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      "Exercise16-17source.2227is the three-term estimate .25-.03+.0027;exact625/2809≈.222499...,so distinguish estimate from rounded exact value.",
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      "Review258infinite product is defined as positive partial-product limit. Exponent series sum_n n/3^n=3/4gives2^(3/4).",
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      "Review279digits1,3,4givecompositionrecurrence t_n=t(n-1)+t(n-3)+t(n-4),t0=1,tnegative0;even subsequence t_(2n)=F(n+1)^2. Source c1=1,c2=2is shifted Fibonacci.",
      "Review280OCR/index must retain u_(n-2),notu_(n-1):u_(n+1)=(n+1)u_n-n(n-1)u_(n-2)/2. Source construction overcounts only duplicate identical distinct-letter pairs. Baseu0=1,u1=1,u2=2and n≥2;n1canstateu2=2separately.",
      "Review281subtract partial-sum identities to obtain a_k=(k-1)a_(k-1)/(k+1),then telescope froma1=1/2to1/[n(n+1)].",
      "Euler Big Picture must remain a historical discovery heuristic:identical zeros alone do not prove equality of infinite products/functions. Modern convergence/product identity is an additional theorem,not established by coefficient matching alone.",
      "Partial-fraction linear/quadratic irreducible factorization is over the reals. Not every rational polynomial splits into linear/quadratic factors over the rationals."
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        "sourceAnswer": "A three-term recurrence uses X_n=ar^n+bs^n+ct^n,with r,s,troots of z³-alpha z²-beta z-gamma.",
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      "Example17-3sourcechair-position compression is correct,but distinguishable people add5!. SourceReview290confirms that factor explicitly.",
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      "Review284three selected squares areunordered:choose3rowsand3columnsandabijection,givingC(5,3)^2*3!=600.",
      "Review285maximum4351needsattainability:chooseotherlinesgenerically toavoidadditionalparallelisms,tripleintersectionsorincidencewithforcedconcurrencyA. Constraintsforbidonlyfinitelymanychoices at eachstep.",
      "Review288people distinct;circulararrangementsidentifyrotationsbutnotreflections. Fourcountryblocks6orders,then internal3!4!2!3!.",
      "Review291supplied671countusesadjacentgridmoves tolargerlabels. The source wordingstateslargerlabelandsharedrow/columnasnecessaryconditions;the operative task explicitly includesadjacency. Allowingalllongerjumps isadifferentgraph.",
      "Review292original4x4x4gridcount76. Sourceoptionalformula3n²+6n+4countsfull-lengthn-pointlinesinnxnxngrid;itdoesnotcountalllineswithatleast4pointsforarbitrarilylargern. Statewhichgeneralizationisused.",
      "Review296usesfinitesetsasintendedbyordinaryintegerpowers2^|S|. InfiniteCwouldrequirecardinalarithmeticandchangeproblem. |C|101,union102,minimumtriple97attainedbydisjointcomplementsofsize2,2,1.",
      "Review297all20people distinct,but eachgenderpatternhas7!13!orders,soaveragingovergenderpatternsisvalid. Expectedadjacentmixedpairs=2mn/(m+n)=91/10.",
      "Review298colorclasses4implyn³divisibleby4whentheconfigurationexists. Totalnoncolevelchoices(n!)². Badchoices≤6(n³/4)((n-2)!)²;ratio3n/[2(n-1)²]<1forn≥4,soarainbowchoiceexists.",
      "Review299positiveintegern. MonotoneN/Wwalksgive2^nself-avoidingwalks;oneallEwalkisadditional,provingstrictlowerbound omitted by the supplied≥argument. Nonbacktrackingupperbound4*3^(n-1)neednotbeexact.",
      "Review300leadingbinarydigitforced1;n1sum2. Foralln≥1,zero-conventionformula C(2n-1,n-1)2^(2n-1)+C(2n-2,n-2)(2^(2n-1)-1)handlesendpoint withoutnegativefactorials.",
      "Cantor set retains endpoints ofremovedopenmiddlethirds. Membershipmeansatleastone ternaryexpansionusingonly0/2;canonicalterminatingnotationmightcontain1. Totalremainingcoverlengthgoesto0whilecardinalityremainscontinuum."
