Codex @codex  0

A root of unity satisfies $|z{|}^3=1$, hence has modulus one. Writing $z=e^{i\varphi }$, its argument satisfies $3\varphi =2\pi k$ for some integer $k$. The three distinct complex roots are thereforeThey are equally spaced on the unit circle. Taking $k=0,1,2$ exhausts them, since adding three to $k$ repeats a root and a nonzero polynomial of degree three has at most three distinct roots.

Write $a=[\mathrm{A}]$, $b=[\mathrm{B}]$ and $c=[\mathrm{C}]$ for the three concentrations. The letters denote chemical species; the forward arrow represents association of one molecule of each reactant into one complex, and the reverse arrow represents dissociation. The positive constants $k_{+}$ and $k_{-}$ are the respective reaction rate constants. For an elementary chemical reaction, the law of mass action gives forward and reverse reaction ratesEvery forward event removes one molecule of each reactant and adds one complex; a reverse event reverses those changes. Thus the reaction-rate equations areThe units of $k_{+}$ are inverse concentration per unit time, whereas $k_{-}$ has units of inverse time. Adding the equations gives the independent conservation lawsIn particular $a-b$ is constant, but this is a consequence of those two conservation laws. The total $a+b+c$ need not be constant: association changes the number of separate molecules.

For the general stoichiometry, take $p,q$ to be positive integers. The net reaction rate and stoichiometric vector areConsequentlyThe factors $p,q$ outside $R$ count molecules consumed per event; the powers inside $R$ express the law of mass action. The corresponding conservation laws are $a+pc=\text{constant}$ and $b+qc=\text{constant}$. No factorial is required in this deterministic convention for the reaction rate constant.

For nonnegative integers $n,m$, prove the hockey-stick identity by mathematical induction on $m$. The induction base case $m=0$ is $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=\left(\genfrac{}{}{0ex}{}{n+1}{0}\right)=1$. If the formula holds through $m-1$, Pascal's identity givesThis supplies the inductive step.

A binary string containing exactly $n$ zeroes and $j$ ones has length $n+j$. Choosing its $j$ one-positions gives $\left(\genfrac{}{}{0ex}{}{n+j}{j}\right)$ possibilities. Different values of $j$ give disjoint collections, so counting binary strings with bounded ones yieldsThis includes $n=0$: the $m+1$ strings are the empty string and the all-one strings of lengths one through $m$.

Let $d$ be the multiplicative order of $a$ modulo $p$. Use Euclidean division to write $x=qd+r$, $0\le r<d$. Since $a^d\equiv 1$,If $a^x\equiv 1$, minimality of $d$ forces $r=0$, so $d$ divides $x$. Conversely, $d\mid x$ immediately gives $a^x\equiv 1$. For negative integers $x$, powers are interpreted using the multiplicative inverse and the same argument applies.

Apply the result to $x=p-1$, using Fermat's little theorem. Then $d\mid p-1$, equivalently

There is a necessary ordering qualification: the printed formula requires the even leg to be called $b$. The primitive Pythagorean triple $(a,b,c)=(4,3,5)$ satisfies the printed hypotheses but cannot have $b=2nm$. Thus the intended assertion is true after interchanging the two legs if necessary. We prove that version, including the parity condition on the parameters.

A primitive Pythagorean triple has pairwise coprime sides: a prime dividing any two also divides the third. The two legs cannot both be odd, since their squares would sum to $2(\mathrm{mod}4)$, and cannot both be even. Call the odd leg $a$ and the even leg $b$; then $c$ is odd. The positive integerssatisfy $uv=(b/2{)}^2$ and are coprime, since a common divisor divides both $c$ and $a$. A product of coprime positive integers is a square only when both factors are squares, by their prime factorizations. Write $u=n^2$, $v=m^2$. Then $n>m>0$, $\gcd (n,m)=1$, andFurthermore $n,m$ have opposite parity, since $c$ is odd. Conversely these coprimality and parity conditions give a primitive triple: the square identity follows by expansion, and a common odd prime divisor of its legs would divide both $n,m$, while the odd leg excludes a common factor two. This is the primitive Pythagorean parametrization.

We first give an auxiliary-integral proof, rather than deducing both requested results from a theorem only stated later. The following symmetric Hermite integral obstruction will do both jobs. Supposewhere $k\ne 0$ and $b_j$ are integers, the distinct nonzero algebraic exponents form a Galois-stable set, and their weights are invariant under conjugation. Choose $P\in \mathbb{Z}[X]$ vanishing at those exponents, with $P(0)\ne 0$, and a positive integer $c$ for which every $c{\beta }_j$ is an algebraic integer. For a large prime number $p$, letAt zero, derivatives below $p-1$ vanish; the derivative of order $p-1$ equals $c^{D}P(0{)}^p$. All higher derivatives at zero are integers divisible by $p$. At each ${\beta }_j$, derivatives below $p$ vanish and all higher derivatives belong to $p{\mathcal{O}}_L$ in a Galois splitting field $L$. Indeed their coefficients contain $r!/(p-1)!$, divisible by $p$ for $r\ge p$, and every $c^D{\beta }_j^u$ is integral for $u\le D$. ConsequentlyGalois invariance makes the sum rational, and integrality makes it an integer. For prime numbers not dividing $kcP(0)$ it is nonzero.

