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

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Bézout coefficients are not unique: $1071 \cdot (- 3) + 462 \cdot 7 = 21$ and $1071 \cdot 19 + 462 \cdot (- 44) = 21$ , and for nonzero $a, b$ every solution has the form $(x_{0} + t b / d, y_{0} - t a / d)$

Example

Bézout's identity: for integers $a, b$ not both zero, $\gcd (a, b)$ is the least positive element of ${a x + b y : x, y \in Z}$ ; in particular $a x + b y = \gcd (a, b)$ has an integer solution asserts that $a x + b y = \gcd (a, b)$ has a solution; it does not assert that the solution is unique, and it is not. For $a = 1071$ and $b = 462$ , where $\gcd (a, b) = 21$ ( $\gcd (1071, 462) = 21$ by the Euclidean algorithm, with the back-substitution giving $21 = 7 \cdot 462 - 3 \cdot 1071$ ),

$1071 \cdot (- 3) + 462 \cdot 7 = 21 and 1071 \cdot 19 + 462 \cdot (- 44) = 21.$

The general statement, for $a$ and $b$ both nonzero, is this. Put $d = \gcd (a, b)$ , $a^{'} = a / d$ and $b^{'} = b / d$ (If $d = \gcd (a, b)$ is nonzero then $a / d$ and $b / d$ are coprime), and let $(x_{0}, y_{0})$ be any solution of $a x + b y = d$ . Then the solutions of that equation are exactly the pairs

$(x, y) = (x_{0} + t b^{'}, y_{0} - t a^{'}) = (x_{0} + t b / d, y_{0} - t a / d), t \in Z .$

For the pair above, $d = 21$ , $a^{'} = 51$ and $b^{'} = 22$ , and $t = 1$ carries $(- 3, 7)$ to $(19, - 44)$ .

Numerals. For $k \in N$ the symbol $k$ written inside $Z$ means $ι (k)$ , the embedding of The naturals embed in the integers; every numerical identity below is the image of the corresponding identity in $N$ .

Facts & Assumptions

Given: Nonzero integers $a$ and $b$ , $d : = \gcd (a, b)$ , and a solution $(x_{0}, y_{0})$ of $a x + b y = d$ ; and, for the numerical part, $a = 1071$ , $b = 462$ , $d = 21$ .

[L1]

$Z$ is a commutative ring: addition and multiplication are associative and commutative, $x \cdot 1 = x$ , $x \cdot 0 = 0$ , multiplication distributes over addition, and every $x$ has an additive inverse; we write $u - v$ for $u + (- v)$ (The integers form a commutative ring, Arithmetic on the integers, The integers as equivalence classes of pairs of naturals).

[L3]

If $d = \gcd (a, b) \neq 0$ then $a = d a^{'}$ and $b = d b^{'}$ for unique integers $a^{'} = a / d$ and $b^{'} = b / d$ , and $\gcd (a^{'}, b^{'}) = 1$ (If $d = \gcd (a, b)$ is nonzero then $a / d$ and $b / d$ are coprime, Common divisor, and the greatest common divisor $\gcd (a, b)$ , with the convention $\gcd (0, 0) : = 0$ , Coprime integers: $\gcd (a, b) = 1$ ).

[L4]

$\gcd (u, v) = \gcd (v, u)$ ( $\gcd$ 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 $a = b = 0$ ).

[L5]

If $\gcd (u, v) = 1$ and $u ∣ v w$ then $u ∣ w$ (If $\gcd (a, b) = 1$ and $a ∣ b c$ then $a ∣ c$ ; and if $a ∣ c$ , $b ∣ c$ and $\gcd (a, b) = 1$ then $a b ∣ c$ ).

[L6]

$u ∣ v$ means $v = u k$ for some $k$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ ).

[L7]

If $x z = y z$ and $z \neq 0$ then $x = y$ ; and a product of nonzero integers is nonzero (The integers have no zero divisors; multiplicative cancellation).

[L8]

$\gcd (1071, 462) = 21$ ( $\gcd (1071, 462) = 21$ by the Euclidean algorithm, with the back-substitution giving $21 = 7 \cdot 462 - 3 \cdot 1071$ ).

