I already got idea of solving gcd with three numbers. But I am wondering how to solve the extended Euclidean algorithm with three, such as:

47x + 64y + 70z = 1

Could anyone give me a hint? Thanks a lot.

  • $\begingroup$ In your example, you could just let $z=0$, and solve $47x+64y=1$. $\endgroup$ – Gerry Myerson Apr 1 '14 at 12:17

Notice that $gcd(x,y,z)=gcd(x,gcd(y,z))$. First we find $a$, $b$ such that $gcd(x,gcd(y,z))=ax+bgcd(y,z)$, then $c$, $d$ such that $gcd(y,z)=cy+dz$. Finally we obtain $gcd(x,y,z)=ax+bcy+bdz$.


This problem can be generalized to finding $x$, $y$, and $z$ in the equation $ax + by + cz = n$ where $a$, $b$, $c$, and $n$ are given and $\gcd(a, b, c) = 1$. If $\gcd(a, b, c) \nmid n$, then there is no solution, otherwise, we can divide both sides by $\gcd(a, b, c)$.

The method to solve for $x$, $y$, and $z$ is essentially the same as solving for $x$ and $y$ given a fixed value of $z$. The equation given is:

$ax + by + cz = n$

This reduces to:

$ax + by = n - cz$

The only restriction on $z$ is that $\gcd(a, b) \mid (n - cz)$.

If we take this equation $\mod b$, then we get:

$ax \equiv n - cz \mod b$

Let $a^{-1} \cdot a \equiv 1 \mod b$. (It is possible to find $a^{-1}$ by using the Extended Euclidean Algorithm on $a$ and $b$).

$a^{-1} \cdot ax \equiv x \equiv a^{-1} \cdot (n - cz) \mod b$

We now have a value for $x$ given $z$. To find the value of $y$ given $z$, all we need to do is solve another equation:

$by = n - ax - cz$

Since $ax \equiv n - cz \mod b$, $n - ax - cz \equiv 0 \mod b$ and $b \mid (n - ax - cz)$. Therefore:

$y = \frac{n - ax - cz}{b}$


If you find $x'$ and $y'$ such that $47x' + 64y' = gcd(47,64)$ then solve $gcd(47,64)\cdot t + 70z = 1$. Then $x = x't$ and $y = y't$. This works since $gcd(47,64,70) = gcd(gcd(47,64),70)$.


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