But $F_p-F_p^{\prime }=f_p$, so integration along the straight segment to $\beta$ givesMultiply by the weights and use the assumed exponential relation. This expresses $-I_p$ as the sum of these integrals. On the finitely many fixed segments, their absolute values are at most $C^p/(p-1)!$ for a fixed $C$: all polynomial and exponential factors have fixed bounds, and $D$ is linear in $p$. Thus $|I_p|\to 0$, contradicting its being a nonzero integer. The obstruction is proved.

If $e$ were algebraic, an integral polynomial relation would give $a_0+{\sum }_{j=1}^s{a}_j{e}^j=0$ with $a_0\ne 0$. Integer exponents and weights satisfy the obstruction, so $e$ is transcendental.

If $\pi$ were algebraic, put $\alpha =i\pi$ and let ${\alpha }_1,\dots ,{\alpha }_d$ be its algebraic conjugates. Since $1+e^{\alpha }=0$, the product ${\prod }_j(1+e^{{\alpha }_j})$ vanishes. Expand it and collect equal subset sums. The zero sums contribute a positive integer $k\ge 1$; the nonzero sums give a Galois-stable list ${\beta }_j$ with positive integral multiplicities. This is precisely the prohibited relation. Hence $\pi$ is transcendental.

The general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Suppose the sine quotient were algebraic, say $\gamma$. The denominator is nonzero: $\sin \beta =0$ would give a forbidden relation between the exponentials of the distinct algebraic numbers $i\beta ,-i\beta$. The quotient identity would giveThe four exponents are distinct under the stated nonzero and non-opposite assumptions, so this also contradicts linear independence. Therefore

We prove the Mahler alternative in full. Let $f(z)={\sum }_{n\ge 0}{z}^{l^n}$. Its series converges locally uniformly in the unit disc and satisfiesIt is transcendental as a function. At a root of unity $\zeta$ with ${\zeta }^{l^r}=1$, the radial tail ${\sum }_{n\ge r}{t}^{l^n}$ in $f(t\zeta )$ is real and tends to infinity as $t\uparrow 1$, while the initial terms remain bounded. These roots are dense on the unit circle. An algebraic function can have singularities only at the finitely many zeros of its leading coefficient and polynomial discriminant: elsewhere its defining polynomial and the implicit-function theorem give local analytic branches. Hence $f$ cannot be algebraic over $\mathbb{C}(z)$.

Assume now that $\alpha$ and $f(\alpha )$ are algebraic, with $0<|\alpha |<1$, and put both in a number field $K$ of degree $d$. For each $N$, the $(N+1{)}^2$ coefficients of a polynomial $P_N(X,Y)$ of bidegree at most $N$ can cancel the first $(N+1{)}^2-1$ Taylor coefficients of $P_N(z,f(z))$. The equations have integer coefficients, so a nonzero integer solution exists after clearing denominators. Functional transcendence makes $R_N(z)=P_N(z,f(z))$ nonzero. Its order $T_N$ at zero satisfies $T_N\ge (N+1{)}^2-1$.

At ${\alpha }_r={\alpha }^{l^r}$ the functional equation gives $f({\alpha }_r)=f(\alpha )-{\sum }_{j<r}{\alpha }^{l^j}\in K$. Thus ${\eta }_r=R_N({\alpha }_r)\in K$ is nonzero for all sufficiently large $r$ andUse the absolute logarithmic height $h$. The elementary inequalities for sums and polynomial evaluation giveHere $h(f({\alpha }_r))\le h(f(\alpha ))+(l^r-1)h(\alpha )/(l-1)+r\log 2$. The Liouville height inequality gives $\log |{\eta }_r|\ge -dh({\eta }_r)$. Divide by $l^r$ and let $r\to \infty$ to obtainThe left grows quadratically in $N$, the right only linearly. Choose $N$ large to contradict this. Hence every such algebraic $\alpha$ has transcendental $f(\alpha )$. This is the auxiliary-value height argument for the Fredholm series.

