{
  "index": "1994-A-4",
  "type": "NT",
  "tag": [
    "NT",
    "ALG"
  ],
  "difficulty": "",
  "question": "Let $A$ and $B$ be $2 \\times 2$ matrices with integer entries such\nthat $A, A+B, A+2B, A+3B$, and $A+4B$ are all invertible matrices whose\ninverses have integer entries. Show that $A+5B$ is invertible and that\nits inverse has integer entries.",
  "solution": "Solution. A square matrix \\( M \\) with integer entries has an inverse with integer entries if and only if \\( \\operatorname{det} M= \\pm 1 \\) : if \\( N \\) is such an inverse, \\( (\\operatorname{det} M)(\\operatorname{det} N)= \\) \\( \\operatorname{det}(M N)=1 \\) so \\( \\operatorname{det} M= \\pm 1 \\); conversely, if \\( \\operatorname{det} M= \\pm 1 \\), then \\( \\pm M^{\\prime} \\) is an inverse with integer entries, where \\( M^{\\prime} \\) is the classical adjoint of \\( M \\). Let \\( f(x)=\\operatorname{det}(A+x B) \\). Then \\( f(x) \\) is a polynomial of degree at most 2 , such that \\( f(x)= \\pm 1 \\) for \\( x=0,1,2 \\), 3 , and 4 . Thus by the Pigeonhole Principle \\( f \\) takes one of these values three or more times. But the only polynomials of degree at most 2 that take the same value three times are constant polynomials. In particular, \\( \\operatorname{det}(A+5 B)= \\pm 1 \\), so \\( A+5 B \\) has an inverse with integer entries.",
  "vars": [
    "M",
    "N",
    "f",
    "x"
  ],
  "params": [
    "A",
    "B"
  ],
  "sci_consts": [],
  "variants": {
    "descriptive_long": {
      "map": {
        "M": "currentmatrix",
        "N": "inversemate",
        "f": "detpolyfunc",
        "x": "scalarparam",
        "A": "basematrix",
        "B": "stepmatrix"
      },
      "question": "Let $basematrix$ and $stepmatrix$ be $2 \\times 2$ matrices with integer entries such\nthat $basematrix, basematrix+stepmatrix, basematrix+2stepmatrix, basematrix+3stepmatrix$, and $basematrix+4stepmatrix$ are all invertible matrices whose\ninverses have integer entries. Show that $basematrix+5stepmatrix$ is invertible and that\nits inverse has integer entries.",
      "solution": "Solution. A square matrix \\( currentmatrix \\) with integer entries has an inverse with integer entries if and only if \\( \\operatorname{det} currentmatrix= \\pm 1 \\) : if \\( inversemate \\) is such an inverse, \\( (\\operatorname{det} currentmatrix)(\\operatorname{det} inversemate)= \\) \\( \\operatorname{det}(currentmatrix inversemate)=1 \\) so \\( \\operatorname{det} currentmatrix= \\pm 1 \\); conversely, if \\( \\operatorname{det} currentmatrix= \\pm 1 \\), then \\( \\pm currentmatrix^{\\prime} \\) is an inverse with integer entries, where \\( currentmatrix^{\\prime} \\) is the classical adjoint of \\( currentmatrix \\). Let \\( detpolyfunc(scalarparam)=\\operatorname{det}(basematrix+scalarparam stepmatrix) \\). Then \\( detpolyfunc(scalarparam) \\) is a polynomial of degree at most 2 , such that \\( detpolyfunc(scalarparam)= \\pm 1 \\) for \\( scalarparam=0,1,2 , 3 , \\) and 4 . Thus by the Pigeonhole Principle \\( detpolyfunc \\) takes one of these values three or more times. But the only polynomials of degree at most 2 that take the same value three times are constant polynomials. In particular, \\( \\operatorname{det}(basematrix+5 stepmatrix)= \\pm 1 \\), so \\( basematrix+5 stepmatrix \\) has an inverse with integer entries."