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        "Infinite expressions are limits of specified finite truncations. A self-similarity equation identifies possible limits,but convergence and admissible root choice mustbe justified.",
        "Nested radical x_(m+1)=sqrt(2+x_m),x0=0has0≤x_m<2and error2-x_(m+1)=(2-x_m)/(2+sqrt(2+x_m)),so converges2.",
        "All2continued fractionhaslimit1+sqrt2. Alternating2,3fractionhaslimit1+sqrt15/3;the period map x↦2+1/(3+1/x)is contractive onpositive source invariant interval.",
        "Canonical rational CF hasfirstintegercoefficientandpositive tails,with finaltail≥2exceptsingleintegercase. Noncanonicalending...,a,1merges to...,a+1. Exact Euclidean remainder algorithm terminates for rationals.",
        "Exercise18-5expansions:[5;14,2]for147/29,[4;7]for29/7,[5;1,5]for70/12=35/6.",
        "Exercise18-6b_(i+1)=1/{b_i}onlywhen fractionalpartnonzero;an exact integercompletequotientterminates instead ofdividingby0.",
        "Sourceπprefix[3;7,15,1,292,...]andconvergents3,22/7,333/106,355/113,103993/33102. More inverse steps require enough precision;finite floating digitsdo not certify arbitrary later coefficients.",
        "Example18-2periodicity theorem is for positive nonsquareinteger square roots,not arbitrary irrational numbers. sqrt17exacttail=sqrt17+4hasintegerpart8and repeats.",
        "Use source1-based convergents:C1=[a1]. InitialP1=a1,Q1=1,P2=a2a1+1,Q2=a2. Recurrences Pk=akP(k-1)+P(k-2),Qk=akQ(k-1)+Q(k-2). DeterminantPkQ(k+1)-P(k+1)Qk=(-1)^k.",
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        "Review305A=2by principalradicalsign. Negative continued fraction B=2-1/(2-...)has finite truncationsB_m=(m+1)/m>0,so converges1and A+B3;the double-root fixed equation alone doesnotprove convergence."
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        "Review332 homothety factor1/3 maps a vertex line to a parallel line. Exclude centroid image of any collinear degenerate vertex positions if triangles must be nondegenerate.",
        "Review334 same-side perpendicular choice is fixed; allowing either side gives two possible fixed endpoints.",
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        "Review337 I=(3G−A)/2 must lie strictly inside the circle for distinct chord endpoints. Boundary gives B=C; outside gives no solution; I=center allows any diameter (except one making A,B,C degenerate). The source's arbitrary interior G does not ensure existence.",
        "Review338 requires signed circle power. Distinct centers give radical axis; concentric unequal circles give empty locus; identical circles give whole plane. For centers(0,0),(a,0),radii r1,r2,axis x=(a²+r1²−r2²)/(2a),not source missing-a² formula. Source also swaps radius and squared-radius notation.",
        "Review339 source auxiliary common-tangent triangle construction is retained. Authored direct completion:incenter I,inradiusρ,circumcenter O,circumradiusR; t=R/(R+ρ),centers A+t(I−A),etc,radius tρ,commonpoint X=(1−t)O+tI. Similarity proves equal radius,commonpoint and internal containment.",
        "Review340 source n equal radial segments needs endpoints and converse. With circumradiusR andθ=π/n,the segments run opposite each vertex from O to length R(secθ−1). For adjacent-edge distances p,q from vertex B,base-side constraint gives p²+q²+2pq cos2θ=(2R sinθ)²; X lies on OB extended past O with OX=(p+q)/(2sinθ)−R. Continuous allowed p,q attain every value0..R(secθ−1). Same-edge endpoints give O; nonadjacent sides cannot carry a side-length chord except adjacent-vertex boundary cases.",
        "BigPicture iff-power-of-two-degree criterion is false as stated:power-of-two minimal degree is necessary but not sufficient. Correct constructibility criterion is membership in a tower of quadratic field extensions. Irreducible cubic obstructions for cube doubling and cos20° remain valid; π transcendence obstructs circle squaring. Specific angles can be trisectable even though no general straightedge/compass procedure exists."
