems/1575), \begin{align*} \mathcal{O}_L = \begin{cases} \mathbb{Z}[\sqrt{d}] & \text{if } d \equiv 2 \text{ or } 3 \pmod 4, \\ \mathbb{Z}\!\left[\tfrac{1}{2}(1 + \sqrt{d})\right] & \text{if } d \equiv 1 \pmod 4. \end{cases} \end{align*} In each case $\mathcal{O}_L = \mathbb{Z}[\alpha]$ for the indicated $\alpha$, so $[\mathcal{O}_L : \mathbb{Z}[\alpha]] = 1$ and the hypothesis $2 \nmid [\mathcal{O}_L : \mathbb{Z}[\alpha]]$ of [Dedekind's Criterion](/theorems/1598) holds for the prime $p = 2$. Since $d$ is square-free and $d \neq 0, 1$, the three remaining cases modulo $4$ are $d \equiv 1, 2, 3 \pmod 4$. (The case $d \equiv 0 \pmod 4$ is impossible because $4 \mid d$ contradicts square-freeness.) We refine the case $d \equiv 1 \pmod 4$ further to $d \equiv 1$ or $5 \pmod 8$ because the reduction of the minimal polynomial modulo $2$ will depend on this finer information.[/step]

[guided]To apply Dedekind's criterion we need a generator $\alpha$ of $\mathcal{O}_L$ as a $\mathbb{Z}$-algebra whose index $[\mathcal{O}_L : \mathbb{Z}[\alpha]]$ is coprime to $p = 2$. The cleanest choice is one for which $\mathcal{O}_L = \mathbb{Z}[\alpha]$, which gives index $1$ (coprime to every prime). For quadratic fields $L = \mathbb{Q}(\sqrt{d})$ with $d$ square-free, [Integers of Quadratic Fields](/theorems/1575) describes $\mathcal{O}_L$ explicitly: \begin{align*} \mathcal{O}_L = \begin{cases} \mathbb{Z}[\sqrt{d}] & \text{if } d \equiv 2 \text{ or } 3 \pmod 4, \\ \mathbb{Z}\!\left[\tfrac{1}{2}(1 + \sqrt{d})\right] & \text{if } d \equiv 1 \pmod 4. \end{cases} \end{align*} So in each congruence class of $d \bmod 4$, $\mathcal{O}_L$ is monogenic (generated by a single element) and we take $\alpha$ to be that generator. Then $[\mathcal{O}_L : \mathbb{Z}[\alpha]] = 1$, Dedekind's hypothesis is satisfied, and we can read off the splitting of $\langle 2 \rangle$ from the factorization of the minimal polynomial of $\alpha$ modulo $2$. A warning about the *wrong* generator: when $d \equiv 1 \pmod 4$, one might be tempted to use $\alpha = \sqrt{d}$ instead of $\frac{1+\sqrt{d}}{2}$. But then $\mathbb{Z}[\sqrt{d}] \subsetneq \mathcal{O}_L$ with index $2$, so $2 \mid [\mathcal{O}_L : \mathbb{Z}[\sqrt{d}]]$ and Dedekind's criterion does **not** apply at $p = 2$. Applying it anyway would yield an incorrect factorization — this is the subtle point flagged at the end of the sketch. Because Dedekind's criterion acts on $g(x) \bmod 2$, the answer depends on $d \bmod 2$ (in the case $d \equiv 2, 3 \pmod 4$) or on $\frac{1-d}{4} \bmod 2$ (in the case $d \equiv 1 \pmod 4$), which in turn depends on $d \bmod 8$. This is the source of the mod-$8$ statement in the theorem.[/guided]

