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What are the integral solutions of $239x-111y=1$

My Try:

$239x-111y=1$

$239x\equiv(1$ mod $111)$

$17x\equiv(13$ mod $111)$

$17(13x)\equiv(13$ mod $111)$

$x\equiv(98$ mod $11)$

So, $x=98$

Is my above attempt correct? Are there any better ways to solve this?

Andrés E. Caicedo
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user982787
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    The standard solution technique is the extended Euclidean algorithm, and there are plenty of other solutions beyond $(98, 211)$. –  Dec 15 '18 at 02:32
  • https://math.stackexchange.com/questions/407478/solving-a-linear-congruence and https://math.stackexchange.com/questions/499939/solve-a-congruence-linear-equation – lab bhattacharjee Dec 15 '18 at 03:07
  • The standard way to write it is $239x\equiv 1\mod 111$ or $230x\equiv 1\pmod {111}.$ That is, 239x\equiv 1 \mod 111 or 239x \equiv 1 \pmod {111}. – DanielWainfleet Dec 15 '18 at 03:43

2 Answers2

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You did find a solution, but the part you're missing is that there are infinitely many such solutions. This is a linear Diophantine equation of two variables, and there's some good information on Wikipedia about this. The most important part is:

$$ax+by= c,\quad \text{(all integers)}$$ [The above equation] has a solution (where $x$ and $y$ are integers) if and only if $c$ is a multiple of the greatest common divisor of $a$ and $b$. Moreover, if $(x, y)$ is a solution, then the other solutions have the form $(x + kv, y − ku)$, where $k$ is an arbitrary integer, and $u$ and $v$ are the quotients of $a$ and $b$ (respectively) by the greatest common divisor of $a$ and $b$.

So, given a single solution, you can find all other solutions to the equation. The usual approach is to use Euclid's algorithm to find $x$ and $y$ such that $\gcd(a, b) = a\cdot x + b\cdot y$, then scale appropriately (i.e. multiply the whole equation by $\frac{c}{\gcd(a, b)}$.

However, you've already found a solution, so let's skip that part. Once you have a single solution $(x_0, y_0)$, you need to describe the solution set. This is:

\begin{equation} \left\{\left(x_0 + \frac{bn}{d}, y_0 + \frac{an}{d}\right) : n \in \mathbb{Z}, d = \gcd(a, b)\right\} \end{equation}

For your case, $a$ and $b$ are coprime, so $d=1$.

apnorton
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  • To the proposer: Suppose $ab\ne 0$ and $ \gcd( a,b)=1$ and $ax+by=1.$....(I). Clearly $a(x+nb)+b(y-na)=1$ for any $n\in \Bbb Z.$ ...(II). If $ax'+by'=1$ then $0=(x+by)-(x'+by')=a(x-x')+b(y-y')$. So $a|b(y-y')$ and $b|a(x-x'),$ and since $\gcd (a,b)=1,$ this implies $a|(y-y')$ and $b|(x-x').$ So let $y'=y+ma$ and $x'=x+nb.$ So $0=a(x-x')+b(y-y')=anb+bma,$ so $m=-n$.... Hence $ {(x+nb,y-na): n\in \Bbb Z}={(x',y'): ax'+by'=1}$ . – DanielWainfleet Dec 15 '18 at 03:37
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Similar to above, but names some important ideas.

The GCD of any two integers can be expressed as a linear combination if those integers, and that in infinite ways.

239 and 111 are relatively prime. The Extended Euclidean Algorithm can be used to find those infinite solutions.

TurlocTheRed
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