More About GCDs

Proposition 1.1   Suppose $a,b\in\mathbb{Z}$ and $\gcd(a,b)=d$. Then there exists $x,y\in\mathbb{Z}$ such that

$$ax + by = d.$$

I won't give a formal proof of this proposition, though there are many in the literature. Instead I will show you how to find $x$ and $y$ in practice, because that's what you will need to do in order to solve equations like $ax\equiv 1\pmod{n}$.

Example 1.2   Let $a=5$ and $b=7$. The steps of the Euclidean $\gcd$ algorithm are:

$$\underline{7} =1\cdot \underline{5} + \underline{2} \underline{2} = \underline{7} - \underline{5}\hfill$$

$$\underline{5} =2\cdot \underline{2} + \underline{1} \underline{1} = \underline{5} - 2\cdot \underline{2} = 3\cdot\underline{5}-2\cdot\underline{7}\hfill$$

On the right, we have written each partial remainder as a linear combination of $a$ and $b$. In the last step, we write $\gcd(a,b)$ as a linear combination of $a$ and $b$, as desired.

That example wasn't too complicated, next we try a much longer example.

Example 1.3   Let $a=130$ and $b=61$. We have

$$\underline{130} = 2\cdot \underline{61} + \underline{8} \underline{8} = \underline{130}-2\cdot\underline{61}$$

$$\underline{61} = 7\cdot \underline{8} + \underline{5} \underline{5} = -7\cdot\underline{130}+15\cdot\underline{61}$$

$$\underline{8} = 1\cdot \underline{5} + \underline{3} \underline{3} = 8\cdot\underline{130}-17\cdot\underline{61}$$

$$\underline{5} = 1\cdot \underline{3} + \underline{2} \underline{2} =-15\cdot\underline{130}+32\cdot\underline{61}$$

$$\underline{3} = 1\cdot \underline{2} + \underline{1} \underline{1} = 23\cdot\underline{130}-49\cdot\underline{61}$$

Thus $x=130$ and $y=-49$.

Remark 1.4   For our present purposes it will always be sufficient to find one solution to $ax+by=d$. In fact, there are always infinitely many solutions. If $x, y$ is a solution to

$$ax + by = d,$$

then for any $\alpha\in\mathbb{Z}$,

$$a\left(x+\alpha\cdot\frac{b}{d}\right) + b\left(y - \alpha\cdot\frac{a}{d}\right) = d,$$

is also a solution, and all solutions are of the above form for some $\alpha$.

It is also possible to compute $x$ and $y$ using PARI.

? ?bezout
bezout(x,y): gives a 3-dimensional row vector [u,v,d] such that 
             d=gcd(x,y) and u*x+v*y=d.
? bezout(130,61)
%1 = [23, -49, 1]