[step:Handle $d \equiv 1 \pmod 8$ by factoring the minimal polynomial of $\frac{1+\sqrt{d}}{2}$ as $x(x-1) \bmod 2$]Suppose $d \equiv 1 \pmod 8$. Then $d \equiv 1 \pmod 4$, so $\mathcal{O}_L = \mathbb{Z}[\alpha]$ with $\alpha = \tfrac{1}{2}(1 + \sqrt{d})$. The minimal polynomial of $\alpha$ is \begin{align*} g(x) = x^2 - x + \tfrac{1-d}{4}. \end{align*} Since $d \equiv 1 \pmod 4$, $\tfrac{1-d}{4} \in \mathbb{Z}$. Since $d \equiv 1 \pmod 8$, $\tfrac{1-d}{4} \equiv 0 \pmod 2$. Reducing $g$ modulo $2$: \begin{align*} \bar{g}(x) = x^2 - x = x(x-1) \in \mathbb{F}_2[x]. \end{align*} The factors $x$ and $x - 1$ are distinct monic irreducibles in $\mathbb{F}_2[x]$ (both of degree $1$). By [Dedekind's Criterion](/theorems/1598), with lifts $\tilde{\varphi}_1(x) = x$ and $\tilde{\varphi}_2(x) = x - 1$, \begin{align*} \langle 2 \rangle = \mathfrak{p}_1 \mathfrak{p}_2, \qquad \mathfrak{p}_1 = \langle 2, \alpha \rangle, \quad \mathfrak{p}_2 = \langle 2, \alpha - 1 \rangle, \end{align*} with $\mathfrak{p}_1 \neq \mathfrak{p}_2$ and $N(\mathfrak{p}_i) = 2^1 = 2$. This is the definition of $2$ splitting in $L$.[/step]

[guided]We specialize to $d \equiv 1 \pmod 8$. The hypothesis $d \equiv 1 \pmod 8$ is stronger than $d \equiv 1 \pmod 4$, so we are in the case $\mathcal{O}_L = \mathbb{Z}[\alpha]$ with $\alpha = \frac{1+\sqrt{d}}{2}$. **Minimal polynomial of $\alpha$.** The number $\alpha = \frac{1+\sqrt{d}}{2}$ satisfies $2\alpha - 1 = \sqrt{d}$, and squaring gives $(2\alpha - 1)^2 = d$, i.e., $4\alpha^2 - 4\alpha + 1 = d$, i.e., $\alpha^2 - \alpha + \frac{1-d}{4} = 0$. So \begin{align*} g(x) = x^2 - x + \tfrac{1-d}{4}. \end{align*} The constant term $\frac{1-d}{4}$ lies in $\mathbb{Z}$ because $d \equiv 1 \pmod 4$ means $4 \mid 1 - d$. Good — $g \in \mathbb{Z}[x]$ as required by Dedekind. **Reduction modulo $2$.** The key observation: $d \equiv 1 \pmod 8 \iff 1 - d \equiv 0 \pmod 8 \iff \frac{1-d}{4} \equiv 0 \pmod 2$. Reducing coefficient-by-coefficient: \begin{align*} \bar{g}(x) = x^2 - x + 0 = x^2 - x = x(x-1) \in \mathbb{F}_2[x]. \end{align*} Note $x - 1 = x + 1$ in $\mathbb{F}_2[x]$, so we could equivalently write $\bar{g}(x) = x(x+1)$. The factors $x$ and $x - 1$ are distinct: they vanish at $0$ and $1$, respectively, which are the two distinct elements of $\mathbb{F}_2$. They are monic irreducibles because both have degree $1$. **Applying Dedekind's criterion.** [Theorem 1598](/theorems/1598) takes $\bar{g}(x) = \varphi_1(x)^{e_1} \cdots \varphi_m(x)^{e_m}$ with $\varphi_i$ distinct monic irreducibles in $\mathbb{F}_p[x]$, lifts each $\varphi_i$ to $\tilde{\varphi}_i \in \mathbb{Z}[x]$, and asserts \begin{align*} \langle p \rangle = \mathfrak{p}_1^{e_1} \cdots \mathfrak{p}_m^{e_m}, \qquad \mathfrak{p}_i = \langle p, \tilde{\varphi}_i(\alpha) \rangle, \qquad N(\mathfrak{p}_i) = p^{\deg \varphi_i}. \end{align*} Verifying hypotheses: $\alpha \in \mathcal{O}_L$, $g \in \mathbb{Z}[x]$ is the minimal polynomial of $\alpha$, and $2 \nmid [\mathcal{O}_L : \mathbb{Z}[\alpha]] = 1$. All satisfied. Here $\varphi_1(x) = x$, $\varphi_2(x) = x - 1$, $e_1 = e_2 = 1$, $\deg \varphi_i = 1$. Choose lifts $\tilde{\varphi}_1(x) = x$, $\tilde{\varphi}_2(x) = x - 1$. Dedekind gives \begin{align*} \langle 2 \rangle = \mathfrak{p}_1 \mathfrak{p}_2, \quad \mathfrak{p}_1 = \langle 2, \alpha \rangle, \quad \mathfrak{p}_2 = \langle 2, \alpha - 1 \rangle, \quad N(\mathfrak{p}_i) = 2. \end{align*} The factorization has two distinct prime factors, each with residue degree $1$ and ramification index $1$. This is precisely the definition: $2$ **splits** in $L$.[/guided]

