# How to find the numbers of Bezout identity for two numbers

I'm having troubles finding two numbers a,b such that $288a+177b=3=gcd(177,288) (1)$

I've been writing the equations of the Euclids algorithm one over another many times to get any pair that verify (1). But, I don't get this yet. I'm trying to solve $288x + 177y = 69$ I understand very well the theorem. But I really, really need help finding the particular solution. If someone can explain me any method, or give me an advice to find a pair (a,b), I would really appreciate it. Thanks for reading,

Greetings!

This is how I do it in my notebook when I have to:

You first have to use the euclidean algorithm to find the greatest common divisor:

$288=(1)177+111$

$177=(1)111+66$

$111=(1)66+45$

$66=(1)45+21$

$45=(2)21+3$

$21=(7)3$ Hence the GCD is $3$. the next step is to pass all of the equations except the last so that the small number is alone:

$111=288-(1)177$

$66=177-(1)111$

$45=111-(1)66$

$21=66-1(45)$

$3=45-2(21)$.

Once I have this I start with the equation $3=45-2(21)$ and I go one by one substituting the smallest factor with the adequate equation and then simplifying:

$3=45-2(21)$

$3=45-2(66-(1)45)=3(45)-2(66)=$

$3(111-(1)66)-2(66)=3(111)-5(66)=$

$3(111)-5(177-111)=8(111)-5(177)=$

$8(288-177)-5(177)=8(288)-13(177)$

Suppose $288 a_1 + 177 b_1 = g$ where $\gcd(288, 177)$. The first step in Euclid's algorithm says: you take $288 = 177 + 111$, and $\gcd(288,177) = \gcd(111, 177)$. Corresponding to this, $288 a_1 + 177 b_1 = 111 a_1 + 177 (a_1 + b_1)$, so $111 a_2 + 177 b_2 = \gcd(111,177)$ where $b_2 = a_1 + b_1$. Each step of Euclid's algorithm gives you an equation in the $a$'s and $b$'s. Keep track of these equations. At the end, you'll have one that's easy to solve, and then you work backwards to get $a_1$ and $b_1$.