Why is Bézout's identity considered an identity?

Question

In most cases, identities seem to take values and give outputs that are irrelevant to the input data.

example: $(a + b)^2 = a^2 + 2ab + b^2$
$[\cos(x)]^2 + [\sin(x)]^2 = 1$

Yet Bézout's identity only seems to assert the existence of something, not a relation of the above kind.

Am I missing something, maybe a way to look at it?

[I know this isn't like a usual rigorous question, but something worth considering for those who enjoy mathematical literature]

Views even if undecisive would be appreciated.

Answer 1

One way to view Bezout as an identity is to define functions $a_{x,y},\ b_{x,y}\,$ so that

$$ a_{x,y}\ x + b_{x,y}\ y\, =\, 1\ \ \ {\rm for\ all\ coprime}\,\ x,y\in\Bbb Z$$

Such functions implicitly exist in most extended Euclidean algorithm software. Compare

$$\bmod xy\!:\ \ \bar a_{x,y}\,x + \bar b_{x,y}\,y\,\equiv\, 1\ \ \ {\rm for\ all\ coprime}\,\ x,y\in\Bbb Z\qquad\qquad\ \ $$

by using the functions $\ \ \bar a_{x,y} = x^{-1}\bmod y,\,\ \bar b_{x,y} = y^{-1}\bmod x,\, $ see $ $ Inverse Reciprocity.

For Bezout we can put: $\ a_{x,y} = x^{-1}\bmod y,\,\ b_{x,y} = (1-a_{x,y}\,x)/y \ $ for example.

The general Bezout identity (for all $\,x,y\neq 0,0)$ reduces to the above by cancelling $\gcd(x,y)$

Remark $ $ We introduced the functions $\,a,\,b\,$ in order to eliminate the existential quatifiers in $\, \forall\, x,y\ \exists\, a,b\!:\ ax + b y = 1,\,$ which leaves only universal quantifiers remaining, i.e. an "identity". The same idea works generally - see Skolemization.

A well known example is inverses: $\,\forall\, x\ \exists\, y\!:\ xy\, = 1,\,$ where we define the inverse function $\,y_x = x^{-1}\,$ which yields the identity $\,\forall x\!:\ x\,x^{-1} = 1\,$ in groups. Notice the close relationship with the modular example above, where the functions actually are (modular) inverses.

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Answer 2

The Wikipedia article Bezout's identity states:

This article is about Bézout's theorem in arithmetic.

and also

In elementary number theory, Bézout's identity (also called Bézout's lemma) is the following theorem:

Already, the result is called an identity, theorem, and lemma. The terminology used in mathematics is not always precise and depends on context. Usually, if there is an equation, then in some cases, it can be called an identity, especially if it is true for all values of the variables. Given any integers $\,a\,$ and $\,b\,$, then the Bezout equation $\, ax+by=d \,$ is true if $\,d\,$ is the GCD of $\,a\,$ and $\,b\,$ for some integers $\,x\,$ and $\,y\,$ which are not specified, but only known to exist and not unique. They can be computed using the extended Euclidean algorithm which is an efficient, effective algorithm. Perhaps a better name would be Bezout's equation, but I have never seen that in print.

What makes it somewhat peculiar is that, unlike the quadratic equation which has a solution using the quadratic formula, there is no similar formula for the solution of Bezout's equation. It would be a nice result if there was a formula, but even though it is a linear equation, there are two unknowns. At best, if we know the value of $\,x,\,$ then the formula $\,y = (d-ax)/b\,$ gives the value of $\,y,\,$ and similarly if we know the value of $\,y,\,$ then there is the formula $\, x = (d-by)/a\,$ for the value of $\,x.$