[step:Handle $d \equiv 5 \pmod 8$ by observing the minimal polynomial becomes irreducible mod $2$]Suppose $d \equiv 5 \pmod 8$. Then $d \equiv 1 \pmod 4$, so again $\mathcal{O}_L = \mathbb{Z}[\alpha]$ with $\alpha = \tfrac{1}{2}(1 + \sqrt{d})$ and $g(x) = x^2 - x + \tfrac{1-d}{4}$. Since $d \equiv 5 \pmod 8$, $\tfrac{1-d}{4} \equiv \tfrac{-4}{4} = -1 \equiv 1 \pmod 2$. Reducing modulo $2$: \begin{align*} \bar{g}(x) = x^2 - x + 1 = x^2 + x + 1 \in \mathbb{F}_2[x], \end{align*} where we used $-x \equiv x \pmod 2$. This polynomial has no root in $\mathbb{F}_2$ (evaluation: $\bar{g}(0) = 1 \neq 0$ and $\bar{g}(1) = 1 + 1 + 1 = 1 \neq 0$) and has degree $2$, so it is irreducible in $\mathbb{F}_2[x]$. Applying [Dedekind's Criterion](/theorems/1598) with $\varphi_1(x) = x^2 + x + 1$, $e_1 = 1$, and lift $\tilde{\varphi}_1(x) = x^2 + x + 1$: \begin{align*} \langle 2 \rangle = \mathfrak{p}, \qquad \mathfrak{p} = \langle 2, \alpha^2 + \alpha + 1 \rangle, \qquad N(\mathfrak{p}) = 2^{\deg \varphi_1} = 2^2 = 4. \end{align*} Since $\langle 2 \rangle$ is itself a prime ideal of $\mathcal{O}_L$, by definition $2$ is **inert** in $L$.[/step]

[guided]The setup is identical to the previous step — we use the same $\alpha$ and the same polynomial $g(x) = x^2 - x + \frac{1-d}{4}$ — but now $d \equiv 5 \pmod 8$ changes the constant term modulo $2$. **Computing $\frac{1-d}{4} \bmod 2$.** If $d \equiv 5 \pmod 8$, write $d = 5 + 8k$ for some $k \in \mathbb{Z}$. Then \begin{align*} \frac{1 - d}{4} = \frac{1 - 5 - 8k}{4} = \frac{-4 - 8k}{4} = -1 - 2k, \end{align*} which is odd, i.e., $\frac{1-d}{4} \equiv 1 \pmod 2$. So \begin{align*} \bar{g}(x) = x^2 - x + 1 \in \mathbb{F}_2[x]. \end{align*} In $\mathbb{F}_2$ we have $-1 = 1$, so $\bar{g}(x) = x^2 + x + 1$. **Irreducibility in $\mathbb{F}_2[x]$.** A degree-$2$ polynomial over a field is irreducible iff it has no root in that field. Test all two elements of $\mathbb{F}_2$: \begin{align*} \bar{g}(0) = 0 + 0 + 1 = 1 \neq 0, \qquad \bar{g}(1) = 1 + 1 + 1 = 1 \neq 0. \end{align*} No roots, hence irreducible. (This is the unique irreducible quadratic in $\mathbb{F}_2[x]$, corresponding to the field $\mathbb{F}_4 = \mathbb{F}_2[x]/(x^2 + x + 1)$.) **Dedekind's criterion gives inertness.** With $\varphi_1 = x^2 + x + 1$ (degree $2$, multiplicity $e_1 = 1$) and lift $\tilde{\varphi}_1 = x^2 + x + 1$: \begin{align*} \langle 2 \rangle = \mathfrak{p}^1 = \mathfrak{p}, \qquad \mathfrak{p} = \langle 2, \alpha^2 + \alpha + 1 \rangle, \qquad N(\mathfrak{p}) = 2^2 = 4. \end{align*} So $\langle 2 \rangle$ is itself prime in $\mathcal{O}_L$ and has norm $4 = 2^n$ with $n = [L : \mathbb{Q}] = 2$. By definition, $2$ is **inert** in $L$: there is a single prime factor, with residue degree equal to $n$ and ramification index $1$. The residue field $\mathcal{O}_L / \mathfrak{p}$ has size $N(\mathfrak{p}) = 4$, which is consistent with it being $\mathbb{F}_2[x]/(x^2 + x + 1) \cong \mathbb{F}_4$ — exactly as Dedekind's criterion predicts.[/guided]

