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Let $T(N)$ count pairs $n,n+2$ both prime with $n+2\le N$. The Selberg sieve replaces primality by the necessary condition that the polynomial $F(n)=n(n+2)$ have no small prime divisor. Its strength is a nonnegative quadratic majorant with optimizable divisor weights; its limitation is that a sifted integer need not be prime.

For a prime $p$, let $\nu (p)$ count the forbidden residues of $n$ modulo $p$. We have $\nu (2)=1$, since the two roots coincide modulo two, and $\nu (p)=2$ for every odd prime, where the roots are zero and $-2$. The Chinese remainder theorem makes $\nu (d)$ multiplicative on squarefree integers. With $g(d)=\nu (d)/d$,These local densities constitute the sieve distribution for the twin-prime problem. Since ${\sum }_{p\le z}g(p)=2\log \log z+O(1)$, it is a dimension-two sieve, which is why the expected upper bound has two logarithms in the denominator.

Let $P(R)={\prod }_{p\le R}p$ and choose real Selberg sieve weights ${\lambda }_d$ supported on squarefree $d\le R$, with ${\lambda }_1=1$. If $\gcd (F(n),P(R))=1$, the only supported divisor of $F(n)$ is one. HenceExpanding and using the distribution of the polynomial roots givesHere $[d,e]$ denotes the least common multiple. The divisor-square support reaches $R^2$, although each individual weight is supported only to $R$; keeping these two levels distinct is essential for estimating the remainder.

The Selberg sieve diagonalization solves the main-term minimization exactly. On squarefree integers putMultiplicativity givesAlso, Möbius inversion gives ${\sum }_{r\le R}\mu (r)Y_r={\lambda }_1=1$. Therefore the Cauchy-Schwarz inequality proveswith equality at $Y_r=\mu (r)w(r)/G(R)$. Inverting the divisor sums gives the optimizing weightsIn particular ${\lambda }_1=1$ and $Q(\lambda )=1/G(R)$.

These weights have $|{\lambda }_d|\le 1$, which supplies a simple rigorous remainder estimate. Indeed, multiply the inner sum by ${\prod }_{p\mid d}(1-g(p){)}^{-1}={\sum }_{e\mid d}w(e)$. The terms are $w(em)$ for distinct squarefree integers $em\le R$ with $e\mid d$ and $(m,d)=1$, hence form a subset of the sum defining $G(R)$. Since $\nu ([d,e])\le \nu (d)\nu (e)$ and $\nu (d)\le \tau (d)$,Thus

For this sieve $w(2)=1$ and $w(p)=2/(p-2)$ for odd primes. Here is a direct proof that $G(R)\asymp {\log }^{2}R$, sufficient for the upper bound. PutThe Mertens theorem for reciprocal primes implies $W(z)\asymp {\log }^{2}z$, by taking logarithms and using the convergence of ${\sum }_{p}g(p{)}^2$. Give each divisor $d\mid P(z)$ probability $w(d)/W(z)$. Its expected logarithm isby the Mertens first theorem. With $z=R^{1/8}$ this is at most $\frac{1}{3}\log R$ for large $R$. The Markov inequality shows that at least two thirds of the total product weight has $d\le R$. These terms occur in $G(R)$, so $G(R)\gg W(R^{1/8})\gg {\log }^{2}R$. Conversely $G(R)\le W(R)\ll {\log }^{2}R$, proving the claim without treating an untruncated Euler product as a truncated divisor sum.

Choose, for example, $R=N^{1/4}$. Any twin-prime pair with smaller member greater than $R$ survives this sieve; pairs with smaller member at most $R$ contribute at most $O(R)$. ConsequentlyThe chosen level is deliberately enough to prove the correct logarithmic order; sharper level and remainder analyses improve the numerical upper-bound constant.

The local correction factor also explains the conjectured main-term constant. Relative to two independent primality conditions, the singular series isThe twin-prime constant $C_2$ is positive and convergent because its logarithm has an absolutely convergent tail. The factor two at $p=2$ records that only one parity is allowed, while each odd-prime factor records the two distinct forbidden residues. The Hardy–Littlewood twin-prime asymptotic conjecture predictsThis is a conjecture, not a consequence of the upper-bound calculation.