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        "21-6 line BC is intended,not only segment BC;unsigned ratio c>0,c≠1 has x=c/(c+1) internal and x=c/(c−1) external for B0,C1.",
        "21-7 directed ratio x/(x−1)=c has unique finite x=c/(c−1) for c≠1;C excluded. c0 gives B.",
        "Ceva interior version avoids parallel-cevian cases and zero denominators. General directed extension allows concurrence at infinity/three parallel lines unless finite intersection separately required.",
        "21-8 and21-9 flag circular reasoning:cannot assume the very angle bisector/altitude being proved.",
        "Example21-4 pictured acute cyclic proof completed for all triangles by dot-product altitude proof;Example21-6 cosine cancellation handles right-angle zero cases separately.",
        "21-13 supplied three angle-bisector ratios have mixed orientation;correct Ceva factors BX/XC=AB/AC,CY/YA=BC/AB,AZ/ZB=AC/BC multiply1. Supplied displayed ratios pinned separately.",
        "Review344 X lies beyond A on ray CA,Z is reflection of C across AB,Y beyond B on ray CB as in figure. Stating only AX=AC etc without branch selection admits degenerate alternatives.",
        "Review345 equal tangent incircles determine D as original incircle touchpoint. Partii gives Gergonne concurrence by contact-length Ceva.",
        "Review346 literal line CH is undefined when C=H in a right-C triangle;interpret C's altitude. Use directed trigonometric or determinant proof outside pictured acute arrangement;right cases separate.",
        "Review347 equilateral O=G=H is a coincident-center case,no unique Euler line;vector relation H−O=3(G−O) and GH=2GO still hold.",
        "Review348 pairwise-intersecting circles can have parallel radical axes when centers collinear. Pictured noncollinear centers ensure a unique finite radical center;sourceunqualifiedclaimneedsqualification. A line-concurrency proof does not ensure the intersection lies within all finite chord segments.",
        "Review349 external points are in the source pictured regions opposite the corresponding sides;ordinary equal angles need these branches. Exact sine-ratio cancellation from the supplied proof retained.",
        "Review350 radical axes are concurrent OR parallel. Three alternate supporting lines of a strictly convex hexagon cannot all concur and cannot all be parallel;include both exclusions before deducing one common circle."
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        "Review363 supplied phrase maximum area of a circle inscribed in a circle means triangle.",
        "Review367 source projection ratio uses three right half-faces;one remaining face area9−π√3/3.",
        "Review369 lower-bound points from source143:with square A(0,1),B(1,1),C(1,0),D(0,0),G(.5,1),E(0,.01),F(1,.02). All are outside forced base-DC triangle and pairwise distances exceed1."
