[#P2] Determinants of the Fibonacci-sum matrix
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Problem. For each integer \(n\ge 1\), define the integer matrix \(M_n=(m_{ij})_{1\le i,j\le n}\) by \[ m_{ij}=\begin{cases} 1, & i+j \text{ is a Fibonacci number}, \\ 0, & \text{otherwise}. \end{cases} \] Prove that \(\det(M_n)\in\{-1,0,1\}\) for every integer \(n\ge 1\).
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Definitions and notation
1Context
This conjecture concerns the determinants of finite indicator matrices whose nonzero entries are selected by Fibonacci sums.
2Remarks
Remark 1. The rows and columns of \(M_n\) are indexed by \(1,\ldots,n\).
1StatusSolution submittedNot Lean-verified
Saved packet · July 26, 2026
Submitted solution (The determinant is always minus one, zero, or one). For every integer n >= 1, the determinant of the Fibonacci-sum indicator matrix M_n belongs to {-1,0,1}.
Proof
Proof. Let
$$ q_0=1,\qquad q_1=2,\qquad q_{s+1}=q_s+q_{s-1}. $$
These are the distinct positive Fibonacci numbers. Since every matrix index sum is at least $2$, this convention gives the same matrices as the standard sequence $F_0=0,F_1=1$.
For $n\geq 1$, let $M_n=(m_{ij})_{1\leq i,j\leq n}$, where
$$ m_{ij}= \begin{cases} 1,&i+j=q_s\text{ for some }s,\\ 0,&\text{otherwise}. \end{cases} $$
We prove the following stronger statement.
Theorem. Every $M_n$ is totally unimodular. Consequently, every square minor of $M_n$, including $\det M_n$, belongs to $\{-1,0,1\}$.
1. The support graph.
Let $Q_n$ be the bipartite graph with row vertices $r_1,\ldots,r_n$, column vertices $c_1,\ldots,c_n$, and an edge $r_i c_j$ exactly when $i+j$ is Fibonacci. The matrix $M_n$ is the biadjacency matrix of $Q_n$. A diagonal entry of $M_n$ becomes the ordinary bipartite edge $r_i c_i$, so loops never arise in $Q_n$.
The ordinary loopless Fibonacci-sum graph has analogous chord and outerplanarity properties [2]. The argument below treats the bipartite row-column support graph directly.
Lemma 1. $Q_n$ is chordal bipartite.
Every cycle of length at least six has a chord.
Take such a cycle and let $m$ be its largest numerical label. Row-column symmetry lets us suppose that $r_m$ lies on the cycle. Choose $k$ with
$$ q_k\leq m<q_{k+1}. $$
If $c_j$ is a neighbor of $r_m$ whose label is at most $m$, then $m+j\in(m,2m]$. The only possible Fibonacci numbers in this interval are $q_{k+1}$ and $q_{k+2}$. If $q_{k+2}>2m$, this gives at most one neighbor, so $r_m$ cannot lie on a cycle. We may therefore assume $q_{k+2}\leq 2m$. The two cycle neighbors of $r_m$ must then be
$$ c_a,\quad c_b,\qquad a=q_{k+1}-m,\quad b=q_{k+2}-m. $$
Put $z=m-q_k$. We have $b>q_k$. Among vertices with labels at most $m$, the vertex $c_b$ therefore has exactly two neighbors, namely $r_m$ and $r_z$, because
$$ b+m=q_{k+2},\qquad b+z=q_{k+1}. $$
Also
$$ a+z=(q_{k+1}-m)+(m-q_k)=q_{k-1}. $$
Thus the cycle contains the path
$$ c_a-r_m-c_b-r_z $$
and $r_zc_a$ is an edge. On a cycle of length at least six, that edge is a chord. The same calculation covers $b=m$. In that case $a=z$, while $r_a,c_a,r_m,c_m$ are still four distinct graph vertices.
Lemma 2. Every edge of $Q_n$ lies in at most two four-cycles.
