2

Bezout’s Identity (but discovered earlier, in a restricted form, by Bachet) is the following result.

Theorem: Let $a$ and $b$ be nonzero integers and let $d$ be their greatest common divisor. Then there exist integers $x$ and $y$ such that $$ax+by=d.$$ In addition, the greatest common divisor $d$ is the smallest positive integer that can be written as $ax + by$, and every integer of the form $ax + by$ is a multiple of the greatest common divisor $d$.

Many theorems in elementary number theory are a result of this one (e.g., Euclid's lemma, the Chinese remainder theorem).

I’m looking for interesting or creative applications of Bezout’s Identity to prove number theory results. One example is Catalan’s “complete integer solution” of the 2.2.3 equation $$X_1^2+X_2^2=Y_1^2+Y_2^2+Y_3^2, \tag{1}$$ which he gave as follows.

Choose relatively prime integers $\alpha$ and $\beta$, and integers $u$ and $v$ such that $\beta u - \alpha v = 1$. Then all integer solutions of (1) are given, without repetition, by the formulas \begin{align*} x_1 &= \tfrac{1}{2}(\alpha^2 + \beta^2 + y_3^2)u + \beta \theta, & y_1 &= x_1 + \alpha, \\ x_2 &= \tfrac{1}{2}(\alpha^2 + \beta^2 + y_3^2)v + \alpha \theta, & y_2 &= x_2 - \beta, \end{align*} where $\theta$ is an arbitrary integer.

[cf. Dickson’s History, Vol II, p. 268; orig: Assoc franç., av. sc. 12, 1883, 98–101]

What other clever applications are there?

Kieren MacMillan
  • 7,889
  • 2
  • 28
  • 70
  • 2
    some here [link]: http://s3.amazonaws.com/aops-cdn.artofproblemsolving.com/resources/articles/olympiad-number-theory.pdf – serg_1 Jun 29 '17 at 14:04
  • 2
    The question is a bit broad, since any result using uniqueness of prime factorizations typically is based on Bezout, since Bezout $\Rightarrow$ atoms (irreducibles) are prime, which is the typical key step used in proving uniqueness of factorizations into atoms (see here for a diagram of logical dependencies). So you may as well as for application of the Fundamental Theorem of Arithmetic or UFD property - which are ubiquitous. – Bill Dubuque Jul 08 '17 at 14:07
  • @BillDubuque: Hmmm… I see what you mean. I guess I'm asking for examples of proofs that explicitly use the property, i.e. they say something like "since $x$ and $y$ are co-prime, there exist $a$ and $b$ such that $ax-by=1$" and go from there (like the Catalan example I gave). – Kieren MacMillan Jul 09 '17 at 13:07
  • @Kieren But you get that also by taking any proof that uses uniqueness and - instead of invoking the uniqueness theorem by name - substituting inline its Bezout-based proof (so then the proof "explicitly" uses Bezout). – Bill Dubuque Jul 09 '17 at 15:28
  • 1
    @BillDubuque: Can you give me an example of such a proof, so I can try doing what you say here? Thanks. – Kieren MacMillan Jul 09 '17 at 23:33
  • Does this answer your question? What is the importance of Bézout's identity? – tryst with freedom Mar 06 '22 at 21:22
  • @Buraian: No, unfortunately it doesn't. I know the fundamentals of why it's important — what I want is concrete examples of its application. – Kieren MacMillan Mar 06 '22 at 23:55

2 Answers2

2

Lemma: If $p$ is a prime number and $k$ is a positive integer not equal to $p$, then $\displaystyle p \mid \dbinom{p}{k}$.

Proof: For $k>p$, $\dbinom{p}{k} = 0$. Hence $p \mid \dbinom{p}{k}$.

For $0<k<p$, we have,

\begin{align*} 1&=ap+bk \ ; \ a, b \in \mathbb{Z} \\ \implies\dbinom{p}{k} &= (ap+bk)\dbinom{p}{k} \\ &= ap\dbinom{p}{k} + bk\dbinom{p}{k} \\ &= ap\dbinom{p}{k} + bp\dbinom{p-1}{k-1} \ \left(\because \dbinom{n}{r} = \dfrac{n}{r} \dbinom{n-1}{r-1}\right) \\ &= p\left(a\dbinom{p}{k} + b\dbinom{p-1}{k-1}\right) \\ \implies p&\mid \dbinom{p}{k} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \square \end{align*}

This can be extended for composite $n$ with $\gcd (n,k) = d$ to show that $n \mid d \dbinom{n}{k}$.

MathGod
  • 5,458
1

Here’s one: Dickson proved that all primitive integral solutions of the equation $$x^2-my^2=zw$$ are given by \begin{align} z &= el^2 + 2flq + gq^2, \\ w &= en^2 - 2fnr + gr^2, \\ x &= \pm (eln+fnq-flr-gqr), \\ y &= lr + nq, \end{align} where $f^2-eg=m$. Dickson says “Since $y=l$ and $w=g$ are relatively prime, there exist integers $f$ and $q$ such that $x=fl+gq$”, and then goes on to prove the theorem.

[@BillDubuque: If you stumble back on this post, I’d love for you to discuss it with respect to your comments under the main post. Thanks!]

Kieren MacMillan
  • 7,889
  • 2
  • 28
  • 70