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Suppose $15\nmid n$ and both $n,-n$ are nonsquares. Each elliptic curve has just one nonidentity rational 2-torsion point, $(0,0)$. The duplication formula shows that halving it requires the corresponding coefficient $\pm n$ to be a square, so neither has rational order-four points. If $3\nmid n$, both good reductions at three have order four, forcing torsion point of an elliptic curve to be exactly order two. Otherwise $5\nmid n$; the reduction orders for coefficients $1,2,3,4$ modulo five are $4,2,10,8$. Among opposite coefficients at least one has order two or four, giving the same conclusion for at least one curve by reduction of torsion points on an elliptic curve.

Over a field of characteristic different from two, three linear projections summing to the identity annihilate one another in both orders. Each pair sum is the complementary projection to the third. The sum of two linear projections criterion therefore proves zero products. Every vector is the sum of its three images; applying each projection to a zero sum proves that these images form a direct sum. Choosing a basis in each image gives a simultaneous eigenbasis, with triples of eigenvalues $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$. This is algebraic orthogonality, and does not assume that the projections are orthogonal projections for a specified inner product.

After fixing axial and vertical angular momenta, the heavy symmetric top has energy $A{\dot{\theta }}^2/2+V_{\mathrm{eff}}(\cos \theta )$. A fixed angle strictly between zero and $\pi$ requires both $E=V_{\mathrm{eff}}(u_0)$ and $V_{\mathrm{eff}}^{\prime }(u_0)=0$. Thus $P(u)=2A(1-u^2)(E-V_{\mathrm{eff}}(u))$ has a repeated interior root. A vertical axis belongs to the excluded endpoints: the upright stationary top has $P(u)=2Amga(1-u{)}^2(1+u)$, whose only repeated root is one. An assertion of an interior common root therefore needs the nonvertical hypothesis.

Decide whether a finite directed graph has an even-length directed walk from a specified source $s$ to a specified target $t$. Vertices may repeat and a zero-length walk is allowed. The NL-completeness proof uses the graph of pairs $(v,b)$ with $b\in \{0,1\}$ tracking integer parity: every original edge flips $b$, and acceptance means reaching $(t,0)$ from $(s,0)$. At most $2|V|-1$ steps are needed in the parity graph. For hardness, subdivide every edge of an ordinary reachability instance into two directed edges. Both constructions are computable by logspace transducers. Requiring a simple path in a graph in the original graph changes the question; deleting an odd closed portion need not preserve integer parity.

The squared-slack support vector machine uses constraints $y_i(w\cdot x_i+b)\ge 1-{\xi }_i$. With an unpenalized intercept, its Lagrangian dual problem maximizes ${\sum }_i{\alpha }_i-\frac{1}{2}{\sum }_{i,j}{\alpha }_i{\alpha }_j{y}_i{y}_{j}K(x_i,x_j)-\frac{1}{4C}{\sum }_i{\alpha }_i^2$ subject to ${\alpha }_i\ge 0$ and ${\sum }_i{\alpha }_i{y}_i=0$. The square adds a diagonal penalty in the dual rather than a finite upper box constraint. The Karush-Kuhn-Tucker conditions give $w={\sum }_i{\alpha }_i{y}_i\varphi (x_i)$ and ${\xi }_i={\alpha }_i/(2C)$ for a feature map realizing the positive-semidefinite kernel $K$.

For a normal sample of size $n\ge 2$, the central Student t confidence interval has width $W=2t_{n-1,1-\alpha /2}S/\sqrt{n}$. Its squared width is therefore a fixed multiple of the sample variance. The exact sampling identity $(n-1)S^2/{\sigma }^2\sim {\chi }_{n-1}^2$ gives the displayed law. It quantifies the random variability in interval precision even when the sample size and coverage level are held fixed.

Replacing an existentially quantified variable by a fresh Skolem function of the preceding universally quantified variables. A witness independent of all universal variables is represented by a fresh constant. The operation preserves satisfiability after expanding the first-order signature; it is not generally logical equivalence for arbitrary interpretations of the new symbols.

For the Segre embedding of ${\mathbb{P}}^n\times {\mathbb{P}}^m$, the homogeneous coordinate ring is the algebra generated by $x_i{y}_j$, assigning degree one to every generator. Its degree-$t$ piece consists of all polynomials bihomogeneous of bidegree $(t,t)$: pair the $t$ factors of an $x$-monomial with the $t$ factors of a $y$-monomial to express any such monomial as a product of generators. Counting the independent monomials gives the displayed Hilbert polynomial, agreeing with the Hilbert function for every integer $t\ge 0$. Its polynomial degree is $n+m$, giving the variety's dimension. Its leading coefficient is $1/(n!m!)$, giving embedded projective degree $(n+m)!/(n!m!)=\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$. The two uses of degree therefore describe different quantities.

