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Bézout's identity — Let $a$ and $b$ be integers with greatest common divisor $d$. Then there exist integers $x$ and $y$ such that $ax + by = d$. More generally, the integers of the form $ax + by$ are exactly the multiples of $d$.

Proof from Wikipedia: https://en.wikipedia.org/wiki/B%C3%A9zout%27s_identity enter image description here

The proof is simple enough, but I was just wondering if anyone had anything more to say about why this theorem is to be expected. Why is a linear combination of two natural numbers a multiple of their greatest common divisor? Can this be understood in terms of the prime factorizations of the two natural numbers? Thank you.

J. W. Tanner
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    In a sense, it cannot be completely understood in terms of the prime factorizations because it holds in contexts where unique factorization fails. See https://en.wikipedia.org/wiki/Bézout_domain – lhf Oct 15 '21 at 16:05
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    For getting a "feel" of what's going on, as opposed to a proof, have you played with the Euclidean Algorithm? I really like the picture with the two sticks that's at the top of that article. – JonathanZ Oct 15 '21 at 16:12
  • The key idea behind the proof is elaborated here in the dupe. – Bill Dubuque May 22 '24 at 17:29

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One thing that I find helpful for building intuition is considering different perspectives until one of them clicks. So here is an alternative look at the matter in case it helps.

One can rearrange Bezout's Identity as: $$ y = \frac{d - ax}{b}, $$ and since $y$ is integer, this means $d-ax$ is divisible by $b$; or equivalently, that adding $d$ to $-ax$ (we will work with $ax'$ instead, where $x'=-x$ is a positive number, which can be assumed without loss of generality given the structure of solutions of Bezout's Identity) makes it divisible by $b$, which in modular arithmetic is expressed as $$ ax' \equiv d \pmod b. $$

In words: it's possible to multiply $a$ by a number $x'$ such that the remainder of the division by $b$ is $d$ (the greatest common divisor of $a$ and $b$). Similarly you can get: $by \equiv d \pmod a$.

How is this meaningful? You cannot get any remainder you want from any division. Think of $a=5$, $b=10$, any remainder of $ax'/b$ is going to be either $5$ or $0$. In a way $0$ is uninteresting as you can always get that by setting $x'=b$, and Bezout's identity is telling you that you can get a $5$ ($d=5$). More generally, in this rearrangement Bezout's Identity can be understood as a statement about the possible remainders of the division $ax'/b$.

To see why Bezout's Identity is not just providing one remainder, note that from the properties for modular arithmetic we have that: given $ax \equiv d \pmod b$ and an integer $k$, we can get: $kax' \equiv kd \pmod b$. That is: if we know the remainder $d$ from dividing $ax'$ by $b$, then we have that any multiple of $d$ ($dk$) is also a remainder. For instance, consider the case of $a=2$, $b=6$, where the possible remainders of the division $2·x'/6$ are $0$, $2$ and $4$. These are all of the form $kd$ ($d=2$).

Juan Martinez
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