    },
    "descriptive_long_confusing": {
      "map": {
        "A": "bluelotus",
        "B": "goldfinch",
        "M": "rainbucket",
        "N": "cloudmirror",
        "f": "mapleforest",
        "x": "dustytrail"
      },
      "question": "Let $bluelotus$ and $goldfinch$ be $2 \\times 2$ matrices with integer entries such\nthat $bluelotus, bluelotus+goldfinch, bluelotus+2goldfinch, bluelotus+3goldfinch$, and $bluelotus+4goldfinch$ are all invertible matrices whose\ninverses have integer entries. Show that $bluelotus+5goldfinch$ is invertible and that\nits inverse has integer entries.",
      "solution": "Solution. A square matrix \\( rainbucket \\) with integer entries has an inverse with integer entries if and only if \\( \\operatorname{det} rainbucket= \\pm 1 \\) : if \\( cloudmirror \\) is such an inverse, \\( (\\operatorname{det} rainbucket)(\\operatorname{det} cloudmirror)= \\) \\( \\operatorname{det}(rainbucket cloudmirror)=1 \\) so \\( \\operatorname{det} rainbucket= \\pm 1 \\); conversely, if \\( \\operatorname{det} rainbucket= \\pm 1 \\), then \\( \\pm rainbucket^{\\prime} \\) is an inverse with integer entries, where \\( rainbucket^{\\prime} \\) is the classical adjoint of \\( rainbucket \\). Let \\( mapleforest(dustytrail)=\\operatorname{det}(bluelotus+dustytrail goldfinch) \\). Then \\( mapleforest(dustytrail) \\) is a polynomial of degree at most 2 , such that \\( mapleforest(dustytrail)= \\pm 1 \\) for \\( dustytrail=0,1,2 \\), 3 , and 4 . Thus by the Pigeonhole Principle \\( mapleforest \\) takes one of these values three or more times. But the only polynomials of degree at most 2 that take the same value three times are constant polynomials. In particular, \\( \\operatorname{det}(bluelotus+5 goldfinch)= \\pm 1 \\), so \\( bluelotus+5 goldfinch \\) has an inverse with integer entries."
    },
    "descriptive_long_misleading": {
      "map": {
        "M": "scalarobj",
        "N": "noninvert",
        "f": "constantfun",
        "x": "fixvalue",
        "A": "floatmatr",
        "B": "subtractor"
      },
      "question": "Let $floatmatr$ and $subtractor$ be $2 \\times 2$ matrices with integer entries such\nthat $floatmatr, floatmatr+subtractor, floatmatr+2subtractor, floatmatr+3subtractor$, and $floatmatr+4subtractor$ are all invertible matrices whose\ninverses have integer entries. Show that $floatmatr+5subtractor$ is invertible and that\nits inverse has integer entries.",
      "solution": "Solution. A square matrix \\( scalarobj \\) with integer entries has an inverse with integer entries if and only if \\( \\operatorname{det} scalarobj= \\pm 1 \\) : if \\( noninvert \\) is such an inverse, \\( (\\operatorname{det} scalarobj)(\\operatorname{det} noninvert)= \\) \\( \\operatorname{det}(scalarobj noninvert)=1 \\) so \\( \\operatorname{det} scalarobj= \\pm 1 \\); conversely, if \\( \\operatorname{det} scalarobj= \\pm 1 \\), then \\( \\pm scalarobj^{\\prime} \\) is an inverse with integer entries, where \\( scalarobj^{\\prime} \\) is the classical adjoint of \\( scalarobj \\). Let \\( constantfun(fixvalue)=\\operatorname{det}(floatmatr+fixvalue subtractor) \\). Then \\( constantfun(fixvalue) \\) is a polynomial of degree at most 2 , such that \\( constantfun(fixvalue)= \\pm 1 \\) for \\( fixvalue=0,1,2 \\), 3 , and 4 . Thus by the Pigeonhole Principle \\( constantfun \\) takes one of these values three or more times. But the only polynomials of degree at most 2 that take the same value three times are constant polynomials. In particular, \\( \\operatorname{det}(floatmatr+5 subtractor)= \\pm 1 \\), so \\( floatmatr+5 subtractor \\) has an inverse with integer entries."