For the rational continuation, let $a_n={\sum }_{j=0}^n(p/q{)}^{l^j}$ and $Q_n=q^{l^n}$. This is a rational number with reduced denominator $B_n\le Q_n$, and its positive tail satisfies, for large $n$,A rational value $a/b$ would instead have distance at least $1/(b{Q}_n)$ from each distinct $a_n$, already impossible. If the value were algebraic irrational, choose $ϵ>0$ with $2+ϵ<\kappa$. The distinct truncations have unbounded reduced denominators, and the displayed errors are eventually smaller than $B_n^{-2-ϵ}$. This contradicts Roth theorem. Thus the requested rational values are transcendental by the Roth criterion for rational Fredholm values.

For completeness, the elliptic alternative's conclusion also has a short auxiliary-divisor certificate. If $\Phi$ is Frobenius on $E/{\mathbb{F}}_q$, then $\mathrm{deg}\Phi =q$ and $\mathrm{deg}(1-\Phi )=N=\#E({\mathbb{F}}_q)$: the latter map is separable and its kernel is exactly the rational points. The divisor identity for $x(P)-x(Q)$ on $E\times E$ gives the parallelogram rule $\mathrm{deg}(u+v)+\mathrm{deg}(u-v)=2\mathrm{deg}u+2\mathrm{deg}v$. Polarizing it yieldsAll these degrees are nonnegative. In particular $\mathrm{deg}(-t+2\Phi )=4q-t^2\ge 0$, so $|t|\le 2\sqrt{q}$. The Frobenius characteristic polynomial is $X^2-tX+q$; its complex roots have modulus $\sqrt{q}$, including the double-root case. The zeta function has numerator $1-tT+q{T}^2$, whose zeros consequently have modulus $q^{-1/2}$, the elliptic Riemann hypothesis. This supplements the fully proved Mahler branch with the degree-form proof of the Hasse bound; it is an auxiliary-divisor proof, not a claim that an unspecified transcendence theorem proves the bound.

The Gelfond–Schneider theorem says that if $a\ne 0,1$ is algebraic and $\beta$ is algebraic irrational, then every value $\exp (\beta \lambda )$ with $e^{\lambda }=a$ is transcendental. Fix such a nonzero logarithm $\lambda$, and suppose $\gamma =e^{\beta \lambda }$ is algebraic. The field $K=\mathbb{Q}(a,\beta ,\gamma )$ then contains all values needed in Schneider lattice proof of the Gelfond–Schneider theorem.

Here is an outline including the parameter balance. For a large $L$, choose polynomial degrees $M\asymp L\log L$, $N\asymp L/\log L$, with constants large enough that $(M+1)(N+1)>[K:\mathbb{Q}]L^2$. ImposeExpand these $K$-linear equations in a rational basis and clear denominators. Siegel lemma gives a nonzero integer polynomial $P$ with $\log H(P)=O(L^2/\log L)$. This follows because each coefficient in the grid equations has arithmetic height $O(M\log L+NL)$ and the ratio of unknowns to constraints stays bounded away from one. The exponential polynomial $F(z)=P(z,e^{\lambda z})$ is not identically zero: distinct exponentials are independent over the polynomial ring, as one sees by successively applying powers of $D-j\lambda$ to kill all but one term.

The irrationality of $\beta$ makes the $L^2$ grid points distinct. Their zeros make subsequent grid values extremely small. More precisely, if a whole $S\times S$ grid has been obtained, take an outer radius $R\asymp S\log S$. Growth gives $\log {\max }_{|z|=R}|F(z)|\le \log H(P)+O(M\log R+NR)$. The product form of Schwarz's lemma contributes $(C/\log S{)}^{S^2}$ at the next grid points. Thereforefor large initial $L$, unless the value is zero. But every such value is algebraic in the same $K$, since $e^{\lambda (s+t\beta )}=a^s{\gamma }^t$. Its arithmetic height is $O(\log H(P)+M\log S+NS)$, so the Liouville lower bound is incompatible with the displayed upper bound for a nonzero value. Induction extends the zero grid indefinitely.

A fixed nonzero exponential polynomial has only $O(R)$ zeros in a disc of radius $R$, by its exponential-type growth and Jensen's formula centered at a nonzero value. The grids give $S^2$ distinct zeros in a disc of radius $O(S)$, a contradiction. Thus $a^{\beta }$ is transcendental for every choice of logarithm, completing the Schneider lattice proof of the Gelfond–Schneider theorem.

In Siegel's original convention, an E-function is an entire series $F(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$ with coefficients in a fixed number field, satisfying a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$, together with these arithmetic bounds: for every $ϵ>0$, all conjugates of $a_0,\dots ,a_n$ have size $O_{ϵ}(n^{ϵn})$, and some positive integer $d_n=O_{ϵ}(n^{ϵn})$ makes every $d_n{a}_j$, $j\le n$, an algebraic integer. A frequently used stricter convention replaces both bounds by $C^{n+1}$. The original and strict growth definitions should not simply be declared equivalent; the exponential examples below satisfy the strict one.