The parity problem in sieve theory is the obstruction to reversing the argument. The nonnegative square majorizes all survivors, including composites with sufficiently large prime factors; small-divisor information does not isolate the desired prime-factor parity or force both factors to be prime. Therefore the Selberg sieve establishes the correct upper-bound scale and the arithmetic local factors, but by itself does not prove a positive lower bound for twin primes, their infinitude, or the conjectured asymptotic.

Use the integer norm lower bound for rational approximation to a quadratic irrational. The integer $2q^2-a^2$ is nonzero, since $\sqrt{2}$ is irrational. If $a/q<2$, thenbecause $\sqrt{2}+a/q<\sqrt{2}+2<4$. If $a/q\ge 2$, the difference is at least $2-\sqrt{2}>1/4\ge 1/(4q^2)$. ThusIn particular, $\sqrt{2}$ is a badly approximable number.

To estimate the prime-weighted exponential sum, let $Q=\lfloor N^{1/2}\rfloor$. The Dirichlet approximation theorem gives a reduced fraction $a/q$ withCombining this with the lower bound gives $q>Q/4$. Thus $q\asymp N^{1/2}$; the approximation cannot have a small denominator.

The relevant analytic input is the prime exponential-sum estimate from Vaughan identity:Here is the structure behind that estimate. With Dirichlet convolution and cutoffs $U,V$, the Vaughan identity isIt follows from $\Lambda =\mu \ast \log$, $\log =\Lambda \ast 1$, and $\mu \ast 1={\delta }_1$. The first two Dirichlet convolution terms become type I sums with one short factor; the last becomes a bilinear type II sum with both factors bounded below. Split these sums into dyadic blocks and use $U=V=N^{2/5}$. For the type I sums, the inner finite geometric series has bound $\min (L,(2\Vert \alpha m\Vert {)}^{-1})$. For the type II sums, the Cauchy-Schwarz inequality followed by expansion introduces analogous sums in differences of indices. Rational approximation separates the phases into blocks of length comparable to $q$, bounding these reciprocal-distance sums. The Dirichlet convolution coefficients are divisor-bounded, producing the logarithmic factors. These steps give the displayed estimate; no unproved cancellation of arbitrary weighted sums is being assumed.

With $\alpha =\sqrt{2}$ and $q\asymp N^{1/2}$, the two denominator-dependent terms are $O(N^{3/4})$. ConsequentlyThe final passage absorbs the logarithm using the positive exponent margin $9/10-4/5=1/10$.

We first give an auxiliary-polynomial proof, rather than assuming a transcendence theorem for exponentials. Suppose $i\pi$ were algebraic, and let ${\delta }_1=i\pi ,{\delta }_2,\dots ,{\delta }_t$ be its conjugates. Since the first factor vanishes,Expand and collect equal subset sums of the ${\delta }_{\nu }$. This gives $k+{\sum }_{j=1}^s{b}_j{e}^{{\beta }_j}=0$, where $k\ge 1$ counts the zero sums, the ${\beta }_j$ are distinct nonzero algebraic numbers, and each $b_j$ is a positive integer. The exponents and their weights are invariant under every Galois permutation.

Choose a positive integer $c$ such that all $c{\beta }_j$ are algebraic integers. Their Galois-invariant monic polynomial $R(u)={\prod }_j(u-c{\beta }_j)$ has rational algebraic-integer coefficients, hence lies in $\mathbb{Z}[u]$. Its constant term $A=R(0)$ is a nonzero integer. For a prime $p$ defineAt each ${\beta }_j$, derivatives below order $p$ vanish. Every higher derivative is $c^{h}h!/(p-1)!$ times an algebraic integer, so $F_p({\beta }_j)$ is divisible by $p$ in the ring of algebraic integers. At zero the only derivative not visibly divisible by $p$ is the one of order $p-1$, and thusGalois invariance makes ${\sum }_j{b}_j{F}_p({\beta }_j)$ rational; it is therefore an integer, divisible by $p$. ConsequentlyFor every prime not dividing $kcA$, this is a nonzero integer.

