Find any two integers that satisfy $19x+47y = 1$

Question

How can I find two integers that satisfy $19x+47y = 1$? Is there some technique to finding 2 numbers? I can't find any 2 numbers in $\mathbb{Z}$ that make this work.

Thank you!

Answer 1

A popular technique is to use the division algorithm.

Notice that

$$47=2 \cdot 19 + 9$$

$$19=2 \cdot 9 +1$$

Now $1$ can be expressed in terms of $19$

$$1=19- 2\cdot 9$$

Also, $$9=47-2\cdot 19$$

Combining both yields,

$$19-2\cdot (47-2\cdot 19)=1$$ $$5\cdot 19-2\cdot 47=1$$

Now notice that, $$lcm(19,47)=19\cdot 47$$ and $19\cdot 47 - 47\cdot 19 =0$

Add $0$ to the above solution yields a general solution

$$(5+47t)\cdot 19-(2+19t)\cdot 47 =0$$

Also for any numbers, $a,b$

We can find solutions for

$$ax+by=k \gcd{(a,b)}$$

Then express $a$ in terms of $b$ $$a=b\cdot q_1+r_1 $$, where $r_1$ is the remainder of $a$ upon division by $b$ and $0\leq r_1\lt b$

Then, similarly express $b$ in terms of $r_1$ $$b=r_1\cdot q_2 + r_2$$.

Continue this until $r_k=\gcd(a,b)$

Answer 2

Consider the continued fraction of $\frac{47}{19}$:

$$ \frac{47}{19}=2+\frac{1}{\frac{19}{9}}=2+\frac{1}{2+\frac{1}{9}}=[2;2,9] $$ truncate it and expand it back: $$ [2;2] = 2+\frac{1}{2} = \frac{5}{2}. $$ We have that $\frac{5}{2}$ and $\frac{47}{19}$ are consecutive convergents of the same continued fraction, hence their difference is $\pm\frac{1}{2\cdot 19}$.
Indeed $$ \frac{47}{19}-\frac{5}{2}=\frac{94-95}{38}=\frac{-1}{38} $$ leads to $$ 47\cdot 2 - 5\cdot 19 = -1 $$ $$ 47\cdot(-2) + 19\cdot 5 = 1 $$ $$ 47(19-2)+19(5-47) = 1 $$ $$ \color{red}{47\cdot 17 - 19\cdot 42 = 1.}$$

Answer 3

Since $\gcd(19,47)=1$ then the Bezout Identity assures that your equation can be solved in $\mathbb Z$.

Of course if $(a,b)$ is a solution, then all the couples $(a+k47,\, b-k19)$ are solutions as well.

$(a,b)$ are given by the Extended Euclidean Algorithm.