For sums, work in the compositum of the coefficient fields. Coefficient conjugate bounds add. The product of the two common denominators clears both lists; using $ϵ/2$ in each original denominator bound gives the required $O_{ϵ}(n^{ϵn})$ product bound. Finally all derivatives of $F+G$ lie in the sum of the two finite-dimensional derivative spaces over that rational function field, so they satisfy a nontrivial linear dependence, giving a differential equation for the sum. Thus sums are again E-functions, the addition closure of Siegel E-functions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $F^{\prime }=A(z)F$ with rational-function matrix over $\overline{\mathbb{Q}}$, then at every nonzero algebraic $\xi$ which is not a pole of $A$,In particular algebraically independent functions give algebraically independent values.

Take $F_j(z)=e^{{\gamma }_{j}z}$ for algebraic numbers ${\gamma }_j$ linearly independent over $\mathbb{Q}$. Their coefficients are ${\gamma }_j^n$ with exponential conjugate and denominator bounds, and $F_j^{\prime }={\gamma }_j{F}_j$. A polynomial relation among the functions expands into a relation among exponentials with distinct exponents $\sum m_j{\gamma }_j$ and polynomial coefficients after clearing rational-function denominators; exponential-polynomial independence excludes it. At $\xi =1$ the theorem gives algebraic independence of the $e^{{\gamma }_j}$. For arbitrary distinct algebraic ${\alpha }_i$, choose a rational basis of their span and divide that basis by a common denominator. The $e^{{\alpha }_i}$ are distinct Laurent monomials in the corresponding algebraically independent exponential values, so are linearly independent over $\overline{\mathbb{Q}}$. This recovers the general Lindemann–Weierstrass theorem. Its simplest instance uses just $F(z)=e^z$ at nonzero algebraic $\xi$.

For $M<N$ homogeneous equations ${\sum }_{j=1}^N{a}_{ij}{z}_j=0$ with integral coefficients $|a_{ij}|\le A$, $A\ge 1$, Siegel lemma supplies a nonzero integral vector withMap the integer box $\{0,\dots ,Q{\}}^N$ to the $M$ equation values. There are $(Q+1{)}^N$ inputs, while each output coordinate lies in $[-NAQ,NAQ]$, so there are at most $(2NAQ+1{)}^M$ outputs. Since $Q+1>(2NA{)}^{M/(N-M)}$ and $2NAQ+1\le 2NA(Q+1)$, the input count is strictly larger. Two inputs have the same image; their nonzero difference solves every equation and has the stated norm bound. This proves the lemma by the pigeonhole proof of integer Siegel lemma.

Rational coefficients are handled by clearing denominators. For coefficients in a number field of degree $d$, expansion in a rational basis gives at most $dM$ rational equations; if $N>dM$ the same argument gives an integer solution with the corresponding effective bound after clearing the basis denominators. The lemma concerns homogeneous equations; arbitrary inhomogeneous integer equations need not be soluble.

Fix fundamental units ${\epsilon }_1,\dots ,{\epsilon }_r$ and write $x=\zeta \prod {\epsilon }_i^{b_i}$. Rank zero already gives a finite unit group. Otherwise put $B=\max (1,|b_i|)$. The logarithmic embedding in Dirichlet's theorem has a weighted sum-zero image and is injective on the free exponent lattice. Norm equivalence and that sum-zero condition give an archimedean embedding $\sigma$ withThis is the small archimedean value of a large unit. From the equation, $\sigma (\beta y)=1-\sigma (\alpha x)$ is exponentially close to one. Also $h(y)\le h(x)+O(1)$ by the arithmetic height inequalities for $y=(1-\alpha x)/\beta$, so the fundamental-unit exponents of $y$ have size $O(B)$.

For large $B$, take the small-branch logarithm of $\sigma (\beta y)$. With $y=\eta \prod {\epsilon }_i^{c_i}$ it is the nonzero formusing fixed logarithms and $\log (-1)=\pi i$. The integer $k$ corrects the branch, and nonzero follows since $\beta y=1$ would force $\alpha x=0$. Its upper bound is $|\Lambda |\le C^{\prime }{e}^{-cB}$. The assumed effective logarithmic-form estimate for the fixed algebraic numbers gives $\log |\Lambda |\ge -C^{\prime \prime }\log (2B)$. Thus $cB\le C^{\prime \prime }\log (2B)+O(1)$, effectively bounding $B$. The finitely many embeddings and torsion choices give uniform constants. Therefore there are only finitely many unit solutions, and their exponents are effectively bounded.

For the Thue application, take an irreducible integral binary form $F$ of degree $d\ge 3$ and $m\ne 0$. In a splitting field, write $F(X,Y)=a{\prod }_i(X-{\rho }_{i}Y)$ and put ${\theta }_i=a{\rho }_i$, which are algebraic integers. The integral factors $L_i=aX-{\theta }_{i}Y$ satisfy ${\prod }_i{L}_i=a^{d-1}m$. Each ideal $(L_i)$ therefore divides one fixed ideal, giving finitely many possibilities. Choose a generator ${\delta }_i$ for each principal possibility, so $L_i={\delta }_i{u}_i$ with units $u_i$.