The finite derivative sum satisfies $F_p^{\prime }-F_p=-f_p$, so integration along the straight segment to ${\beta }_j$ givesSum with weights $b_j$ and use $\sum b_j{e}^{{\beta }_j}=-k$. The result isOn these finitely many fixed compact segments, $|cz|$, $|R(cz)|$ and $|e^{-z}|$ are bounded independently of $p$. Hence $|S_p|\le C_1{C}_2^p/(p-1)!\to 0$. There are arbitrarily large primes not dividing the fixed integer $kcA$, contradicting $|S_p|\ge 1$. Therefore $\boxed{\pi \text{ is transcendental}}$. This is the Hermite auxiliary-polynomial proof for pi.

The more general Lindemann–Weierstrass theorem states that exponentials of distinct algebraic numbers are linearly independent over $\overline{\mathbb{Q}}$. Equivalently, exponentials of rationally independent algebraic numbers are algebraically independent over $\overline{\mathbb{Q}}$. In particular, the exponential of a nonzero algebraic number is transcendental.

Suppose $C=\cos \alpha \cos \beta$ were algebraic. The exponential formula for cosine givesAfter combining equal exponents this is an algebraic-coefficient linear relation among exponentials of distinct algebraic numbers. At least one exponent is nonzero, since $\alpha +\beta$ and $\alpha -\beta$ cannot both vanish when $\alpha ,\beta \ne 0$. Its coefficient is a positive integer, even when two exponents coincide. The relation is therefore nontrivial and contradicts the theorem. Thus $\boxed{\cos \alpha \cos \beta \text{ is transcendental}}$, including the cases $\alpha =\pm \beta$.

An integer version of Siegel lemma is as follows. For $M<N$ homogeneous linear equations $Az=0$, with integer matrix entries bounded in absolute value by $A_0\ge 1$, there is a nonzero integer solution satisfyingPut the displayed integer bound equal to $Q$ and evaluate the equations at every vector in $\{0,\dots ,Q{\}}^N$. There are $(Q+1{)}^N$ inputs and at most $(2N{A}_{0}Q+1{)}^M$ outputs. But $2N{A}_{0}Q+1\le 2N{A}_0(Q+1)$ and $(Q+1{)}^{N-M}>(2N{A}_0{)}^M$, so the number of inputs is larger. Two distinct inputs have identical outputs; their difference proves the claim. This is the pigeonhole proof of integer Siegel lemma. Rational coefficients are handled by clearing denominators. Over a number field of degree $d$, expanding in a rational basis gives at most $dM$ rational equations, so the same construction applies when $N>dM$, with the height of the resulting integer matrix controlling the bound.

Siegel's original arithmetic definition starts with an entire series $E(z)={\sum }_{n\ge 0}{a}_n{z}^n/n!$, with all $a_n$ in a fixed number field. For every $ϵ>0$ there must be positive integers $d_n$ such that $d_n{a}_0,\dots ,d_n{a}_n$ are algebraic integers, and both $d_n$ and the absolute values of every conjugate of $a_n$ are $O_{ϵ}(n^{ϵn})$. The growth condition ensures convergence everywhere. The stronger, now customary strict E-function definition uses bounds $C^{n+1}$ for a fixed $C$ and also includes the requirement of a nonzero linear differential equation over $\overline{\mathbb{Q}}(z)$. In the original formulation, that differential condition is explicitly part of the system hypothesis in the theorem below. These definitions should be distinguished; the exponentials used here satisfy even the stronger conditions.

The Siegel–Shidlovsky theorem states that if a vector of E-functions satisfies $f^{\prime }=A(z)f$, with $A(z)$ a matrix of rational functions over $\overline{\mathbb{Q}}$, then for every nonzero algebraic $\xi$ which is not a pole of $A$,In particular, algebraic independence of the functions implies algebraic independence of the values. The exclusion of zero is necessary: their Taylor values there are algebraic.

Take rationally independent algebraic ${\gamma }_1,\dots ,{\gamma }_s$ and $f_j(z)=e^{{\gamma }_{j}z}$. Their coefficients are ${\gamma }_j^n$, their conjugates grow exponentially, and powers of a fixed common denominator clear all coefficients through degree $n$. They satisfy the nonsingular diagonal system $f_j^{\prime }={\gamma }_j{f}_j$.