First classify the four-cycles in the infinite version of the graph. Suppose a four-cycle uses rows $x<X$ and columns $y<Y$. Set
$$ A=x+y,\quad B=x+Y,\quad C=X+y,\quad D=X+Y. $$
All four numbers are Fibonacci, $A<B,C<D$, and
$$ A+D=B+C. $$
Assume $B\leq C$, and write
$$ A=q_\alpha,\quad B=q_\beta,\quad C=q_\gamma,\quad D=q_\delta. $$
Then
$$ q_\delta-q_\gamma=q_\beta-q_\alpha. $$
If $\delta\geq\gamma+2$, the left side is at least $q_{\gamma+1}$, while the right side is smaller than $q_\beta\leq q_\gamma$. Hence $\delta=\gamma+1$, and the common difference is $q_{\gamma-1}$. If $\beta\leq\gamma-1$, the right side is smaller than $q_\beta\leq q_{\gamma-1}$. Hence $\beta=\gamma$, followed by $\alpha=\gamma-2$.
Writing $t=\gamma$, every four-cycle has corner sums
$$ q_{t-2},\quad q_t,\quad q_t,\quad q_{t+1}, $$
and its row and column increments are both
$$ q_t-q_{t-2}=q_{t-1}. $$
Now fix an edge $r_xc_y$ with $x+y=q_s$. It can occur in a classified four-cycle in three ways:
1. As the $q_{t-2}$ corner. This determines at most one four-cycle.
2. As the $q_{t+1}$ corner. This determines at most one four-cycle and requires $x,y>q_{s-2}$.
3. As one of the two $q_t$ corners. The two orientations require, respectively, $x>q_{s-1}$ or $y>q_{s-1}$.
The two orientations in the third case cannot both occur, since $x+y=q_s<2q_{s-1}$. The second and third cases cannot occur together, since their inequalities would give
$$ x+y>q_{s-2}+q_{s-1}=q_s. $$
There is therefore at most one four-cycle from the first case and at most one from the other two cases combined.
Lemma 3. $Q_n$ is outerplanar.
We construct an outerplane embedding by induction on $n$. The claim is immediate for $n=1$. Assume $Q_{m-1}$ has an outerplane embedding, and choose $k$ with $q_k\leq m<q_{k+1}$. Set
$$ a=q_{k+1}-m. $$
If $2m<q_{k+2}$, the new vertices $r_m,c_m$ add the two pendant edges $r_mc_a$ and $r_ac_m$. Draw them in the outer face.
Suppose $2m=q_{k+2}$. Then
$$ a=m-q_k,\qquad 2a=q_{k-1}. $$
The old graph contains $r_ac_a$, and the two new vertices complete the four-cycle
$$ r_a-c_a-r_m-c_m-r_a. $$
Suppose $2m>q_{k+2}$. Put
$$ b=q_{k+2}-m,\qquad z=m-q_k. $$
Here $b<m$. In $Q_{m-1}$, $c_b$ is pendant at $r_z$, and $r_b$ is pendant at $c_z$. The new vertices complete two four-cycles, attached along the old edges
$$ r_zc_a\qquad\text{and}\qquad r_ac_z. $$
In both the equality and strict cases, each attachment edge borders the outer face of the chosen embedding. To see this, suppose an attachment edge bordered two bounded faces. It lies in a two-connected block. Each of those faces has an induced cycle as its boundary, and Lemma 1 makes each boundary a four-cycle. The new four-cycle would place the attachment edge in three four-cycles, contrary to Lemma 2.
We can therefore draw each new four-cycle in the outer face beside its attachment edge. The old pendant vertices $c_b$ and $r_b$ can be moved within that face into the required positions. In the last case the two attachment edges are vertex-disjoint, since
$$ z-a=2m-q_{k+2}>0. $$
This completes the induction.
2. The divisibility condition.
We need one plane-graph observation.
Lemma 4. Eulerian outerplane quadrangulations have a multiple of four edges.
Let $H$ be an outerplane graph in which every vertex has even degree and every bounded face is a four-cycle. Then $|E(H)|$ is divisible by four.
It is enough to work one connected component at a time. A connected graph in which every vertex has even degree has no bridges. Its plane dual is bipartite, so its faces admit a black-white coloring with adjacent faces receiving opposite colors. Color the outer face white. Every edge then borders exactly one black face. All black faces are bounded four-faces, and hence
$$ |E(H)| =\sum_{\text{black faces }f}|\partial f| =4\cdot\#\{\text{black faces}\}. $$
3. Camion's criterion.
Camion's criterion [3] says that a $0,\pm1$ matrix is totally unimodular if and only if the sum of the entries in every square submatrix with even row sums and even column sums is divisible by four.
