Timeless Theorems of Mathematics/Bézout's Identity

Bézout's Identity is a theorem of Number Theory and Algebra, which is named after the French mathematician, Étienne Bézout (31 March 1730 – 27 September 1783). The theorem states that the greatest common divisor, ${\displaystyle }$ of the integers, ${\displaystyle a}$ and ${\displaystyle b}$ can be written in the form, ${\displaystyle ax+by=d,}$ where ${\displaystyle x}$ and ${\displaystyle y}$ are integers. Here, ${\displaystyle x}$ and ${\displaystyle y}$ are called Bézout coefficients for ${\displaystyle (a,b)}$.

There are infinite number of pairs of ${\displaystyle (x,y)}$ which satisfies the equation ${\displaystyle ax+by=d,}$. A general formula can be developed to compute pairs as much as you want. To do that, first of all, it's required to calculate one pair of ${\displaystyle (x,y)}$. One simple way to calculate a pair is using the extended Euclidean algorithm.

Once you have one pair ${\displaystyle (x_{0},y_{0}),}$ you can apply the formula: $${\displaystyle (x_{k},y_{k})=(x_{0}-k{\frac {b}{d}},y_{0}+k{\frac {a}{d}})}$$where ${\displaystyle k\in \mathbb {Z} }$, that means ${\displaystyle k}$ is an integer.

Proof: As ${\displaystyle (x,y)=(x_{0},y_{0})}$ satisfies the equation ${\displaystyle ax+by=d,}$ then,

${\displaystyle ax_{0}+by_{0}=d}$

Or, ${\displaystyle {\frac {d(ax_{0}+by_{0})}{d}}=d}$

Or, ${\displaystyle {\frac {dax_{0}+dby_{0}}{d}}=d}$

Or, ${\displaystyle {\frac {dax_{0}-kba+kba+dby_{0}}{d}}=d}$

Or, ${\displaystyle ax_{0}-ak{\frac {b}{d}}+by_{0}+bk{\frac {a}{d}}=d}$

Or, ${\displaystyle a(x_{0}-k{\frac {b}{d}})+b(y_{0}+k{\frac {a}{d}})=d}$

Or, ${\displaystyle a(x_{0}-k{\frac {b}{d}})+b(y_{0}+k{\frac {a}{d}})=ax_{k}+by_{k}}$

Therefore, the coefficients of ${\displaystyle a}$ are equal and the coefficients of ${\displaystyle b}$ are also equal, ${\displaystyle (x_{k},y_{k})=(x_{0}-k{\frac {b}{d}},y_{0}+k{\frac {a}{d}})}$

[Note: The formula only works when ${\displaystyle d\neq 0}$. Also, as ${\displaystyle x\in \mathbb {Z} }$, then ${\displaystyle k\in \mathbb {Z} }$.]

Example: The greatest common divisor of ${\displaystyle a=8}$ and ${\displaystyle b=12}$ is ${\displaystyle \gcd(8,12)=4.}$ According to the identity, there exists integers ${\displaystyle x}$ and ${\displaystyle y}$, so that ${\displaystyle 8x+12y=4}$ ${\displaystyle \Rightarrow 2x+3y=1}$. If you try to solve the equation, you may soon come up with a pair of solutions like ${\displaystyle (x_{0},y_{0})=(-1,1)}$. So, ${\displaystyle (x_{k},y_{k})=(-(1+k{\frac {b}{d}}),1+k{\frac {a}{d}})}$. By using this formula, you may find pairs as much as you want.

Assume ${\displaystyle S=\{ax+by:x,y\in \mathbb {Z} {\text{ and }}ax+by>0\}}$ where ${\displaystyle a}$ and ${\displaystyle b}$ are non zero integers. The set is not an empty set as it contains either ${\displaystyle a}$ or ${\displaystyle -a}$ when ${\displaystyle x=\pm 1}$ and ${\displaystyle y=0}$. Since ${\displaystyle S}$ is not an empty set, by the well-ordering principle, the set has a minimum element ${\displaystyle d=as+bt}$.

The Euclidean division for ${\displaystyle {\frac {a}{d}}}$ may be written as ${\displaystyle a=dq+r}$, where ${\displaystyle q=\left\lfloor {\frac {a}{d}}\right\rfloor }$, ${\displaystyle r=a-d\left\lfloor {\frac {a}{d}}\right\rfloor }$ and ${\displaystyle 0\leq r<d}$.

Here, ${\displaystyle r=a-dq}$ ${\displaystyle =a-q(as+bt)}$ ${\displaystyle =a-qas+qbt}$ ${\displaystyle =a(1-qs)+b(qt)}$

Thus, ${\displaystyle r}$ is in the form ${\displaystyle ax+by}$, and hence ${\displaystyle r\in S\cup \{0\}}$. But, ${\displaystyle 0\leq r<d}$ and ${\displaystyle d}$ is the smallest positive integer in ${\displaystyle S}$. So, ${\displaystyle r}$ must be ${\displaystyle 0}$. Thus, ${\displaystyle d}$ is a divisor of ${\displaystyle a}$. ${\displaystyle q={\frac {a}{d}}>0}$ as the remainder is zero and a is a non zero integer. Similarly, ${\displaystyle d}$ is also a divisor of ${\displaystyle b}$. Therefore, ${\displaystyle d}$ is a common divisor of ${\displaystyle a}$ and ${\displaystyle b}$.

Assume ${\displaystyle c}$ as any common divisor of ${\displaystyle a}$ and ${\displaystyle b}$; ${\displaystyle a=cu}$, ${\displaystyle b=cv}$. Again, ${\displaystyle d=as+bt}$ ${\displaystyle =cus+cvt}$ ${\displaystyle =c(us+vt)}$

Thus, ${\displaystyle c}$ is a divisor of d. Since ${\displaystyle d>0}$ ${\displaystyle \implies c\leq d}$. Therefore, any common divisor of ${\displaystyle a}$ and ${\displaystyle b}$ is less than or equals to ${\displaystyle d}$.

${\displaystyle \therefore d}$ is the same as ${\displaystyle gcd(a,b)}$ and ${\displaystyle d}$ can be expressed as ${\displaystyle ax+by=d}$. [Proved]