To check functional independence, a polynomial relation among these exponentials, after clearing rational-function denominators, would be ${\sum }_m{P}_m(z)e^{{\lambda }_{m}z}=0$ with distinct ${\lambda }_m={\sum }_j{m}_j{\gamma }_j$. Such an exponential polynomial cannot vanish identically. To isolate a nonzero term with exponent ${\lambda }_0$, apply ${\prod }_{\lambda \ne {\lambda }_0}(D-\lambda {)}^{\mathrm{deg}{P}_{\lambda }+1}$. All other terms vanish, while the remaining polynomial is acted on by operators $D+{\lambda }_0-\lambda$, each injective on polynomials because its nonzero constant preserves a nonzero leading coefficient. Thus the remaining term is nonzero, a contradiction.

The theorem at $\xi =1$ now makes $e^{{\gamma }_j}$ algebraically independent. For arbitrary distinct algebraic exponents ${\alpha }_i$, choose a rational basis of their span and divide its elements by a common denominator so that every ${\alpha }_i$ is an integer combination of the resulting ${\gamma }_j$. Then $e^{{\alpha }_i}$ are distinct Laurent monomials in $e^{{\gamma }_j}$. A linear relation among them becomes a nonzero polynomial relation after multiplying by a common monomial, contradicting independence. These particular E-functions therefore yield the full Lindemann–Weierstrass theorem.

Write $f(z)={\sum }_{n\ge 0}{z}^{l^n}$. The authoritative PDF has exponent $l^n$; the geometric-series exponent in the converted TeX is a transcription error. On the unit disk this Fredholm series satisfiesFirst it is transcendental as a function. If ${\zeta }^{l^s}=1$, the terms with $n\ge s$ in $f(\rho \zeta )$ are the positive terms ${\rho }^{l^n}$ and their sum tends to infinity as $\rho \uparrow 1$; the finite preceding sum stays bounded. Thus every root of unity of $l$-power order is a singularity. These roots are dense on the unit circle. An algebraic function has only finitely many possible branch points or poles, as seen from its defining polynomial's leading coefficient and discriminant. Hence $f$ cannot be algebraic over $\mathbb{C}(z)$.

Suppose now that $\alpha$ is algebraic, $0<|\alpha |<1$, and $a=f(\alpha )$ is algebraic. For a positive integer $N$, a polynomial $P(X,Y)\in \mathbb{Z}[X,Y]$, nonzero and of degree at most $N$ in each variable, can be chosen so thatIndeed the first $T$ Taylor coefficients impose $T$ homogeneous integer equations on $(N+1{)}^2$ coefficients. There is a nonzero rational solution, which can be cleared to an integer one. Functional transcendence ensures $G$ is not identically zero.

Iterating the functional equation givesThus ${\beta }_k=G({\alpha }^{l^k})$ belongs to the fixed field $K=\mathbb{Q}(\alpha ,a)$ and is nonzero for all sufficiently large $k$, since a nonzero analytic $G$ has no sufficiently small nonzero zeros. Its analytic upper bound isChoose an integer $q\ge 1$ making $q\alpha ,qa$ algebraic integers, let $d=[K:\mathbb{Q}]$, and choose $M\ge 2$ bounding every conjugate of $\alpha$ and $a$. Then $q^{l^k}{\alpha }^{l^k}$ and $q^{l^{k-1}}f({\alpha }^{l^k})$ are algebraic integers. ThereforeFor every embedding $\sigma$ of $K$, the coefficient sum $L(P)$ bounds the polynomial and givesTaking the norm of the nonzero integral $D_k{\beta }_k$ yields a nonzero rational integer, of absolute value at least one. ConsequentlyOn the other hand the upper bound is $\log |{\beta }_k|\le -T\lambda l^k+O_P(1)$, where $\lambda =-\log |\alpha |>0$. Choose $N$ so large that $T\lambda >N{C}_0$, possible because $T=N^2+2N$. Letting $k\to \infty$ contradicts the two bounds. This proves, by the Mahler method, $\boxed{f(\alpha )\text{ is transcendental for }0<|\alpha |<1}$ at every algebraic $\alpha$.