Take any square submatrix $B=M_n[I,J]$ whose row and column sums are even. Its support graph is the subgraph of $Q_n$ induced by
$$ \{r_i:i\in I\}\cup\{c_j:j\in J\}. $$
This graph remains outerplanar and chordal bipartite. Every vertex has even degree. In an outerplane embedding, each bounded face is an induced cycle inside a two-connected block. Lemma 1 makes every such face a four-cycle. Lemma 4 now gives
$$ \sum_{i\in I,\ j\in J} B_{ij}=|E(B)|\equiv0\pmod4. $$
Camion's criterion applies to every such $B$. Thus $M_n$ is totally unimodular for every $n$, which proves
$$ \boxed{\det M_n\in\{-1,0,1\}\quad\text{for every }n\geq1.} $$□
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How the 48 records connect
ProblemDeterminants of the Fibonacci-sum matrix
- Proposition 2Every Fibonacci-sum matrix is totally unimodularclaims to resolveReview pending
- Computation 1Exact determinant range verified through n=120supportsReproduced
- Artifact 5Exact Bareiss determinant sweep through n=120checksExecutable material
- Proposition 1Determinant equals a signed matching imbalancereformulatesSupported
- Claim 2Cancellation at n=33 is exactsupportsSupported
- Artifact 1Allowed-permutation parity count at n=33checksSupported
- Route 2Construct a sign-reversing involution on allowed permutationsaddressesConjectured
- Route 7Use the signed perfect-matching graphaddressesConjectured
- Formalization 1Lean definition of the Fibonacci-sum matrixformalizesReported
- Formalization 7Total unimodularity of the Fibonacci-sum matrixdepends onReported
- Formalization 8Lean determinant-range reductiondepends onReported
- Formalization 2Fibonacci support four-cycle classificationdepends onReported
- Formalization 5Fibonacci odd square coverdepends onReported
- Formalization 6Fibonacci support divisibility conditiondepends onReported
- Route 8Classify sequences with totally unimodular sum matricesgeneralizesConjectured
- Computation 3Fibonacci growth rate does not force the determinant propertyconstrainsReproduced
- Artifact 6Exact growth-rate counterexample at n=5checksExecutable material
- Question 1Characterize the nonzero determinant indicessubproblem ofSupported
- Conjecture 2Primary gaps in nonzero support follow a Fibonacci formulasupportsSupported
- Artifact 2Nonzero determinant support and Zeckendorf tablechecksSupported
- Conjecture 3Secondary support gaps repeat the primary sizessupportsSupported
- Conjecture 4A tertiary support-gap family has its own recurrencesupportsSupported
- Computation 2Observed nonzero determinant indices through n=120supportsReproduced
- Claim 9Dated status and exact unresolved remainderaddressesSupported
- Route 3Search for an explicit unimodular eliminationaddressesConjectured
- Route 6Sweep Smith normal formstestsConjectured
- Route 4Close a recursion on the peeled coreaddressesSupported
- Claim 4Residual cores appear on Zeckendorf-sparse symmetric index setssupportsReported
- Artifact 8Residual-core index samplessupportsExecutable material
- Route 5Infer the core index map in Zeckendorf coordinatesenablesSupported
- Claim 5Support-difference rows replicate four levels latersupportsSupported
- Artifact 3Colored block self-similarity diagramchecksSupported
- Claim 6Support blocks exhibit mirror symmetriessupportsSupported
- Artifact 4Planned core sweep through n=500implementsConjectured
- Claim 7A boundary effect interrupts the mirror ruleconstrainsSupported
- Route 9Search matrix and Fibonacci literature for a known reductionaddressesConjectured
- Route 10Formalize the conjecture and finite reductions in LeanaddressesConjectured
- Claim 1The determinant is always minus one, zero, or oneresolvesReported
- Route 1Reduce by forced singleton rows and columnsaddressesIncomplete method
- Proposition 3Forced peeling reduces the determinant to a core determinantsalvagesSupported
- Computation 4Singleton peeling usually leaves a residual coreruled out as a complete methodReproduced
- Artifact 7Singleton-peeling residual-core sweepchecksExecutable material
- Conjecture 1Lucas sequences may share total unimodularityin this packetReported
- Claim 3Several non-Lucas sequences also pass initial testsbroadensReported
- Formalization 3Almost-TU determinant and nonvanishing cofactorsin this packetReported
- Formalization 4Camion's total-unimodularity criteriondepends onReported
2 records outside this overview
All 52 recorded relations between these records and the problem
- Characterize the nonzero determinant indices is a subproblem of Determinants of the Fibonacci-sum matrix
- Several non-Lucas sequences also pass initial tests broadens Lucas sequences may share total unimodularity