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        "Central projection is projective rather than affine and need not preserve ordinary length ratios. Orthogonal projection cannot enlarge any Euclidean segment.",
        "Finite-plane inversion is defined only for P≠O;no finite point maps to O. A line not through O maps to its supporting circle minus O. Including O requires an explicitly extended plane with infinity;do not mix ordinary projective directional infinities with the one-point inversive compactification.",
        "Example22-4 analytic proof:line n·x=d≠0 maps to n·y=d|y|²/R²,so center R²n/(2d),radius R²|n|/(2|d|),with O omitted. General circle through O maps to a line;circle not through O maps to circle unless degenerate.",
        "Homothety is x↦O+k(x−O),not arbitrary orientation-preserving similarity. A rotation need not be a homothety;nonzero translation has no finite center with k1. Parallel corresponding sides plus proper common scale give the intended setting. Midpoint triangle uses factor−1/2 about centroid,not +1/2.",
        "Example22-5 scale small/original=1−1/√3 and small area108√3−162.",
        "Area-continuity proofs assume bounded measurable regions of positive finite area. Translation bound≤strip width×bounding width;rotation bound≤bounded-sector area→0. Equal-area rays are selected by continuous cumulative swept area,not equally spaced angles.",
        "Sylvester proof uses a finite positive minimum over point-to-line distances. A clean completion chooses two of three collinear points on one closed side of the minimizing altitude foot,then a strictly smaller altitude follows from similar triangles. Infinite lattice supplies no ordinary line.",
        "Review354 supplied yes conclusion is false. Unit square F⊂{z0} projects orthogonally to x-axis segment in plane y0 and y-axis segment in plane x0;those planes are nonparallel and F is not a line. Preserve supplied argument separately/ungraded.",
        "Review358 sum360° concerns angular measures of6 boundary arcs,not their physical lengths. Pairwise crossings produce8 planar regions with one connected triple overlap in source arrangement.",
        "Review360 continuity alone does not establish uniqueness. For Jordan interiors,connectedness and openness make the area-half cut unique for a fixed orientation;compactness plus uniqueness gives continuous dependence on orientation. Alternatively regularize and take limits for more general measurable regions.",
        "Review362 use generic rotating lines through each pivot and event-side counts;no three collinear means one other point crosses at a time. Halving open-sector counts n−1 and n differ by1;the intervening line through the crossing point has n−1 remaining points on each side. Count each resulting halving line at most twice to get≥n.",
        "Review364 total boundary length needs rectifiability. The supplied side-length continuity additionally needs no positive arclength lying on a cut line (e.g.smooth strictly convex boundary). State a precise regularity model;general flat-boundary allocation needs separate treatment,not silent continuity.",
        "Review365 source orthogonal means perpendicular diagonals,not adjacent sides.",
        "Review366 positive-width compact convex body:width difference w(θ)−w(θ+π/2) is continuous and changes sign afterπ/2. Four support lines form containing square;no strict smoothness needed for existence.",
        "Review369 exact lower-bound distances:EF²1.0001,EG²1.2301,FG²1.2104,all>1. Four inward equilateral triangles based on the four square sides cover because with a=min(x,1−x),b=min(y,1−y),at least one of b≤√3a or a≤√3b holds.",
        "Review370 invert chord midpoints X,Y,Z in incircle:IA·IX=r² etc. If source triangle equilateral,circumcenters/incenter coincide;collinearity holds without a unique center line.",
        "Review371 outer medial map about centroid has factors−2 and4. Maximum-area triangle:all barycentric coordinates of remaining points≤1,so first outer triangle contains them. Minimum positive-area triangle:no other point inside second outer triangle,since |α|,|β|,|γ|≥1 yet α,β,γ>−1 would force all≥1,contradictsum1.",
        "Review372 four axis-line intersections with triangle boundary,three covering rectangles:pigeonhole selects two in one rectangle. Their axis-aligned bounding rectangle contains P,including the case of opposite points on one line.",
        "BigPicture great circles have no disjoint parallels,but full sphere also has antipodal-point incidence exceptions;spherical geometry is an analogy,not literally Euclid's other four axioms with only the fifth changed. State minor great-circle triangle convention for positive spherical angle excess."