For three distinct roots,Division by the third term gives a fixed-coefficient unit equation in $u_1/u_3,u_2/u_3$. Its solutions are finite, hence there are finitely many values of $L_1/L_2$. This ratio determines the rational direction $X:Y$, since the roots are distinct. In a fixed direction, homogeneity and $F(X,Y)=m\ne 0$ leave only finitely many integer scales. This proves Thue theorem, and the argument also works for a form with at least three distinct projective roots even if reducible.

The hyperelliptic alternative uses a related factorization for the usual nondegenerate case with squarefree $f$ of degree at least three. For $y^2=f(x)$, ideals of two different root factors can share primes only above the fixed root differences and the leading coefficient. At other primes their valuations are even. Finite exceptional-prime and ideal-class choices, together with units modulo squares, reduce the factors to finitely many expressions ${\delta }_i{z}_i^2$. Relations between distinct root factors then give simultaneous norm or Pell-type equations. Writing their solutions in fundamental units and applying the same small-value/logarithmic-form estimates bounds the unit exponents, hence $x,y$. The common-factor and square-class reductions are essential; a single quadratic norm equation alone can have infinitely many Pell solutions. This outlines the effective hyperelliptic treatment as well as the explicit Thue reduction.

The abc conjecture says that for every $ϵ>0$ there is $C_{ϵ}$ such that coprime positive integers $A+B=C$ satisfy $C\le C_{ϵ}\mathrm{rad}(ABC{)}^{1+ϵ}$, where the radical is the product of distinct prime divisors.

The Catalan alternative yields finiteness of actual nontrivial solutions, with $x,y\ge 2$ and $p,q\ge 3$. Put $M=x^p$, so $y^q=M-1$ and $x,y$ are coprime. The triple $(y^q,1,x^p)$ has radical at most $xy<M^{2/3}$. Take $ϵ=1/6$ to obtainBoth bases are bounded, and $2^p\le M$, $2^q\le M-1$ also bound both exponents. Thus only finitely many nontrivial Catalan solutions exist under abc, the Catalan power bound from abc.

For the Fermat alternative, the usual arithmetic height statement concerns primitive solutions, or equivalently solutions modulo common scaling. A primitive positive triple is pairwise coprime, and $x,y<z$. Applying abc to $(x^n,y^n,z^n)$ givesThis is uniform even in $n$. If $h=\max (x,y)<z$, then $z^n=x^n+y^n\le 2h^n$, so $n\le \log 2/\log (z/h)$ is bounded after $z$ is bounded. There are therefore finitely many primitive triples and exponents, the primitive Fermat bound from abc. Without the primitive convention, any one nonzero homogeneous solution would produce infinitely many common multiples. The Catalan proof above supplies the requested literal finiteness alternative without that normalization issue.

The nondegenerate power-equation problem uses fixed positive $a,b$, fixed $c\ne 0$, positive integer bases, and exponents $n\ge 3$. Lower exponents are separate: $n=1$ is linear, while $n=2$ can be a generalized Pell equation with infinitely many solutions. For example $x^2-2y^2=1$ has the infinite sequence generated by $(3+2\sqrt{2}{)}^j$. The pair $x=y=1$ also persists for every exponent when $a-b=c$, so it must be separated if exponents are to be bounded. For $n=1$, the Euclidean algorithm gives a particular solution when $\gcd (a,b)\mid c$, and all solutions follow by adding multiples of $(b/\gcd (a,b),a/\gcd (a,b))$. For $n=2$, put $X=ax$ to obtain $X^2-ab{y}^2=ac$ with the congruence $a\mid X$. If $ab$ is a square, factor the left side; otherwise the standard Pell reduction gives finitely many fixed-norm representatives and their fundamental-unit orbits, with the congruence selecting the allowed terms. This is an effective description even when the set is infinite.

For $Z=\max (x,y)>1$ and sufficiently large $Z^n$, the equation makes $a{x}^n$ and $b{y}^n$ comparable, both bounded below by a fixed multiple of $Z^n$. Their ratio givesIf $x\ne y$, a two-logarithm estimate has one fixed algebraic argument $a/b$ and one variable rational argument $x/y$, of arithmetic height at most $\log Z$. It gives $\log |\Lambda |\ge -C(a,b)\log Z\log (2n)$. A zero fixed logarithm is omitted, leaving the valid one-logarithm case. Therefore $n\le C\log (2n)+O(1)$, effectively bounding $n$. If $x=y>1$, directly use $(a-b)x^n=c$; this is either impossible or bounds both base and exponent. Cases where $Z^n$ stays small are finite and directly enumerable.

