ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passverified 2026-08-02 (claude-opus-5)

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Consecutive Fibonacci numbers are coprime, and for every $m \in N$ the Euclidean algorithm on $(F_{m + 3}, F_{m + 2})$ takes exactly $m + 1$ divisions, with quotient $1$ in the first $m$ of them and quotient $2$ in the last

Example

The sequence. Write $k + 1$ for $σ (k)$ and $k + 2$ for $σ (σ (k))$ (Addition of natural numbers). By the recursion theorem (The recursion theorem) applied to the set $Z \times Z$ , the element $(0, 1)$ and the function $(u, v) \mapsto (v, u + v)$ , there is exactly one $G : N \to Z \times Z$ with $G (0) = (0, 1)$ and $G (k + 1) = (v, u + v)$ whenever $G (k) = (u, v)$ . Define $F_{k}$ to be the first coordinate of $G (k)$ . Then $G (k) = (F_{k}, F_{k + 1})$ for every $k$ , so

$F_{0} = 0, F_{1} = 1, F_{k + 2} = F_{k + 1} + F_{k},$

and the sequence begins $0, 1, 1, 2, 3, 5, 8, 13$ . The indexing starts at $F_{0} = 0$ , and the statements below depend on that choice.

Coprimality. For every $k \in N$ ,

$\gcd (F_{k + 1}, F_{k}) = 1,$

so consecutive Fibonacci numbers are coprime (Coprime integers: $\gcd (a, b) = 1$ ).

The division count. For $m \in N$ let $R$ be the remainder descent of The Euclidean algorithm: for $a \in Z$ and $b > 0$ the remainder descent from $(a, b)$ terminates, and the last nonzero remainder is $\gcd (a, b)$ from $(F_{m + 3}, F_{m + 2})$ , which is legitimate because $F_{m + 2} > 0$ , and for $k$ below the terminating index $N$ let $q_{k}$ be the quotient used at step $k$ , so that $a_{k} = q_{k} b_{k} + b_{k + 1}$ with $0 \leq b_{k + 1} < b_{k}$ . Then

$N = m + 1, q_{k} = 1 for k < m, q_{m} = 2.$

Equivalently, on $(F_{n + 2}, F_{n + 1})$ with $n = m + 1 \geq 1$ the algorithm performs exactly $n$ divisions. At $m = 0$ the pair is $(F_{3}, F_{2}) = (2, 1)$ , there is a single division $2 = 2 \cdot 1 + 0$ , and the list of quotients equal to $1$ is empty.

What is not claimed. Nothing here says these pairs are the worst case for their size; that is Lamé's theorem, and no bound on the number of divisions in terms of the size of the inputs is available at this point in the reading order.

Facts & Assumptions

Given: The sequence $F : N \to Z$ described above, with $F_{0} = 0$ , $F_{1} = 1$ and $F_{k + 2} = F_{k + 1} + F_{k}$ .

[L1]

$Z$ is a commutative ring; its order is total, antisymmetric and transitive and is compatible with addition; positives are closed under multiplication; $x < y$ means $x \leq y$ together with $x \neq y$ (The integers form a commutative ring, Arithmetic on the integers, The integers form a totally ordered ring, Order on the integers, The integers as equivalence classes of pairs of naturals).

[L2]

Induction on $N$ (The principle of mathematical induction); $σ (k) = k + 1$ (Addition of natural numbers, The natural numbers $N$ (von Neumann)); $m < n$ iff $σ (m) \leq n$ on $N$ (Discreteness: $σ (n)$ is the immediate successor, Order on the natural numbers).

[L3]

For a set $X$ , an $α \in X$ and $f : X \to X$ there is exactly one $g : N \to X$ with $g (0) = α$ and $g (σ (k)) = f (g (k))$ (The recursion theorem).

[L4]

$ι : N \to Z$ is injective, preserves order, and has image the nonnegative integers (The naturals embed in the integers); hence $0 < x$ in $Z$ implies $1 \leq x$ , since $x = ι (k)$ with $k \neq 0$ , so $σ (0) \leq k$ by the discreteness of [L2] and order preservation gives $1 = ι (1) \leq ι (k) = x$ (Discreteness: $σ (n)$ is the immediate successor).