    },
    "garbled_string": {
      "map": {
        "A": "tndghswe",
        "B": "rckvmpoa",
        "M": "zqplmstn",
        "N": "hvdcrlga",
        "f": "kmrwqsop",
        "x": "bgtvlkqe"
      },
      "question": "Let $tndghswe$ and $rckvmpoa$ be $2 \\times 2$ matrices with integer entries such\nthat $tndghswe, tndghswe+rckvmpoa, tndghswe+2rckvmpoa, tndghswe+3rckvmpoa$, and $tndghswe+4rckvmpoa$ are all invertible matrices whose\ninverses have integer entries. Show that $tndghswe+5rckvmpoa$ is invertible and that\nits inverse has integer entries.",
      "solution": "Solution. A square matrix \\( zqplmstn \\) with integer entries has an inverse with integer entries if and only if \\( \\operatorname{det} zqplmstn= \\pm 1 \\) : if \\( hvdcrlga \\) is such an inverse, \\( (\\operatorname{det} zqplmstn)(\\operatorname{det} hvdcrlga)= \\) \\( \\operatorname{det}(zqplmstn hvdcrlga)=1 \\) so \\( \\operatorname{det} zqplmstn= \\pm 1 \\); conversely, if \\( \\operatorname{det} zqplmstn= \\pm 1 \\), then \\( \\pm zqplmstn^{\\prime} \\) is an inverse with integer entries, where \\( zqplmstn^{\\prime} \\) is the classical adjoint of \\( zqplmstn \\). Let \\( kmrwqsop(bgtvlkqe)=\\operatorname{det}(tndghswe+bgtvlkqe rckvmpoa) \\). Then \\( kmrwqsop(bgtvlkqe) \\) is a polynomial of degree at most 2 , such that \\( kmrwqsop(bgtvlkqe)= \\pm 1 \\) for \\( bgtvlkqe=0,1,2, 3 ,\\) and \\( 4 \\). Thus by the Pigeonhole Principle \\( kmrwqsop \\) takes one of these values three or more times. But the only polynomials of degree at most 2 that take the same value three times are constant polynomials. In particular, \\( \\operatorname{det}(tndghswe+5 rckvmpoa)= \\pm 1 \\), so \\( tndghswe+5 rckvmpoa \\) has an inverse with integer entries."
    },
    "kernel_variant": {
      "question": "Let n ≥ 4 be an integer.  \nLet  \n\n A , B₁ , B₂ , … , Bₙ  \n\nbe n × n matrices with integer entries.  For every n-tuple  \n\n (k₁ , k₂ , … , kₙ) with 0 ≤ kᵢ ≤ 2n (i = 1,…,n)\n\nassume that the matrix  \n\n M(k₁,…,kₙ) := A + k₁B₁ + k₂B₂ + ⋯ + kₙBₙ  \n\nis invertible and that M(k₁,…,kₙ)⁻¹ again has integer entries.\n\n1.  Prove that the determinant  \n\n  f(x₁,…,xₙ) := det(A + x₁B₁ + ⋯ + xₙBₙ)  \n\nis the constant polynomial ±1 in the n indeterminates x₁,…,xₙ.\n\n2.  Deduce that for every integer n-tuple (t₁,…,tₙ) the matrix  \n\n  A + t₁B₁ + ⋯ + tₙBₙ  \n\nis invertible and that its inverse again has integer entries.",
      "solution": "Throughout we use the elementary fact\n\n(★) An n × n matrix with integer entries is invertible over ℤ (i.e. admits an inverse whose entries are integers) iff its determinant equals ±1.\n\nStep 1.  Bounding the individual degrees of f  \n\nWrite  \n\n f(x₁,…,xₙ)=det(A+Σ_{i=1}^{n} xᵢBᵢ) ∈ ℤ[x₁,…,xₙ].\n\nView the determinant as an alternating n-linear function of the n column\nvectors.  Fix an index i.  In the j-th column we have\n\n col_j (A+Σ xᵢBᵢ) = col_j(A) + Σ_{i=1}^{n} xᵢ col_j(Bᵢ),\n\nwhich is an affine-linear polynomial in each xᵢ.  Expanding the\ndeterminant via its multilinearity, a monomial in xᵢ is obtained by\nchoosing, for every column, either the constant term col_j(A) or the\nxᵢ-term xᵢ col_j(Bᵢ).  Hence the exponent of xᵢ in any resulting\nmonomial equals the number of columns for which the second choice is\nmade; at most n columns exist, so\n\n deg_{xᵢ} f ≤ n for every i = 1,…,n.  (1)\n\nConsequently the polynomial\n\n h := f² − 1 ∈ ℤ[x₁,…,xₙ]\n\nsatisfies  \n\n deg_{xᵢ} h ≤ 2n  for every i.  (2)\n\nStep 2.  The polynomial f must be constant ±1  \n\nPut S := {0,1,…,2n}.  For every K=(k₁,…,kₙ)∈Sⁿ the hypothesis says\nM(K) ∈ GLₙ(ℤ); by (★)\n\n f(K)=det M(K)=±1.  (3)\n\nHence h vanishes on the whole grid Sⁿ.  \nBecause |S| = 2n+1 and by (2) deg_{xᵢ} h ≤ 2n < |S|, the following\ngeneral interpolation lemma applies.\n\nInterpolation Lemma.  \nLet d≥0 and let S⊂ℚ with |S| = d+1.  If P ∈ ℚ[x₁,…,xₙ] satisfies\ndeg_{xᵢ} P ≤ d for all i and P vanishes on Sⁿ, then P is the zero\npolynomial.\n\nProof.  For n = 1 the Vandermonde matrix built from S is nonsingular, so\nevaluation on S determines a degree ≤ d polynomial uniquely.  For\ngeneral n take the tensor product of these univariate evaluation maps;\ninjectivity follows immediately. ∎\n\nTaking d = 2n and S = {0,…,2n}, the lemma forces h ≡ 0, whence\n\n f(x₁,…,xₙ)² ≡ 1 in ℤ[x₁,…,xₙ].\n\nTherefore f is the constant polynomial ±1, completing Part 1.\n\nStep 3.  Unimodularity for arbitrary integral parameters  \n\nLet (t₁,…,tₙ) ∈ ℤⁿ be arbitrary and set  \n\n M := A + t₁B₁ + ⋯ + tₙBₙ.\n\nBy Part 1 we have det M = ±1, so M is unimodular by (★).\nThe adjugate identity\n\n M · adj M = (det M) Iₙ = ± Iₙ\n\ngives M⁻¹ = ± adj M.  Because adj M is a matrix whose entries are\ninteger polynomials in the entries of M, all entries of adj M—and hence\nthose of M⁻¹—are integers.\n\nThus both statements are proved.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T19:09:31.736369",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension and multiple variables:  The determinant now depends simultaneously on n independent variables, not just on one.  \n•  Far more hypotheses must be synthesised:  (n+1)ⁿ distinct matrices are assumed unimodular; their information must be woven together via multivariate-polynomial arguments instead of a single-variable interpolation.  \n•  Proof demands repeated reduction in each coordinate, an iterative application of univariate polynomial reasoning inside an n-dimensional setting; careless counting of zeros is no longer sufficient.  \n•  The solver must recognise that degree bounds hold separately in every variable, invoke them iteratively, and keep track of constants carefully—a conceptual and technical leap beyond the single-variable case.  \n•  Showing that integral inverses exist for all integer parameters relies on structural properties of the adjugate matrix in the multivariate setting.  Handling these polynomial–matrix identities substantially deepens the algebraic content compared with the original deterministic-only argument.\n\nHence the enhanced variant is significantly harder: it replaces one variable by n independent variables, raises both the combinatorial and algebraic complexity of the interpolation argument, and obliges the solver to master polynomial identities in several variables together with matrix-adjugate considerations."
      }
    },
    "original_kernel_variant": {
      "question": "Let n ≥ 4 be an integer.  \nLet  \n\n A , B₁ , B₂ , … , Bₙ  \n\nbe n × n matrices with integer entries.  For every n-tuple  \n\n (k₁ , k₂ , … , kₙ) with 0 ≤ kᵢ ≤ 2n (i = 1,…,n)\n\nassume that the matrix  \n\n M(k₁,…,kₙ) := A + k₁B₁ + k₂B₂ + ⋯ + kₙBₙ  \n\nis invertible and that M(k₁,…,kₙ)⁻¹ again has integer entries.\n\n1.  Prove that the determinant  \n\n  f(x₁,…,xₙ) := det(A + x₁B₁ + ⋯ + xₙBₙ)  \n\nis the constant polynomial ±1 in the n indeterminates x₁,…,xₙ.\n\n2.  Deduce that for every integer n-tuple (t₁,…,tₙ) the matrix  \n\n  A + t₁B₁ + ⋯ + tₙBₙ  \n\nis invertible and that its inverse again has integer entries.",