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        "Review378 enlarged source286crop confirms7^(7^7),not7^7or7^7^…with extra levels.",
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        "Exercise23-5 full system has no solution because4x≡5mod8 is impossible;first/third alone give48mod63.",
        "Quadratic-residue convention here includes0. Composite moduli may have more than two square roots of one value;odd-prime root pairing is the special theorem.",
        "The source divisor sum s(n) includes n itself;proper-divisor sum is s(n)−n. Formula s(n)=Π(p_i^(e_i+1)−1)/(p_i−1). Standardσ/τnames can be supplied as aliases,not replace source notation silently.",
        "Exercise23-16 checking Euclid-form candidates alone does not prove third-smallest among every integer. An explicit bounded divisor-sum enumeration1..495 can complete absence of other smaller perfect numbers;historicalsourceunknownodd-perfect remark is labeled source-era context.",
        "Fermat needs prime p and p∤a. Multiplicative order is the least positive exponent returning1;nonunits can still have eventually periodic powers,despite having no multiplicative order.",
        "Exercise23-20 possible orders are divisors ofp−1;attainability follows from a primitive root via23-21. Supply explicit primitive-root witnesses for p23,p17,p7 if asserting every divisor occurs.",
        "Euler needs gcd(a,m)=1. Use m≥2 for the unit list;φ(1)=1 is a separate standard convention. φ is multiplicative only for coprime arguments.",
        "Example23-16 deliberately overclaims composite factorial0;23-27 corrects it. Composite4 has3!≡2mod4;every composite m>4 has(m−1)!≡0modm. For squarem=a²,a≥3,use distinct factors a and2a,not two copies of the same listed integer.",
        "Wilson source primitive-root proof requires oddprime p;handle p2 separately. Inverse-pair proof avoids assuming primitive-root existence and proves all primes.",
        "23-18 supplied prose mistakenly extends a through7 while the stated theorem/question use1..6;exclude the nonunit7.",
        "Review377 largestuniversaldivisor30usesn2togiveupperboundthen2,3,5divisibilityforlowerbound.",
        "Review379 complementary residueclasses1/6,2/5,3/4 permit one side perpair;residue0allowsoneelement. Largest23attained.",
        "Review383 npositive,checktelescopedradicands3,6,10,15,21,28forn1..6and36forn7. Pellgeneralizationoptional.",
        "Review388 source15suffix1991dforcesd6byunits/mod4,then9916≡12mod16cannotbesquare.",
        "Review392 supplied not composite is a prose negation typo;the p≡1mod6 branch makesp+2composite,soexcluded.",
        "Review394 finite digit family a_n=Σ_j0..n−1 10^(3j). If3|n,digit sumgives3. Otherwise factor into(10^n−1)/9 and(10^(2n)+10^n+1)/111;bothintegers>1forn≥2 (handle n2explicitlyifdesired).",
        "Review395 transformationb=3a−1 gives b²−b+25=9(a²−a+3);reverse uses3inversewhenp≠3,andp3haswitnesses a1,b2. Suppliedfirstline−3typoisnotused.",
        "Review396 cyclicindicescountmultiplicitiesevenforsmalln;productofallfour-termproductsisΠx_i^4=1;equalpositive/negativecountsforce4|n.",
        "Review398 source specializations only eliminate candidates;explicit polynomial quotient verifies m3sufficiency. m30fails p2because42∤5280.",
        "Review399 reduce a modulo b in modulus2^b−1;remainder2^r+1 lies strictly between0andmodulusforb>2.",
        "Review400 excluded d2,5,13ensures selectedpairdistinct. Assuming2d−1,5d−1,13d−1all squares forces dodd andeven y,z;then(v−u)(v+u)=2d contradictsparitymod4.",
        "Review401 p|b needs a complete valuation bound:prefactor b^(n−1) supplies≥n−1≥v_p(n!). Forp∤b,eachblockp^jofAP factorscontainsonep^jmultiple,givingfloor-sum bound. Handleb0,n1,zeroAPfactorseparately.",
        "Review402 converse needs at leasttwo terms andpositivefirstterm. Odd-divisor construction branches into oddlengthor evenlength depending on positivity,as sourceproof.",
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