For the shorter rational-approximation argument at $1/2$, truncate after $n=m$ and write $s_m=p_m/q_m$ with $q_m=2^{l^m}$. The numerator is odd because its last summand is one and all previous summands are even, so this fraction is reduced. The tail satisfiesThe value is irrational: if it were $P/Q$, the positive difference from a truncation would be at least $1/(Q{q}_m)$, contradicting the tail bound for large $m$. If it were algebraic irrational and $l\ge 3$, choose $0<ϵ<l-2$. Eventually the displayed upper bound is below $q_m^{-2-ϵ}$ for infinitely many reduced fractions. The Roth theorem says that an algebraic irrational admits only finitely many such approximations. This contradiction proves transcendence directly. The argument does not cover $l=2$, because its exponent is not strictly larger than two; the Mahler proof does cover that case.

The Gelfond–Schneider theorem says that if $a\ne 0,1$ is algebraic and $b$ is algebraic irrational, every number $e^{b\lambda }$ with $e^{\lambda }=a$ is transcendental. Thus the statement includes every determination of the power, not merely a chosen principal branch.

Here is an auxiliary-function outline with the arithmetic and analytic steps specified. Assume $v=e^{b\lambda }$ were algebraic, and put $K=\mathbb{Q}(a,b,v)$, $d=[K:\mathbb{Q}]$. For a large integer $T$, choose degrees $L_1\asymp T\log T$ and $L_2\asymp T/\log T$, with constants large enough that $(L_1+1)(L_2+1)>2d(T+1{)}^2$. Apply Siegel lemma to imposeon a nonzero polynomial $P(X,Y)\in \mathbb{Z}[X,Y]$ of these two degrees. Each field equation gives at most $d$ rational equations. Clearing denominators and bounding conjugates gives equation-coefficient logarithmic size $O(L_1\log T+L_{2}T)$; the surplus of variables therefore permits coefficient height $H$ with

The entire function $F(z)=P(z,e^{\lambda z})$ is nonzero: its polynomial-exponential terms have distinct exponents $j\lambda$, and their independence follows by the differential-operator argument in question 2. Here $\lambda \ne 0$. Its maximum modulus satisfies $\log M_F(R)=O(\log H+L_1\log R+L_{2}R)$. The grid points $s+tb$ are distinct because $b$ is irrational, and lie in a disk of radius $(1+|b|)T$.

Suppose the vanishing grid has been extended to $0\le s,t\le B$, for some $B\ge T$. Divide $F$ by the product of its $(B+1{)}^2$ known linear zero factors and use the maximum principle on a disk of radius $CB$, with $C$ large and fixed. At a point of the doubled grid, the quotient estimate givesif the value is nonzero. But this value equals the algebraic number $P(s+tb,a^s{v}^t)$. Clearing its denominators and taking its norm gives the opposite boundFor sufficiently large initial $T$, the terms on the right are $o(B^2)$ uniformly for every $B\ge T$: $L_2/B=O(1/\log T)$ and $L_1\log B/B^2=O((\log T{)}^2/T)$. Thus every doubled-grid value must be zero. Iterating doubles the grid indefinitely. This supplies quadratically many distinct zeros in disks of linearly growing radius, whereas Jensen's formula and the exponential-polynomial growth bound allow only $O(R)$ zeros for a nonzero $F$. The contradiction proves the theorem. This is the Schneider lattice proof of the Gelfond–Schneider theorem.

Now fix any logarithms of algebraic $\alpha ,\beta \ne 0,1$. Their values are nonzero. If $q=\log \alpha /\log \beta$ were algebraic irrational, then $e^{q\log \beta }=\alpha$ would be algebraic, contradicting the theorem with base $\beta$. HenceFor arbitrarily many logarithms the qualitative Baker theorem on algebraic linear combinations of logarithms states that logarithms of algebraic numbers which are linearly independent over $\mathbb{Q}$ remain independent over $\overline{\mathbb{Q}}$, even after adjoining one to the list. Equivalently a nonzero homogeneous algebraic-coefficient linear combination of such logarithms is transcendental. Rational relations are therefore the only source of homogeneous algebraic relations among the chosen logarithms.