For the standard Gödel sentence and Rosser sentence, the formalized Rosser unprovability theorem and $G\leftrightarrow \mathrm{Con}(T)$ yield $T⊢G\to R$. If $T+\mathrm{Con}(T)$ is consistent, the converse is unprovable under the usual formalized incompleteness hypotheses. A proof of $R\to G$ would give $\square \neg G\to \square \neg R$. Formalized negative unprovability identifies $\neg \square \neg R$ with $\mathrm{Con}(T)$ and $\neg \square \neg G$ with $\mathrm{Con}(T+\mathrm{Con}(T))$. The strengthened theory would then prove its own consistency, contradicting the Gödel second incompleteness theorem. The consistency assumption belongs to this strictness conclusion.

For disjoint sets numeralwise enumerated by $P,R$, finite-initial-segment and total-comparison axioms let $H$ separate the first set from the second in the theory. For a member of the first set use a positive $P$ stage and finitely many refutations of $R$. For a member of the second set use its positive $R$ stage: earlier proposed $P$ stages are all refuted by a finite bounded proof, while later stages lose to that known $R$ witness. This proves a separator without inferring arbitrary negative facts from the mere Sigma-1 form of the enumerators.

The internal derivability conditions of a Rosser provability predicate depend on the proof numbering, even when the ordinary provability predicate is unchanged up to provable equivalence. Joining proofs of $A$ and $A\to B$ can increase their codes beyond the two bounds in the Rosser hypotheses, leaving the absence of earlier proofs of $\neg B$ unestablished. Internal modus ponens rule therefore needs a separate argument for the chosen numbering. Proof predicates can be constructed with different combinations of closure and introspection properties. ^[ref]. ^[ref].

If a standard Young tableau of shape $(n-1,1)$ has second-row entry $a$, then its self-pair under RSK corresponds to the adjacent transposition $(a-1,a)$. Inserting the word $1,\dots ,a-2,a,a-1,a+1,\dots ,n$ first appends increasing entries to the first row. At time $a$, inserting $a-1$ bumps $a$ into the unique second-row cell. Every later entry appends to the first row. Thus the insertion tableau is the stated tableau, and the recording tableau has its second-row label $a$ as well. This proves the correspondence directly, including $n=2$, without invoking a separate theorem about involutions.

Let a removable rim hook occupy rows $i$ through $j$. Connectedness and absence of a two-by-two square imply that its pieces in consecutive rows overlap in exactly one column. Therefore ${\mu }_l={\lambda }_{l+1}-1$ for $i\le l<j$, and summing the row lengths gives $|\lambda /\mu |={\lambda }_i-{\mu }_j+j-i$. The column $c={\mu }_j+1$ has height $j$, since lower rows are unchanged and have length at most ${\mu }_j$. The hook length at $(i,c)$ is exactly that sum. Thus a diagram with only odd hook lengths has no removable even rim hook. Applying the Murnaghan–Nakayama rule shows that its ordinary character vanishes on every cycle type containing an even part.

For Brownian motion started at $i$ in the upper half-plane, the real-axis hit $T$ is finite and its Cauchy distribution is nonzero almost surely. The continuous path on $[0,T]$ therefore stays in a compact annulus avoiding zero. The derivative of $z^{\alpha }$ is bounded on that annulus, proving finite changed exit time from its wedge image. The exponent zero is excluded because the map is constant and its image is not an open domain.

For a bounded positive operator $T$, the form $B(x,y)=⟨Tx,y⟩$ is positive semidefinite and Hermitian. Expanding $B(x+ty,x+ty)\ge 0$ for complex $t$ and minimizing its quadratic polynomial gives the displayed inequality. If $B(y,y)=0$, varying $t$ along the negative direction of $B(x,y)$ forces $B(x,y)=0$, covering the degenerate case. In particular $⟨Tx,x⟩=0$ implies $Tx=0$, by testing against arbitrary $y$.

For bounded positive operators $S,T$, membership $x\in \ker (S+T)$ gives $0=⟨Sx,x⟩+⟨Tx,x⟩$. Both terms are nonnegative, so both vanish. The Cauchy-Schwarz inequality for a positive operator makes $Sx=Tx=0$. The reverse inclusion is immediate. Commutativity is not required.

A standard polytabloid is the polytabloid associated with a standard Young tableau. The standard polytabloids of one shape form a basis of the corresponding Specht module. For linear independence, sort each row of a tabloid increasingly and compare row-reading words lexicographically. In the column antisymmetrization of a standard tableau, the identity term is its unique least row word: in the first row affected by a nontrivial column permutation, each moved-in entry is larger than the entry it replaces. Hence its coefficient is one, and all other terms are larger. Distinct standard tableaux have distinct least tabloids, so a linear relation is impossible by taking its least nonzero term. Spanning is the straightening part of the standard polytabloid basis theorem.