[step:Handle $d \equiv 2, 3 \pmod 4$ by showing the minimal polynomial of $\sqrt{d}$ becomes a square mod $2$]Suppose $d \equiv 2 \pmod 4$ or $d \equiv 3 \pmod 4$. Then $\mathcal{O}_L = \mathbb{Z}[\alpha]$ with $\alpha = \sqrt{d}$, and the minimal polynomial is \begin{align*} g(x) = x^2 - d. \end{align*} We compute $\bar{g}(x) \in \mathbb{F}_2[x]$ in both subcases. *Case $d \equiv 2 \pmod 4$.* Then $d$ is even, so $d \equiv 0 \pmod 2$, giving \begin{align*} \bar{g}(x) = x^2 - 0 = x^2 = (x - 0)^2 \in \mathbb{F}_2[x]. \end{align*} *Case $d \equiv 3 \pmod 4$.* Then $d$ is odd, so $d \equiv 1 \pmod 2$, giving \begin{align*} \bar{g}(x) = x^2 - 1 = x^2 + 1 = (x + 1)^2 \in \mathbb{F}_2[x], \end{align*} where the last equality uses $(x+1)^2 = x^2 + 2x + 1 \equiv x^2 + 1 \pmod 2$. In both subcases, $\bar{g}(x) = \varphi_1(x)^2$ for a single monic linear irreducible $\varphi_1$: $\varphi_1(x) = x$ when $d \equiv 2 \pmod 4$, and $\varphi_1(x) = x + 1$ when $d \equiv 3 \pmod 4$. Applying [Dedekind's Criterion](/theorems/1598): \begin{align*} \langle 2 \rangle = \mathfrak{p}^2, \qquad N(\mathfrak{p}) = 2^1 = 2, \end{align*} where \begin{align*} \mathfrak{p} = \begin{cases} \langle 2, \sqrt{d} \rangle & \text{if } d \equiv 2 \pmod 4, \\ \langle 2, \sqrt{d} + 1 \rangle & \text{if } d \equiv 3 \pmod 4. \end{cases} \end{align*} Since the ramification index is $2 > 1$, by definition $2$ **ramifies** in $L$.[/step]

[guided]Here $d \equiv 2$ or $3 \pmod 4$, so $\mathcal{O}_L = \mathbb{Z}[\sqrt{d}]$ and the generator is $\alpha = \sqrt{d}$ with minimal polynomial $g(x) = x^2 - d$. The index $[\mathcal{O}_L : \mathbb{Z}[\sqrt{d}]] = 1$, so Dedekind's criterion applies at $p = 2$. **Reducing $x^2 - d$ modulo $2$.** Only the parity of $d$ matters. We handle the two subcases separately. *Subcase $d \equiv 2 \pmod 4$.* In particular, $d$ is even, so $d \equiv 0 \pmod 2$ and \begin{align*} \bar{g}(x) = x^2 - 0 = x^2 = (x)^2 \in \mathbb{F}_2[x]. \end{align*} This is the square of the irreducible linear polynomial $\varphi_1(x) = x$. *Subcase $d \equiv 3 \pmod 4$.* Then $d$ is odd, so $d \equiv 1 \pmod 2$, and \begin{align*} \bar{g}(x) = x^2 - 1 \in \mathbb{F}_2[x]. \end{align*} To write this as a power of an irreducible, note that over $\mathbb{F}_2$ the Frobenius identity $(a + b)^2 = a^2 + b^2$ holds (since the characteristic is $2$), so $(x + 1)^2 = x^2 + 1$. Also $-1 \equiv 1 \pmod 2$, so $x^2 - 1 \equiv x^2 + 1 \pmod 2$. Therefore \begin{align*} \bar{g}(x) = x^2 + 1 = (x + 1)^2 \in \mathbb{F}_2[x], \end{align*} which is the square of the irreducible linear polynomial $\varphi_1(x) = x + 1$. **Applying Dedekind's criterion.