Use the following fixed-number form of a logarithmic estimate: for chosen logarithms of fixed nonzero algebraic numbers $a_1,\dots ,a_s$, a nonzero form $\Lambda =\sum b_j\log a_j$ with integer coefficients $|b_j|\le B$ satisfies $|\Lambda |\ge c(2B{)}^{-C}$, for fixed positive constants $c,C$. This follows from the Baker lower bound for a homogeneous linear form in logarithms.

By the Dirichlet unit theorem, write units as $x=\zeta \prod {ϵ}_j^{m_j}$ and $y=\xi \prod {ϵ}_j^{n_j}$, with finitely many roots of unity and fixed fundamental units. If the unit rank is zero, there are already only finitely many choices. Otherwise put $B=\max (1,|m_j|,|n_j|)$. The real logarithmic embedding of the free unit group has full rank and lies in the product-formula hyperplane. Equivalence of norms in that finite-dimensional space shows that a unit with exponent size $B$ has some archimedean conjugate of modulus at most $e^{-c_{0}B}$: a large positive logarithmic coordinate forces a comparably large negative one, and a large negative coordinate itself suffices.

Interchanging the two terms if necessary, choose such a conjugate for $y$. At this embedding the equation makes $\sigma (\alpha x)=1-\sigma (\beta y)$ with $|\sigma (\beta y)|\le C_0{e}^{-c_{0}B}$. For large $B$ define the logarithm near one. It is nonzero because $\beta y\ne 0$, andOn fixed branches it isfor an integer $k=O(B)$. The bound on $k$ follows from summing the fixed imaginary parts and then reducing to the near-one branch. There are only finitely many possible embeddings and roots of unity, so the stated lower bound applies uniformly with coefficients $O(B)$. It contradicts the exponential upper bound when $B$ is large. Hence the exponents are bounded and $\boxed{\alpha x+\beta y=1\text{ has finitely many unit solutions}}$ in ${\mathcal{O}}_K^{\times }$. This is the unit equation over a number field proof; “unit” here means an algebraic-integer unit, not an arbitrary nonzero field element.

For the Thue theorem, let $F\in \mathbb{Z}[X,Y]$ be irreducible homogeneous of degree $d\ge 3$ and let $m\ne 0$ be fixed. In a splitting field $L$, write $F(X,Y)=a{\prod }_i(X-{\theta }_{i}Y)$ with distinct roots. Choose a fixed integer $c$ so that each $c{\theta }_i$ is integral and $c^{d}m/a$ is a nonzero integer. For a solution $(u,v)$ the nonzero algebraic integers $t_i=c(u-{\theta }_{i}v)$ satisfy ${\prod }_i{t}_i=c^{d}m/a$. Their principal ideals therefore divide a fixed ideal and have only finitely many possibilities. For each occurring principal ideal choose a generator ${\gamma }_i$, so $t_i={\gamma }_i{\epsilon }_i$ with ${\epsilon }_i$ a unit.

Any three distinct roots give the identityDivide by the third term and absorb the finitely many generator ratios into the coefficients. This is a fixed-coefficient unit equation in ${\epsilon }_1/{\epsilon }_3$ and ${\epsilon }_2/{\epsilon }_3$, so it has finitely many solutions for each generator choice. Thus $\rho =t_1/t_3$ has finitely many values. The relation $(1-\rho )u=({\theta }_1-\rho {\theta }_3)v$ fixes a rational direction whenever an integer solution exists; $\rho =1$ simply gives $v=0$. On each such direction, $(u,v)=k(u_0,v_0)$ for a primitive integer pair, and $k^{d}F(u_0,v_0)=m$ allows finitely many $k$. This proves the desired finiteness. When $m=0$, irreducibility excludes a nonzero rational root direction, leaving only $(0,0)$.