      "solution": "Throughout we use the elementary fact\n\n(★) An n × n matrix with integer entries is invertible over ℤ (i.e. admits an inverse whose entries are integers) iff its determinant equals ±1.\n\nStep 1.  Bounding the individual degrees of f  \n\nWrite  \n\n f(x₁,…,xₙ)=det(A+Σ_{i=1}^{n} xᵢBᵢ) ∈ ℤ[x₁,…,xₙ].\n\nView the determinant as an alternating n-linear function of the n column\nvectors.  Fix an index i.  In the j-th column we have\n\n col_j (A+Σ xᵢBᵢ) = col_j(A) + Σ_{i=1}^{n} xᵢ col_j(Bᵢ),\n\nwhich is an affine-linear polynomial in each xᵢ.  Expanding the\ndeterminant via its multilinearity, a monomial in xᵢ is obtained by\nchoosing, for every column, either the constant term col_j(A) or the\nxᵢ-term xᵢ col_j(Bᵢ).  Hence the exponent of xᵢ in any resulting\nmonomial equals the number of columns for which the second choice is\nmade; at most n columns exist, so\n\n deg_{xᵢ} f ≤ n for every i = 1,…,n.  (1)\n\nConsequently the polynomial\n\n h := f² − 1 ∈ ℤ[x₁,…,xₙ]\n\nsatisfies  \n\n deg_{xᵢ} h ≤ 2n  for every i.  (2)\n\nStep 2.  The polynomial f must be constant ±1  \n\nPut S := {0,1,…,2n}.  For every K=(k₁,…,kₙ)∈Sⁿ the hypothesis says\nM(K) ∈ GLₙ(ℤ); by (★)\n\n f(K)=det M(K)=±1.  (3)\n\nHence h vanishes on the whole grid Sⁿ.  \nBecause |S| = 2n+1 and by (2) deg_{xᵢ} h ≤ 2n < |S|, the following\ngeneral interpolation lemma applies.\n\nInterpolation Lemma.  \nLet d≥0 and let S⊂ℚ with |S| = d+1.  If P ∈ ℚ[x₁,…,xₙ] satisfies\ndeg_{xᵢ} P ≤ d for all i and P vanishes on Sⁿ, then P is the zero\npolynomial.\n\nProof.  For n = 1 the Vandermonde matrix built from S is nonsingular, so\nevaluation on S determines a degree ≤ d polynomial uniquely.  For\ngeneral n take the tensor product of these univariate evaluation maps;\ninjectivity follows immediately. ∎\n\nTaking d = 2n and S = {0,…,2n}, the lemma forces h ≡ 0, whence\n\n f(x₁,…,xₙ)² ≡ 1 in ℤ[x₁,…,xₙ].\n\nTherefore f is the constant polynomial ±1, completing Part 1.\n\nStep 3.  Unimodularity for arbitrary integral parameters  \n\nLet (t₁,…,tₙ) ∈ ℤⁿ be arbitrary and set  \n\n M := A + t₁B₁ + ⋯ + tₙBₙ.\n\nBy Part 1 we have det M = ±1, so M is unimodular by (★).\nThe adjugate identity\n\n M · adj M = (det M) Iₙ = ± Iₙ\n\ngives M⁻¹ = ± adj M.  Because adj M is a matrix whose entries are\ninteger polynomials in the entries of M, all entries of adj M—and hence\nthose of M⁻¹—are integers.\n\nThus both statements are proved.",
      "metadata": {
        "replaced_from": "harder_variant",
        "replacement_date": "2025-07-14T01:37:45.570119",
        "was_fixed": false,
        "difficulty_analysis": "•  Higher dimension and multiple variables:  The determinant now depends simultaneously on n independent variables, not just on one.  \n•  Far more hypotheses must be synthesised:  (n+1)ⁿ distinct matrices are assumed unimodular; their information must be woven together via multivariate-polynomial arguments instead of a single-variable interpolation.  \n•  Proof demands repeated reduction in each coordinate, an iterative application of univariate polynomial reasoning inside an n-dimensional setting; careless counting of zeros is no longer sufficient.  \n•  The solver must recognise that degree bounds hold separately in every variable, invoke them iteratively, and keep track of constants carefully—a conceptual and technical leap beyond the single-variable case.  \n•  Showing that integral inverses exist for all integer parameters relies on structural properties of the adjugate matrix in the multivariate setting.  Handling these polynomial–matrix identities substantially deepens the algebraic content compared with the original deterministic-only argument.\n\nHence the enhanced variant is significantly harder: it replaces one variable by n independent variables, raises both the combinatorial and algebraic complexity of the interpolation argument, and obliges the solver to master polynomial identities in several variables together with matrix-adjugate considerations."
      }
    }
  },
  "checked": true,
  "problem_type": "proof"
}