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Given two numbers, find the gcd, then use Bézout's algorithm to find $s$ and $t$ such that $as+bt=\\operatorname{gcd}(a,b)$.

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Finally, find all solutions of an equation $\\mod b$.

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Given two numbers, find the gcd, then use Bézout's algorithm to find $s$ and $t$ such that $as+bt=\\operatorname{gcd}(a,b)$. Finally, find all solutions of an equation $\\mod b$.

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Find $d = \\operatorname{gcd}(\\var{a},\\var{b})$, the greatest common divisor of $\\var{a}$ and $\\var{b}$:

\n\n\n\n

$d = \\phantom{{}}$[[0]]

\n\n\n\n

Now find integers $s$ and $t$ such that:

\n\n\n\n

\$\\var{a}s+\\var{b}t = d\$

\n\n\n\n

$s = \\phantom{{}}$[[1]] $, t = \\phantom{{}}$[[2]]$.$

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Find all solutions $x$ of the following congruence:

\n\n\n\n

\$\\var{a}x \\equiv \\var{n} \\mod \\var{b} \$

\n\n\n\n

The least solution $x$ such that $0 \\lt x \\lt \\var{b}$ is: [[0]]

\n\n\n\n

What is the difference between two successive solutions? [[1]]

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#### a)

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On applying the standard method for finding the $\\operatorname{gcd}$ of two numbers we have the following sequence:

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{describegcd(a,b)}

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The last non-zero remainder is $\\var{d}$, so this is the $\\operatorname{gcd}$ of $\\var{a}$ and $\\var{b}$.

\n

Now work backwards through those steps, rearranging them to find the remainders as linear combinations of the other numbers.

\n

When you reach the last line you will have found $s$ and $t$ such that
\$\\var{a}s+\\var{b}t = \\var{d}\$

\n

{describebezout(a,b)}

\n

So $s = \\var{s}$ and $t = \\var{t}$.

\n
\n

\n

### First Step

\n

We would like to find all the solutions $x$ of the equation
\$\\var{a}x \\equiv \\var{n} \\mod \\var{b}.\$

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In order to solve such an equation first determine if there is a solution at all.

\n
\n
##### Lemma
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An equation of the form $ax \\equiv n \\mod b$ has a solution if and only if $\\operatorname{gcd}(a,b)|n$.

\n
\n

So the first task is to find $\\operatorname{gcd}(a,b)$ and see if it divides $n$. If not, there is no solution.

\n

In the first part, we found that $d = \\operatorname{gcd}(\\var{a},\\var{b}) = \\var{d}$.

\n

$\\var{n} = \\var{d} \\times \\var{k}$, so the equation $\\var{a}x \\equiv \\var{n} \\mod \\var{b}$ has a solution.

\n

### Finding a solution

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Note that because $d|n$, then $n=kd$ for some integer $k$.

\n

In order to find a solution of $x$ go back to the first part and use the integers $s$ and $t$ such that

\n

\$\\var{a}s + \\var{b}t = d. \$

\n

If we multiply both sides of this equation by $k$ we find that

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\$\\var{a}ks + \\var{b}kt = kd = n, \$

\n

and by taking both sides modulo $\\var{b}$ we see that

\n

\$\\var{a}ks \\equiv n \\mod \\var{b}. \$

\n

So $x = ks = \\var{k*s}$ is a solution of the equation.

\n

The difference between successive terms is $\\frac{\\var{b}}{\\var{d}} = \\var{diff}$.

\n

{describeLeast}

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 First of all, the biggest number needs to go first: $\\\\operatorname{gcd}(\\\\var{'+a+'},\\\\var{'+b+'}) = \\\\operatorname{gcd}(\\\\var{'+b+'},\\\\var{'+a+'}),$ Finally, $\\\\var{'+b+'}$ divides into $\\\\var{'+a+'}$ exactly $\\\\var{'+a/b+'}$ times: $\\\\var{'+a+'} = \\\\var{'+b+'} \\\\times \\\\var{'+a/b+'}.$ Divide $\\\\var{'+a+'}$ by $\\\\var{'+b+'}$ with remainder $\\\\var{'+(a%b)+'}$: $\\\\var{'+a+'} = \\\\var{'+b+'} \\\\times \\\\var{'+Math.floor(a/b)+'} + \\\\var{'+(a%b)+'},$