[L5]

For $u \in Z$ and $v > 0$ there is exactly one pair $(q, r)$ with $u = q v + r$ and $0 \leq r < v$ (Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ ).

[L6]

If $u = q v + r$ then $\gcd (u, v) = \gcd (v, r)$ (If $a = q b + r$ then $a, b$ and $b, r$ have exactly the same common divisors, so $\gcd (a, b) = \gcd (b, r)$ ).

[L7]

$\gcd (u, 0) = ∣ u ∣$ and $∣ u ∣ = u$ for $u \geq 0$ ( $\gcd$ is symmetric and unchanged by signs: $\gcd (a, b) = \gcd (b, a) = \gcd (∣ a ∣, ∣ b ∣)$ ; moreover $\gcd (a, 0) = ∣ a ∣$ , $\gcd (a, 1) = 1$ , $\gcd (a, a) = ∣ a ∣$ , and $\gcd (a, b) \geq 1$ unless $a = b = 0$ , The absolute value $∣ a ∣$ of an integer, Absolute value in $Z$ : $∣ a ∣ \geq 0$ ; $∣ a ∣ = 0$ exactly when $a = 0$ ; $∣ - a ∣ = ∣ a ∣$ ; $∣ a b ∣ = ∣ a ∣ ∣ b ∣$ ; $- ∣ a ∣ \leq a \leq ∣ a ∣$ ; and $∣ a ∣ \leq c$ exactly when $- c \leq a \leq c$ , Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ ).

[L8]

The descent of The Euclidean algorithm: for $a \in Z$ and $b > 0$ the remainder descent from $(a, b)$ terminates, and the last nonzero remainder is $\gcd (a, b)$ from $(a, b)$ with $b > 0$ satisfies $R (0) = (a, b)$ and $R (k + 1) = (b_{k}, a_{k} - q_{k} b_{k})$ whenever $b_{k} > 0$ , with $q_{k}$ the quotient of $a_{k}$ by $b_{k}$ ; it is the unique such sequence, and $N$ is the least index with $b_{N} = 0$ .

[L9]

$u$ and $v$ are coprime exactly when $\gcd (u, v) = 1$ (Coprime integers: $\gcd (a, b) = 1$ ).

Verification

technique · induction

1.1

$G (k) = (F_{k}, F_{k + 1})$ for every $k$ , by induction: it holds at $k = 0$ since $G (0) = (0, 1) = (F_{0}, F_{1})$ , and if $G (k) = (F_{k}, F_{k + 1})$ then $G (k + 1) = (F_{k + 1}, F_{k} + F_{k + 1})$ , whose first coordinate is $F_{k + 1}$ , so the second is $F_{k + 2}$ by definition of $F$ as the first coordinate one step later. Hence $F_{k + 2} = F_{k + 1} + F_{k}$ .

L1L2L3

1.2

Induction hypothesis: fix $m$ and assume that the descent from $(F_{m + 3}, F_{m + 2})$ terminates at index $m + 1$ with quotients $q_{k} = 1$ for $k < m$ and $q_{m} = 2$ .

2.1

For every $k$ : $F_{k} \geq 0$ and $F_{k + 1} \geq 1$ . By induction, at $k = 0$ this is $F_{0} = 0 \geq 0$ and $F_{1} = 1 \geq 1$ ; and if it holds at $k$ then $F_{k + 1} \geq 1 \geq 0$ and $F_{k + 2} = F_{k + 1} + F_{k} \geq 1 + 0 = 1$ , using compatibility of the order with addition.

step 1.1L1L2L4

2.2

Coprimality. By induction on $k$ : at $k = 0$ , $\gcd (F_{1}, F_{0}) = \gcd (1, 0) = ∣ 1 ∣ = 1$ by [L7]. If $\gcd (F_{k + 1}, F_{k}) = 1$ , then $F_{k + 2} = 1 \cdot F_{k + 1} + F_{k}$ and [L6] give $\gcd (F_{k + 2}, F_{k + 1}) = \gcd (F_{k + 1}, F_{k}) = 1$ . So $\gcd (F_{k + 1}, F_{k}) = 1$ for every $k$ , and consecutive Fibonacci numbers are coprime.