The argument also proves finiteness for any binary form with at least three distinct projective roots: irreducibility was used only to guarantee distinct roots and to dispose of the zero-value case. A rational invertible change of variables can move every projective root away from infinity; after clearing its fixed denominators, the resulting pairs lie in an integer lattice and the same proof applies. This stronger nonzero-value version will apply to the binomial forms in question 6.

The abc conjecture asserts that for every $ϵ>0$ there is $K_{ϵ}\ge 1$ such that coprime positive integers $A+B=C$ satisfywhere the radical is the product of the distinct prime divisors.

For a primitive positive Fermat triple, the three integers are pairwise coprime. Apply the conjecture to $A=x^n,B=y^n,C=z^n$. Since $x,y<z$, $\mathrm{rad}(ABC)=\mathrm{rad}(xyz)\le xyz<z^3$. Taking $ϵ=1/6$ yields $z^{n-7/2}\le K_{1/6}$. For $n\ge 4$ and $z\ge 2$ this simultaneously givesThus there are finitely many primitive triples and exponents. The primitivity restriction is essential to this abc argument: a nonprimitive triple is a common multiple of a primitive one, and the conjecture normalizes that factor away rather than bounding it. Consequently the intended finiteness conclusion counts primitive triples, or scaling classes. The literal unrestricted Fermat assertion is true for the independent reason that Fermat's last theorem gives no positive solutions for $n>2$; it does not follow from the displayed primitive height bound alone.

There is a genuine exception to the second printed assertion. If $a-b=c$, then $x=y=1$ solves the equation for every $n>2$; for example $(a,b,c)=(2,1,1)$ gives an infinite family. The correct uniform finiteness statement excludes $\max (x,y)=1$.

For that corrected statement put $H=\max (x,y)\ge 2$ and $d=\gcd (a{x}^n,b{y}^n)$. Since their difference is $c$, $d\mid c$. Apply abc to $b{y}^n/d+c/d=a{x}^n/d$. Its radical is at most $\mathrm{rad}(abc)xy$, and its largest term is at least $\min (a,b)H^n/c$. ThereforeWith $ϵ=1/4$, enlarge the fixed right side to $D\ge 1$. For $n\ge 3$ the exponent is at least $1/2$, soThis proves finiteness of all nonconstant-power solutions, without requiring $x,y$ coprime.

For an unconditional proof, the diagonal case $x=y=H\ge 2$ satisfies $(a-b)H^n=c$ and is elementary. If $a=b$, the unequal-variable gap is at least $2^n-1$, also bounding $n$. In the remaining case interchange variables and coefficients, changing the sign of $c$ if necessary, to have $1\le X<Y$ andThe nonzero real logarithmic form $\Lambda =\log (A/B)+n\log (X/Y)$ satisfies $e^{\Lambda }-1=\delta /(B{Y}^n)$. Once $B{Y}^n>2c$, $|\Lambda |\le 2c/(B{Y}^n)$, so $\log |\Lambda |\le -n\log Y+O_{a,b,c}(1)$. The two-logarithm Baker lower bound for a homogeneous linear form in logarithms, with the fixed height of $A/B$ and height of $X/Y$ at most $\log Y$, givesHence $n\le C(a,b)\log (2n)+O_{a,b,c}(1)$ because $Y\ge 2$, and $n$ is bounded. The omitted case $B{Y}^n\le 2c$ has bounded exponent and variables directly. For each of the finitely many exponents, $a{X}^n-b{Y}^n$ has at least three distinct projective roots. The stronger nonzero-value Thue theorem argument from question 5 therefore gives finitely many pairs, whether or not that binomial form is irreducible. This establishes the exponent bound for a binomial power equation and the corrected finiteness statement with no unproved hypothesis. The constant-power family remains the unavoidable exception.

A useful strong form of the Hensel lemma is the following. Let $F$ be a complete discretely valued field, with valuation ring $\mathcal{O}$, normalized valuation $v$, and $P\in \mathcal{O}[T]$. If $a_0\in \mathcal{O}$ satisfiesthen there is a unique root $a\in \mathcal{O}$ in the ball $v(a-a_0)>v(P^{\prime }(a_0))$. When $P(a_0)=0$ the assertion is immediate; otherwise the condition includes $P^{\prime }(a_0)\ne 0$.