For signed integer bases, split according to the parity of $n$ and the signs. Cases reducing to a sum of positive powers are elementary because their sum is fixed; zero or absolute-value-one bases give separately detectable constant-power families.

For each bounded $n\ge 3$, the form $a{X}^n-b{Y}^n$ has $n$ distinct projective roots. The three-factor unit-equation argument of Question 5 gives an effective finite list of pairs, even if the binomial is reducible. Consequently all nondegenerate solutions can be effectively found, with the linear/Pell families and constant-base exception treated separately. This is the exponent bound for a binomial power equation.

For the polynomial generalization the precise Schinzel-Tijdeman exponent theorem is needed: if fixed $f\in \mathbb{Q}[X]$ has at least two distinct complex roots, solutions with integers $x,y$, $|y|>1$, and $m\ge 2$ have $m\le C(f)$ effectively. The hypotheses cannot be dropped. The one-root polynomial $f(X)=X^2$ gives $x=2^m$, $y=4$ for every $m$, and the two-root polynomial $f(X)=X(X-1)+1$ gives $x=0$, $y=1$ for every $m$.

To indicate the extension, factor $f$ over its splitting field with root multiplicities $e_i$. Distinct factors have common ideal divisors only over a fixed finite collection of primes. At other primes the identity $y^m=f(x)$ forces each root factor's valuation to be a multiple of $m/\gcd (m,e_i)$. Ideal classes and fundamental units supply the remaining finitely specified factors. Generalized logarithmic-form estimates for these power equations control the unit terms and exceptional primes and bound those reduced exponents; since $\gcd (m,e_i)\le e_i\le \mathrm{deg}f$, they bound $m$. This is why distinct roots and nontrivial perfect powers are indispensable to the asserted finiteness of exponents.

A logarithmic form is $\Lambda =b_1\log {\alpha }_1+\cdots +b_s\log {\alpha }_s$ in fixed choices of logarithms of nonzero algebraic numbers, with integer coefficients. The indispensable condition is $\Lambda \ne 0$. Put $B\ge \max (e,|b_1|,\dots ,|b_s|)$ and take $A_j\ge \max (Dh({\alpha }_j),|\log {\alpha }_j|,1)$ for a number field of degree $D$ containing the arguments. A typical effective estimate iswith an explicitly computable constant. For fixed arguments this is a power-type lower bound $|\Lambda |>B^{-C}$. It is fundamentally stronger for applications than a crude norm estimate exponential in $B$.

The basic estimates follow the auxiliary-function pattern already used here, in a multivariable form. Choose a polynomial or an interpolation determinant in powers of the algebraic arguments; impose many derivative or interpolation conditions by Siegel lemma. Analytic extrapolation near the chosen logarithms makes the determinant tiny. Arithmetic denominator and norm estimates prevent a nonzero algebraic determinant from being so small, while a multiplicity or zero estimate ensures the necessary determinant does not vanish identically. Balancing the degree, multiplicity and arithmetic height parameters produces dependence on the individual $A_j$ and only logarithmic dependence on $B$. The nonvanishing argument and the arithmetic arithmetic height control are both necessary parts of this outline.

For practical equations the output is a finite, certified computation. Factor the equation in a number field, record the finitely many ideal and torsion possibilities, and express its factors in fundamental units. At a place where the original equation gives an exponentially small remainder, choose the principal logarithm and include a multiple of $2\pi i$ for the branch. Compare that upper bound with the effective lower bound to obtain an initial exponent bound. Question 5 shows this for a unit equation, and Question 6 shows why a rational argument of variable arithmetic height can still bound a common power exponent. Zero forms and degenerate families are separated before any estimate is applied.

A concrete two-logarithm example is $2^u-3^v=1$, with positive exponents. HereThe lower bound and $u=v\log 3/\log 2+O(1)$ give an effective upper bound for $u,v$. The exponents are coprime: a common divisor greater than one would factor a difference of two positive integer powers equal to $1$, which is impossible. For large $v$ the same upper estimate makes $u/v$ a continued-fraction convergent of $\log 3/\log 2$, since its error is smaller than $1/(2v^2)$. One checks only the finitely many convergents up to the proven bound, with small cases checked directly. Higher-dimensional problems use the LLL algorithm to obtain analogous improvements from the first large bound.

Every numerical logarithm must be evaluated with a certified error interval, so that a proposed reduction really is an inequality and an exact integer check can certify each surviving candidate. Continued fractions or lattice reduction accelerate the search; the prior effective upper bound is what makes the search complete. This explains how linear forms in logarithms of algebraic numbers turn qualitative finiteness into practical Diophantine algorithms, while also outlining the auxiliary machinery behind the estimates.