step 1.1L1L2L6L7L9

2.3

Base case of the division count, $m = 0$ . Here $F_{2} = F_{1} + F_{0} = 1$ and $F_{3} = F_{2} + F_{1} = 1 + 1 = 2$ , so the descent starts at $(2, 1)$ with $b_{0} = 1 > 0$ . Dividing, $2 = q \cdot 1 + r$ with $0 \leq r < 1$ forces $r = 0$ , since $0 < r$ would give $1 \leq r$ by [L4]; hence $q = 2$ and $R (1) = (1, 0)$ . So $N = 1 = m + 1$ , the single quotient is $q_{0} = 2$ , and there is no index $k < m = 0$ .

basestep 1.1L1L4L5L8

3.1

For every $m$ : $0 \leq F_{m + 1}$ , $0 < F_{m + 2}$ and $F_{m + 2} < F_{m + 3}$ . The first two are step 2.1; and $F_{m + 3} = F_{m + 2} + F_{m + 1}$ with $F_{m + 1} \geq 1 > 0$ gives $F_{m + 2} < F_{m + 3}$ .

step 1.1step 2.1L1

4.1

Inductive step. Consider the descent $R^{'}$ from $(F_{m + 4}, F_{m + 3})$ ; it is legitimate since $F_{m + 3} > 0$ by step 3.1. Its first division is $F_{m + 4} = 1 \cdot F_{m + 3} + F_{m + 2}$ , and $0 \leq F_{m + 2} < F_{m + 3}$ by step 3.1, so by the uniqueness in [L5] the quotient is $q_{0}^{'} = 1$ and $R^{'} (1) = (F_{m + 3}, F_{m + 2})$ .

step 1.1step 3.1L5L8

5.1

The shifted sequence $k \mapsto R^{'} (k + 1)$ starts at $(F_{m + 3}, F_{m + 2})$ and obeys the same recursion as $R^{'}$ , hence, by the uniqueness in [L3], equals the descent from $(F_{m + 3}, F_{m + 2})$ . Therefore its terminating index is $m + 1$ by step 1.2, so $R^{'}$ terminates at index $1 + (m + 1) = (m + 1) + 1$ ; and its quotients are $q_{0}^{'} = 1$ followed by the quotients of step 1.2, that is $1$ repeated $m + 1$ times and then $2$ . This is the claim at $m + 1$ .

step 1.2step 4.1L3L8

6.1

By induction the division count holds for every $m \in N$ : the descent from $(F_{m + 3}, F_{m + 2})$ takes exactly $m + 1$ divisions, with quotient $1$ in the first $m$ and quotient $2$ in the last; together with step 2.2 this is the whole example.

step 2.2step 2.3step 5.1L2discharge-induction∎

Remarks

The last quotient is $2$ , not $1$ , and the reason is the repeated value $F_{1} = F_{2} = 1$ . The chain of quotient- $1$ divisions $F_{k + 2} = 1 \cdot F_{k + 1} + F_{k}$ is a valid division only while $F_{k} < F_{k + 1}$ , which fails exactly at $k = 1$ . The descent therefore ends at the pair $(F_{3}, F_{2}) = (2, 1)$ with the single division $2 = 2 \cdot 1 + 0$ . A statement of the form "quotient $1$ at every step" is false for that reason, and the count would also be wrong at the first index if the sequence were indexed from $1$ .
Coprimality does not need the division count, and the count does not need coprimality; they are recorded together because both are read off the same identity $F_{k + 2} = F_{k + 1} + F_{k}$ , once through If $a = q b + r$ then $a, b$ and $b, r$ have exactly the same common divisors, so $\gcd (a, b) = \gcd (b, r)$ and once through Division with remainder in $Z$ : for $a \in Z$ and $b > 0$ there are unique $q, r \in Z$ with $a = q b + r$ and $0 \leq r < b$ .
No worst-case claim. Lamé's theorem — that the Fibonacci pairs minimise the size of the inputs for a given number of divisions — is a genuinely different statement, and nothing above establishes or assumes it.

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

Fibonacci sequence (Wikipedia) (standard reference, not scraped)
Euclidean algorithm (Wikipedia) (standard reference, not scraped)

Consecutive Fibonacci numbers are coprime, and for every m∈Nm∈N the Euclidean algorithm on (Fm+3,Fm+2)(Fm+3​,Fm+2​) takes exactly m+1m+1 divisions, with quotient 11 in the first mm of them and quotient 22 in the last