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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.

Ethan Bolker
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John_Nash
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    The term identity is often used to describe an equality. For example, Euler's identity. – copper.hat Jun 23 '19 at 18:45
  • People are quirky and inconsistent in their language use. That's really all that can be said. I personally don't like the term "Bezout's Identity" for the exact reasons you state so I'd tell the world to stop using it. But apparently the world doesn't seem to revolve around me so ... no one pays attention when I do. – fleablood Jun 24 '19 at 00:11

2 Answers2

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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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Bill Dubuque
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  • This needs something extra I think. Namely a specification of the representative for $x^{-1} \mod y$?

    For example, take $x = 2, y = 3$. Then $2^{-1} \equiv 2 \mod 3$ and $3^{-1} \equiv 1 \mod 2$. But $2 \cdot 2 + 1 \cdot 3 = 7$. Maybe I misunderstand your notation? Do you mean that there exists a representative such that the identity holds?

     – Ruben Jun 23 '19 at 18:59
  • I see you've made an edit. This makes more sense. – Ruben Jun 23 '19 at 19:00
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    @RubenduBurck Yes, was editing it while you were replying – Bill Dubuque Jun 23 '19 at 19:01
  • @BillDubuque is this the usual modulo notation, a = b (mod n) implies n divides a-b ? if so, what does x(x inverse mod y) mean, just finished high school, would appreciate the extra time of yours I'm taking – John_Nash Jun 23 '19 at 19:05
  • @John_Nash Yes, but I also use the binary operation mod (= remainder) – Bill Dubuque Jun 23 '19 at 19:17
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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 $\,x.$

Somos
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  • I agree that an explicit formula is missing. That is what makes it useful to Cryptography from what I've read. – John_Nash Jun 23 '19 at 19:18
  • @John There are in fact formulas for the Bezout coefficients (in terms of modular inverses) - see the edit to my answer. – Bill Dubuque Jun 23 '19 at 19:57
  • @BillDubuque Well, how is the modular inverse computed? My understanding is that Euclidean Algorithm is used as in Bezout's identity, and so it is a circular formula. – Somos Jun 23 '19 at 20:28
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    @Somos No,there are plenty of other ways to compute inverses without (explicitly) compting the Bezout identity, e.g. by Gauss's algorithm, or little Fermat and repeated squaring, etc. But we can easily compute one from the other, so they are equivalent in that sense. – Bill Dubuque Jun 23 '19 at 20:33
  • @BillDubuque Thanks for that comment! I just did not immediately think about the other ways to compute inverses because I depend on CAS to do the calculations, although I knew about some of them. – Somos Jun 23 '19 at 20:49