- Residual cores appear on Zeckendorf-sparse symmetric index sets supports Close a recursion on the peeled core
- Primary gaps in nonzero support follow a Fibonacci formula supports Characterize the nonzero determinant indices
- Secondary support gaps repeat the primary sizes supports Characterize the nonzero determinant indices
- A tertiary support-gap family has its own recurrence supports Characterize the nonzero determinant indices
- Support-difference rows replicate four levels later supports Infer the core index map in Zeckendorf coordinates
- Support blocks exhibit mirror symmetries supports Infer the core index map in Zeckendorf coordinates
- A boundary effect interrupts the mirror rule constrains Infer the core index map in Zeckendorf coordinates
- Allowed-permutation parity count at n=33 is evidence for Cancellation at n=33 is exact
- Nonzero determinant support and Zeckendorf table is evidence for Primary gaps in nonzero support follow a Fibonacci formula
- Nonzero determinant support and Zeckendorf table is evidence for Secondary support gaps repeat the primary sizes
- Nonzero determinant support and Zeckendorf table is evidence for A tertiary support-gap family has its own recurrence
- Colored block self-similarity diagram is evidence for Support-difference rows replicate four levels later
- Colored block self-similarity diagram is evidence for Support blocks exhibit mirror symmetries
- Planned core sweep through n=500 implements Infer the core index map in Zeckendorf coordinates
- Search for an explicit unimodular elimination addresses Determinants of the Fibonacci-sum matrix
- Close a recursion on the peeled core addresses Determinants of the Fibonacci-sum matrix
- Infer the core index map in Zeckendorf coordinates enables Close a recursion on the peeled core
- Sweep Smith normal forms tests Search for an explicit unimodular elimination
- Classify sequences with totally unimodular sum matrices generalizes Determinants of the Fibonacci-sum matrix
- Search matrix and Fibonacci literature for a known reduction addresses Determinants of the Fibonacci-sum matrix
- Formalize the conjecture and finite reductions in Lean addresses Determinants of the Fibonacci-sum matrix
- Lean definition of the Fibonacci-sum matrix formalizes Determinants of the Fibonacci-sum matrix
- Lean determinant-range reduction formalizes Determinants of the Fibonacci-sum matrix
- Lean determinant-range reduction depends on Total unimodularity of the Fibonacci-sum matrix
- Total unimodularity of the Fibonacci-sum matrix depends on Lean definition of the Fibonacci-sum matrix
- Total unimodularity of the Fibonacci-sum matrix depends on Camion's total-unimodularity criterion
- Camion's total-unimodularity criterion depends on Almost-TU determinant and nonvanishing cofactors
- Total unimodularity of the Fibonacci-sum matrix depends on Fibonacci support divisibility condition
- Fibonacci support divisibility condition depends on Fibonacci odd square cover
- Fibonacci odd square cover depends on Fibonacci support four-cycle classification
- Fibonacci support four-cycle classification depends on Lean definition of the Fibonacci-sum matrix
- Determinant equals a signed matching imbalance reformulates Determinants of the Fibonacci-sum matrix
- Exact determinant range verified through n=120 supports Determinants of the Fibonacci-sum matrix
- Cancellation at n=33 is exact supports Determinant equals a signed matching imbalance
- Observed nonzero determinant indices through n=120 supports Characterize the nonzero determinant indices
- The determinant is always minus one, zero, or one resolves Determinants of the Fibonacci-sum matrix
- Every Fibonacci-sum matrix is totally unimodular claims to resolve Determinants of the Fibonacci-sum matrix
- Fibonacci growth rate does not force the determinant property constrains Classify sequences with totally unimodular sum matrices
- Singleton peeling usually leaves a residual core refutes as a complete method Reduce by forced singleton rows and columns
- Forced peeling reduces the determinant to a core determinant salvages Reduce by forced singleton rows and columns
- Exact Bareiss determinant sweep through n=120 is evidence for Exact determinant range verified through n=120
- Nonzero determinant support and Zeckendorf table is evidence for Observed nonzero determinant indices through n=120
- Exact growth-rate counterexample at n=5 is evidence for Fibonacci growth rate does not force the determinant property
- Singleton-peeling residual-core sweep is evidence for Singleton peeling usually leaves a residual core
- Residual-core index samples supports Residual cores appear on Zeckendorf-sparse symmetric index sets