Here is a proof by Newton iteration over a valued field. Write $c=v(P^{\prime }(a_0))$ and $t_0=v(P(a_0))>2c$, and setSuppose $v(P^{\prime }(a_j))=c$ and $t_j=v(P(a_j))>2c$. The correction has valuation $t_j-c>c\ge 0$, so $a_{j+1}\in \mathcal{O}$. Taylor expansion over $\mathcal{O}$ cancels the linear term and givesThus the derivative still has valuation $c$, and $t_{j+1}-2c\ge 2(t_j-2c)$. Unless an exact root is reached earlier, the corrections tend to zero with exponentially increasing valuations. Completeness gives $a_j\to a$, and continuity gives $P(a)=0$ and $v(a-a_0)=t_0-c>c$. For uniqueness, if $a,b$ are in this ball, polynomial expansion around $a_0$ givesThe second factor has valuation exactly $c$ and is nonzero, so two roots in the ball must be equal. In particular, a simple residue root lifts uniquely: $v(P(a_0))\ge 1$ and $v(P^{\prime }(a_0))=0$ satisfy the hypothesis.

For the square-class group of a p-adic field, write every element as $p^{r}u$ with $u\in {\mathbb{Z}}_p^{\times }$. Multiplication by a square changes $r$ by an even integer, so valuation parity gives one independent factor $C_2$.

For odd $p$, a unit $u$ is a square if and only if its residue is a square in ${\mathbb{F}}_p^{\times }$. Necessity follows by reduction; sufficiency applies the simple-root Hensel lemma to $T^2-u$, whose derivative is a unit at a nonzero residue root. The cyclic group ${\mathbb{F}}_p^{\times }$ has square subgroup of index two. If $u_0$ is a unit with nonsquare residue, thenFor $p=2$, every odd square is one modulo eight. Conversely, if $u\equiv 1(\mathrm{mod}8)$, take $P(T)=T^2-u$ and $a_0=1$. Then $v_2(P(1))\ge 3>2v_2(P^{\prime }(1))=2$, so the strong Hensel lemma gives a square root. Thus unit square classes are exactly their residues $1,3,5,7$ modulo eight. The classes $-1$ and $5$ generate this group of order four independently, and valuation parity supplies the class $2$. Hence

Put $t=\sqrt{-30}$ and $K=\mathbb{Q}(t)$. The ring of integers is $\mathbb{Z}[t]$, and the fundamental discriminant is $D_K=-120$. For a reduced positive definite binary quadratic form $(a,b,c)$ of discriminant $-120$, reduction givesThe coefficient $b$ is even; checking $1\le a\le 6$ and $|b|\le a$, with the usual nonnegative sign choice at boundary equalities, leaves preciselyThe correspondence between proper quadratic forms and ideal classes therefore gives $h_K=4$. All four classes equal their inverses, since inversion replaces $b$ by $-b$. HenceThe three nontrivial classes may be represented by the ramified prime ideals ${\mathfrak{p}}_2=(2,t)$, ${\mathfrak{p}}_3=(3,t)$ and ${\mathfrak{p}}_5=(5,t)$. They square to principal ideals and satisfy ${\mathfrak{p}}_2{\mathfrak{p}}_3{\mathfrak{p}}_5=(t)$. None is principal: an integral generator of norm $2$, $3$ or $5$ would require $a^2+30b^2$ to equal that number, which is impossible. The relation then makes the three classes distinct, consistently with the four-form calculation.

ConsiderThe three rational square classes are independent, so $[H:\mathbb{Q}]=8$. Their radical product is $\sqrt{-30}$, so $K\subset H$ and $[H:K]=4$; the extension $H/K$ is abelian. To verify that it is unramified, rather than assuming that radical adjunctions are harmless at $2,3,5$, use the conductor-discriminant formula. The seven quadratic subfields have fundamental discriminantsThe formula givesThe discriminant tower identity now givesThus the relative discriminant is the unit ideal and no finite prime ramifies. The base has no real places, so there is no infinite ramification to check. This unramified abelian degree-four extension has degree $h_K$, and part (i) provesThis is the Hilbert class field of Q of square root minus thirty.