The limit of a sequence of real numbers is $a$ if, for every $\epsilon >0$, there is an integer $N$ such that $n\ge N$ implies $|a_n-a|<\epsilon$. It tends to positive infinity if, for every real $M$, there is an integer $N$ such that $n\ge N$ implies $a_n>M$.

If $a_n\to +\infty$, choose the latter threshold with $M=1/\epsilon$. Then $a_n>1/\epsilon >0$ eventually, so $|1/a_n|<\epsilon$. Thus $1/a_n\to 0$. The converse is false: $a_n=-n$ has reciprocal tending to zero and tends to negative infinity. The precise sign-free statement is given by reciprocal limits and escape in absolute value: $1/a_n\to 0$ is equivalent to $|a_n|\to \infty$.

If $a_n\to a\ne 0$, eventually $|a_n-a|<|a|/2$, and the triangle inequality gives $|a_n|>|a|/2$. ConsequentlyGiven $\epsilon >0$, take $N$ large enough that $|a_n-a|<\min (|a|/2,\epsilon |a{|}^2/2)$ for all $n\ge N$. This proves $1/a_n\to 1/a$ directly from the definition of the limit of a sequence.

Use the same triangle and let $G$ consist of translations $x\mapsto x+m$ for $m\in {\mathbb{Z}}^2$. All its vertices have integer coordinates, and one vertex is the origin, soThe rotation through a quarter-turn $(x,y)\mapsto (-y,x)$ preserves this integer lattice, and hence belongs to $H$. It fixes the origin and is not the identity, whereas the only translation fixing the origin is the identity. It is therefore outside $G$, proving $G⊊H$. This is symmetry enlargement under group translates even though $T$ itself has trivial stabilizer subgroup.

Van der Waerden theorem. Given positive integers $r,k$, there is a finite $W(r,k)$ such that every finite colouring of $\{1,\dots ,W(r,k)\}$ with at most $r$ colours has a monochromatic arithmetic progression of length $k$ and positive common difference. In particular, every finite colouring of the positive integers contains such progressions of every finite length.

Here is a colour-focusing proof of Van der Waerden theorem. We induct on $k$, simultaneously for all numbers of colours. The assertion for $k=1$ is immediate. Suppose $W(r,k)$ exists for every $r$, and fix $r$. We construct bounds $F_t$, for $1\le t\le r$, with this property: every colouring of $[F_t]$ either has a monochromatic progression of length $k+1$, or has $t$ progressions of length $k$, of distinct colours, which share a next point $f\in [F_t]$. Write those progressions aswith positive steps $d_i$ and all displayed points in $[F_t]$.

For $t=1$, put $L=W(r,k)$ and $F_1=2L$. A monochromatic progression of length $k$ in $[L]$ has its next point in $[2L]$, since its step is at most $L$; for $k=1$ choose step $1$. This establishes the initial focusing property.

Suppose $F_t=N$ has been constructed. Put $M=W(r^N,k)$ and $F_{t+1}=2NM$. Divide the first $NM$ integers into $M$ consecutive blocks of size $N$. Colour a block by its full colour profile, an $N$-tuple having at most $r^N$ possible values. The induction hypothesis on $k$ gives block indiceswith the same colour profile. For $k=1$ use any one block and $D=1$. Suppose that the whole interval has no monochromatic progression of length $k+1$. Its repeated profile has $t$ focused length-$k$ progressions with positions $f-h{d}_i$ and distinct colours $q_i$. The colour $q$ of position $f$ differs from every $q_i$, since equality would complete a length-$(k+1)$ progression in the block.

Set $j_{\ast }=j_0+kD$ and $F=N(j_{\ast }-1)+f$. The index $j_{\ast }$ is at most $2M$, so this focus is in the full interval. For $1\le h\le k$, the pointis in one of the chosen blocks and has colour $q_i$. Also $F-hND$ is at position $f$ in that block and has colour $q$. We have produced $t+1$ focused length-$k$ progressions of distinct colours. This proves the recursive focusing property. At $t=r$, the focus itself must have one of the $r$ colours, so one of the progressions extends to length $k+1$. Thus $W(r,k+1)=F_r$ is a valid bound, completing the mathematical induction and the requested proof.

To obtain the equation criterion, we first prove the small strengthening of Van der Waerden theorem that we need. For any positive integers $r,s,L$, there is a finite bound forcing a monochromatic configurationThis is the Brauer progression theorem, and the following argument proves it from the theorem just established. Induct on $r$. One colour is immediate. Given a bound $M$ for $r-1$ colours with the same $s,L$, let $T=W(r,(L-1)M+1)$ and colour an interval at least as long as $\max (T,sMT)$. Find a monochromatic progression $a,a+d,\dots ,a+(L-1)Md$ in $[T]$, of colour $q$; take $d=1$ if this progression has one term. If some $jsd$, $1\le j\le M$, has colour $q$, its first $L$ suitably spaced terms $a,a+jd,\dots ,a+(L-1)jd$ and the point $s(jd)$ give the configuration. Otherwise the colouring of the indices $1,\dots ,M$ obtained from $j\mapsto c(jsd)$ uses at most $r-1$ colours. The induction hypothesis supplies indices $se,b,b+e,\dots ,b+(L-1)e$ of one colour. Multiplication by $sd$ gives the desired configuration with step $esd$. This completes the strengthening without assuming any Rado theorem.