- Construct a sign-reversing involution on allowed permutations addresses Determinant equals a signed matching imbalance
- Reduce by forced singleton rows and columns addresses Determinants of the Fibonacci-sum matrix
- Close a recursion on the peeled core uses Forced peeling reduces the determinant to a core determinant
- Use the signed perfect-matching graph addresses Determinant equals a signed matching imbalance
- Dated status and exact unresolved remainder addresses Characterize the nonzero determinant indices
2See also
- Characterize the nonzero determinant indicessame packet
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“Determinants of the Fibonacci-sum matrix.” TheoremDB. P2. Problem statement; statement text SHA-256 91312d1aa020718968fced2c0d3e2d44bb353b99505661afc56cb1c53398c700. https://theoremdb.org/statement/?ref=P2
@misc{theoremdb-problem-91312d1aa020718968fced2c0d3e2d44bb353b99505661afc56cb1c53398c700,
title = {{Determinants of the Fibonacci-sum matrix}},
howpublished = {TheoremDB},
note = {Problem statement; statement text SHA-256 91312d1aa020718968fced2c0d3e2d44bb353b99505661afc56cb1c53398c700},
url = {https://theoremdb.org/statement/?ref=P2}
}Plain text: Built Markdown snapshot
This problem includes 48 records joined by 103 typed links, sourced from mathoverflow.net[1], current as of July 26, 2026.
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- fibSumMatrix_totallyUnimodular is a prerequisite of fibSumMatrix_det_range
- fibSumMatrix is a prerequisite of fibSumMatrix_totallyUnimodular
- isTotallyUnimodular_of_camion is a prerequisite of fibSumMatrix_totallyUnimodular
- minimal_bad_square_minor_det_and_adjugate_nonzero is a prerequisite of isTotallyUnimodular_of_camion
- fibSumMatrix_camion_divisibility is a prerequisite of fibSumMatrix_totallyUnimodular
- fibSubmatrixSupport_has_odd_square_cover is a prerequisite of fibSumMatrix_camion_divisibility
- fibonacci_support_four_corner_classification is a prerequisite of fibSubmatrixSupport_has_odd_square_cover
- fibSumMatrix is a prerequisite of fibonacci_support_four_corner_classification
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1References
- Packet source. Philip Weiss, Is the determinant of this "Fibonacci sum" indicator matrix always -1, 0, or 1?, MathOverflow question 513340 (2026), with answers and comments. Question lines 57-60. ↗forum · reference source · web version checked 2026-08-01 · checked 2026-07-26Source use: citation only.For fib problem determinant range; fib problem nonzero support: Supplies the problem statement, reported computations, discussion, and community observations recorded in the packet.Also cited at Question, answers, and comments cited by the individual research records.Also cited at question line 60.Also cited at question line 64.Also cited at comment line 135.Also cited at answer lines 304-306.Also cited at answer lines 305-306.Also cited at answer line 306.Also cited at answer lines 327-371.Also cited at comment lines 109-135.Also cited at answer line 303.Also cited at question lines 65-67.Also cited at MathOverflow comments.Also cited at answer lines 169-306.Also cited at answer image after line 304.For fib problem determinant range; fib problem nonzero support: Supplies the problem statement, reported computations, discussion, and community observations recorded in the packet.Source named by the research packet.Source used to formulate or check the problem record.
- Andrii Arman, David S. Gunderson, and Pak Ching Li, “Properties of the Fibonacci-sum graph”. arXiv:1710.10303 (2017). The outerplanarity result for the Fibonacci-sum graph. ↗preprint · primary source · arXiv:1710.10303v1 · checked 2026-08-01Source use: original summary.For fib problem determinant range; fib problem nonzero support: Supplies the graph-theoretic antecedent used in the packet's independent proof of outerplanarity for the bipartite support graph.Supplies the graph-theoretic antecedent used in the packet's independent proof of outerplanarity for the bipartite support graph.
- Paul Camion, “Characterization of totally unimodular matrices”. Proceedings of the American Mathematical Society 16(5) (1965), 1068-1073. DOI 10.1090/S0002-9939-1965-0180568-2. Characterization theorem, pp. 1068-1073. ↗journal article · primary source · publisher version · checked 2026-08-01Source use: original summary.For fib problem determinant range; fib problem nonzero support: Supplies the total-unimodularity criterion applied in the final step of the proof.Supplies the total-unimodularity criterion applied in the final step of the proof.
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