Verification

technique · direct

1.1

Since $a \neq 0$ and $b \neq 0$ we have $(a, b) \neq (0, 0)$ , so $d \geq 1$ and in particular $d \neq 0$ ; fix $a^{'}, b^{'}$ with $a = d a^{'}$ , $b = d b^{'}$ and $\gcd (a^{'}, b^{'}) = 1$ . Both $a^{'}$ and $b^{'}$ are nonzero, since $d a^{'} = a \neq 0$ and $d b^{'} = b \neq 0$ .

givenL1L3

2.1

Every pair of the stated form is a solution: for $t \in Z$ , $a (x_{0} + t b^{'}) + b (y_{0} - t a^{'}) = (a x_{0} + b y_{0}) + t (a b^{'} - b a^{'}) = d + t ((d a^{'}) b^{'} - (d b^{'}) a^{'}) = d + t \cdot 0 = d$ .

step 1.1givenL1

2.2

Conversely, let $(x, y)$ satisfy $a x + b y = d$ . Subtracting $a x_{0} + b y_{0} = d$ gives $a (x - x_{0}) = - b (y - y_{0})$ , that is $d a^{'} (x - x_{0}) = - d b^{'} (y - y_{0})$ ; cancelling $d \neq 0$ gives $a^{'} (x - x_{0}) = - b^{'} (y - y_{0})$ .

step 1.1givenL1L7

2.3

The numerical instance. Here $d = 21$ by [L8], and $21 \cdot 51 = 1071$ , $21 \cdot 22 = 462$ , so $a^{'} = 51$ and $b^{'} = 22$ . The pair $(- 3, 7)$ is a solution, since $1071 \cdot (- 3) + 462 \cdot 7 = - 3213 + 3234 = 21$ .

step 1.1L1L8algebra

3.1

Hence $b^{'} ∣ a^{'} (x - x_{0})$ , and $\gcd (b^{'}, a^{'}) = \gcd (a^{'}, b^{'}) = 1$ , so $b^{'} ∣ x - x_{0}$ by [L5]: write $x - x_{0} = t b^{'}$ for some $t \in Z$ .

step 2.2L4L5L6

4.1

Substituting into step 2.2, $a^{'} t b^{'} = - b^{'} (y - y_{0})$ ; cancelling $b^{'} \neq 0$ gives $a^{'} t = - (y - y_{0})$ , that is $y = y_{0} - t a^{'}$ . So $(x, y) = (x_{0} + t b^{'}, y_{0} - t a^{'})$ .

step 1.1step 2.2step 3.1L1L7

5.1

With step 2.1 this shows the solutions are exactly the pairs $(x_{0} + t b^{'}, y_{0} - t a^{'})$ , $t \in Z$ .

step 2.1step 4.1

6.1

Taking $t = 1$ in step 5.1 gives $(- 3 + 22, 7 - 51) = (19, - 44)$ , and directly $1071 \cdot 19 + 462 \cdot (- 44) = 20349 - 20328 = 21$ . So the same equation has at least the two solutions $(- 3, 7)$ and $(19, - 44)$ , and they are distinct because $- 3 \neq 19$ .

step 5.1step 2.3L1algebra∎

Remarks

The family is infinite. Distinct values of $t$ give distinct pairs, since $b^{'} \neq 0$ and $x_{0} + t b^{'} = x_{0} + t^{'} b^{'}$ forces $t = t^{'}$ by cancellation. So a Bézout equation with $a, b$ both nonzero never has a unique solution.
Why the statement is restricted to $a, b$ both nonzero. With $b = 0$ and $a \neq 0$ the quotient $b / d$ is $0$ and the family collapses to $(x_{0}, y_{0} - t a / d)$ , which is still the complete solution set but for a different reason: $x$ is then forced and $y$ is free. The uniform statement above is the one used elsewhere, and the degenerate case is recorded here so that its absence from the claim is deliberate rather than an oversight.

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Sources

Bézout's identity (Wikipedia) (standard reference, not scraped)
Diophantine equation (Wikipedia) (standard reference, not scraped)

Bézout coefficients are not unique: 1071⋅(−3)+462⋅7=211071⋅(−3)+462⋅7=21 and 1071⋅19+462⋅(−44)=211071⋅19+462⋅(−44)=21, and for nonzero a,ba,b every solution has the form (x0+tb/d, y0−ta/d)(x0​+tb/d,y0​−ta/d)