** In both subcases, $\bar{g}(x) = \varphi_1(x)^{e_1}$ with $e_1 = 2$, a single irreducible factor of degree $1$. With a choice of lift $\tilde{\varphi}_1 \in \mathbb{Z}[x]$ (either $x$ or $x + 1$), [Theorem 1598](/theorems/1598) gives \begin{align*} \langle 2 \rangle = \mathfrak{p}^2, \qquad \mathfrak{p} = \langle 2, \tilde{\varphi}_1(\sqrt{d}) \rangle, \qquad N(\mathfrak{p}) = 2^{\deg \varphi_1} = 2. \end{align*} Concretely: \begin{align*} \mathfrak{p} = \begin{cases} \langle 2, \sqrt{d} \rangle & \text{if } d \equiv 2 \pmod 4, \\ \langle 2, \sqrt{d} + 1 \rangle & \text{if } d \equiv 3 \pmod 4. \end{cases} \end{align*} **Interpretation.** The exponent $e = 2$ of $\mathfrak{p}$ in $\langle 2 \rangle$ is the *ramification index* of $\mathfrak{p}$ over $2$. Since it is strictly greater than $1$, by definition $2$ **ramifies** in $L$, with a single prime above it of ramification index $2$ and residue degree $1$. Note the norm check: $N(\mathfrak{p}^2) = N(\mathfrak{p})^2 = 4 = 2^n$, as required. **Why using the wrong generator fails.** It is instructive to note that the sketch's warning applies here in the opposite direction too: if we had erroneously used $\alpha' = \sqrt{d}$ as the generator when $d \equiv 1 \pmod 4$ (where $\mathcal{O}_L$ is strictly larger than $\mathbb{Z}[\sqrt{d}]$), we would have computed $\bar{g}(x) = x^2 - d \equiv x^2 - 1 = (x+1)^2 \pmod 2$ (since $d$ is odd). Dedekind's criterion would then falsely predict $2$ ramifies in **every** case $d \equiv 1 \pmod 4$. This is wrong: when $d \equiv 1 \pmod 8$, we have shown above that $2$ actually splits. The discrepancy is precisely what the hypothesis $p \nmid [\mathcal{O}_L : \mathbb{Z}[\alpha]]$ protects against; here the index $[\mathcal{O}_L : \mathbb{Z}[\sqrt{d}]] = 2$ is divisible by $p = 2$, so the hypothesis fails and Dedekind's criterion is not applicable.[/guided]

[step:Assemble the three cases into the biconditionals in the theorem statement]The three cases $d \equiv 1 \pmod 8$, $d \equiv 5 \pmod 8$, and $d \equiv 2, 3 \pmod 4$ partition the square-free integers $d \neq 0, 1$: indeed, every integer is congruent modulo $4$ to exactly one of $1, 2, 3$ (excluding $0$ by square-freeness for $d \not\in \{0\}$; the case $d = 1$ is excluded by hypothesis, and $d \equiv 1 \pmod 4$ splits further into $d \equiv 1$ or $5 \pmod 8$). Steps 2-4 establish, case-by-case: \begin{align*} d \equiv 1 \pmod 8 &\implies 2 \text{ splits}, \\ d \equiv 5 \pmod 8 &\implies 2 \text{ is inert}, \\ d \equiv 2 \text{ or } 3 \pmod 4 &\implies 2 \text{ ramifies}. \end{align*} Since the cases are mutually exclusive and collectively exhaustive, and since a rational prime in a quadratic extension must be exactly one of split, inert, or ramified (by $\sum_i e_i f_i = n = 2$, the only decompositions of $\langle 2 \rangle$ are $\mathfrak{p}_1 \mathfrak{p}_2$ with distinct $\mathfrak{p}_i$ of residue degree $1$, a single $\mathfrak{p}$ of residue degree $2$, or $\mathfrak{p}^2$ with $\mathfrak{p}$ of residue degree $1$), each forward implication is also a biconditional. This yields the three statements of the theorem.[/step]