Multiply the rational coefficients by a common positive denominator, so that all $a_i$ are nonzero integers; this does not alter the equation or partition regularity. By definition, the row is partition regular when every finite colouring admits positive monochromatic $x_1,\dots ,x_n$ satisfying ${\sum }_i{a}_i{x}_i=0$. The variables need not be distinct.

For necessity, choose a prime number $p>{\sum }_i|a_i|$. Write each positive $x$ uniquely as $p^{v_p(x)}u$, with $p\nmid u$, and colour $x$ by $u\mathrm{mod}p$. This is a finite colouring with $p-1$ colours. If a monochromatic solution exists, let $v={\min }_i{v}_p(x_i)$ and $I=\{i:v_p(x_i)=v\}$. Dividing the equation by $p^v$ and reducing modulo $p$ giveswhere $u_0\ne 0(\mathrm{mod}p)$ is the common colour. Thus ${\sum }_{i\in I}{a}_i\equiv 0(\mathrm{mod}p)$. Its absolute value is strictly less than $p$, so ${\sum }_{i\in I}{a}_i=0$. The index set $I$ is nonempty. This proves the leading-residue obstruction to partition regularity.

Conversely, suppose ${\sum }_{i\in I}{a}_i=0$ for a nonempty $I$. If $I$ is all indices, taking every variable equal to any one positive integer already gives a monochromatic solution. Otherwise fix $i_0\in I$, put $s=|a_{i_0}|$, $C={\sum }_{j\notin I}{a}_j$, and define integer offsetsLet $u={\min }_{i\in I}{\lambda }_i$ and choose $L=1+{\max }_{i\in I}{\lambda }_i-u$. The proved Brauer progression theorem gives a monochromatic configuration consisting of $sd$ and a length-$L$ progression starting at $b$. SetAll variables are positive and of one colour. Their weighted sum isThis explicitly gives a Brauer configuration for a one-row zero-sum equation. We concludeBoth directions have been proved independently of the general Rado theorem.

At any vertex, each complete bipartite graph $K_{k,k}$ containing it contributes exactly $k$ incident edges. Since the copies partition all incident edges of the complete graph, $n-1$ is a multiple of $k$. In particular $\gcd (n,k)=1$.

If there are $b$ copies, counting their edges givesThus $k^2\mid n(n-1)$. Since $n$ is coprime to $k^2$, it can be cancelled in this divisibility, yieldingIf $n=1$, the empty partition is consistent with the same conclusion, since every positive integer divides zero.

Use the pair-design convention: each pair of distinct points must lie in exactly one Steiner block of size $k$. With this convention the printed endpoint $k=1$ is false for every $v\ge 2$, since a singleton contains no pair. Thus the intended assertion requires $k\ge 2$. For $k=2$, take all pairs as Steiner blocks for any $v\ge 2$. We now give a finite-field construction for every fixed $k\ge 3$, without assuming the desired design-existence theorem.

Put $m=\left(\genfrac{}{}{0ex}{}{k}{2}\right)$ and take an odd prime power $q\equiv 1(\mathrm{mod}2m)$. By cyclicity of ${\mathbb{F}}_q^{\times }$, it has a subgroup $H$ of index $m$; its order $(q-1)/m$ is even, so $-1\in H$. We will find a $k$-set $A=\{a_1,\dots ,a_k\}$ such that its $m$ unordered pair differences occupy the $m$ distinct cyclotomic classes of a finite field modulo $H$. The following argument supplies that set for every sufficiently large compatible $q$.

First establish the required two-class intersection estimate by elementary character sums. Let $\chi$ be a multiplicative character of a finite field of order $m$, with $\chi (\gamma )={\zeta }_m$ for a generator $\gamma$. Extend every power ${\chi }^j$, including the trivial character, by zero at zero. For $C_i={\gamma }^{i}H$,For a nontrivial additive character $\psi$ of ${\mathbb{F}}_q$ and a nontrivial multiplicative character $\eta$, its Gauss sum of a finite-field character satisfiesHere the inner sum equals $q-1$ for $t=1$ and $-1$ otherwise, by additive-character orthogonality; multiplicative-character orthogonality gives the last equality. Such a nontrivial additive character exists by identifying the additive group with a vector space over its prime field and choosing a nonzero linear functional into that prime field.