[guided]We have established three forward implications: \begin{align*} d \equiv 1 \pmod 8 \implies 2 \text{ splits}, \quad d \equiv 5 \pmod 8 \implies 2 \text{ is inert}, \quad d \equiv 2, 3 \pmod 4 \implies 2 \text{ ramifies}. \end{align*} The theorem asserts the converse implications as well. We package both directions into biconditionals via a trichotomy argument. **Trichotomy on the left-hand side.** Every square-free integer $d \neq 0, 1$ falls into exactly one of the three congruence classes: - $d \equiv 1 \pmod 8$, - $d \equiv 5 \pmod 8$, - $d \equiv 2 \pmod 4$ or $d \equiv 3 \pmod 4$. Together with $d \equiv 0 \pmod 4$ (which would force $4 \mid d$, contradicting square-freeness), these classes partition $\mathbb{Z} \setminus \{0, 1\}$. Indeed, modulo $8$ we have eight residues, and the odd ones are $1, 3, 5, 7$; the residues $1, 5 \pmod 8$ are exactly those with $d \equiv 1 \pmod 4$, while $3, 7 \pmod 8$ satisfy $d \equiv 3 \pmod 4$, which is included in the third class. The even residues modulo $8$ are $0, 2, 4, 6$; of these $0$ and $4$ are impossible (they give $4 \mid d$), and $2, 6$ are $d \equiv 2 \pmod 4$, also included in the third class. **Trichotomy on the right-hand side.** For a rational prime $p$ in a quadratic extension $L/\mathbb{Q}$ of degree $n = 2$, the fundamental identity from [Prime Ideals Lie Above Rational Primes](/theorems/1597) gives \begin{align*} \sum_{\mathfrak{p} \mid \langle p \rangle} e_{\mathfrak{p}} f_{\mathfrak{p}} = n = 2, \end{align*} where $e_{\mathfrak{p}}$ is the ramification index and $f_{\mathfrak{p}}$ is the residue degree. The only way to write $2$ as a sum of positive integers $e_i f_i$ is: - two terms with $e_i = f_i = 1$ (two distinct primes, residue degree $1$ each) — **split**; - one term with $e = 1$, $f = 2$ (single prime, residue degree $2$) — **inert**; - one term with $e = 2$, $f = 1$ (single prime squared, residue degree $1$) — **ramified**. These three outcomes are mutually exclusive. **Converting forward implications to biconditionals.** We have three mutually exclusive classes on the left, three mutually exclusive outcomes on the right, and forward implications between matching pairs. Standard logic (or a cleanup: if $P_1, P_2, P_3$ partition and $Q_1, Q_2, Q_3$ partition, and $P_i \implies Q_i$, then each $Q_i \implies P_i$ by contradiction — if $P_j$ held with $j \neq i$, then $Q_j$ would hold, contradicting the exclusivity of $Q_i$ and $Q_j$) shows each forward implication is a biconditional: \begin{align*} 2 \text{ splits} &\iff d \equiv 1 \pmod 8, \\ 2 \text{ is inert} &\iff d \equiv 5 \pmod 8, \\ 2 \text{ ramifies} &\iff d \equiv 2 \text{ or } 3 \pmod 4. \end{align*} This is the theorem statement.[/guided]