Does there exists more than one set of numbers $(m,n)$ so that $132m+34n=\gcd(132,34)$?
I was comparing to, and thinking about how and why euclids algorithm solves for one set.
How might one find the other sets?
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1When you say $m132$ do you mean $m\cdot132$ or the digit $m$ followed by $132$, so, say, $m=2$ gives $2132$? – Dan Uznanski Nov 29 '16 at 23:45
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Are you asking how to find all solutions once you have found one solution? – Improve Nov 29 '16 at 23:46
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i meant m⋅132. I'm asking, how you can find any solution other than the one via the extended euclidean algorithm – Estalito Nov 29 '16 at 23:50
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suppose you have found $m,n$ that satisfy the equality.
notice that the other solutions can be found via
$$\left(m+k\frac{34}{\gcd(132,34)}\right)132+\left( n-k\frac{132}{\gcd(132,34)}\right)34= \gcd(132,34)$$
Edit:
It is clear that solutions are those forms are indeed a solution. However, I have not justified I have covered all the solutions.
suppose $$132m+34n=132m'+34n'$$ $$34(n-n')=132(m'-m)$$ $$\frac{34}{\gcd(132,34)}(n-n')=\frac{132}{\gcd(132,34)}(m'-m)$$
Since $\frac{34}{\gcd(132,34)}$ and $\frac{132}{\gcd(132,34)}$ are both coprime,
$$\exists k \in \mathbb{Z}, m'-m=\frac{34}{\gcd(132,34)}k$$ As a result, we also have
$$n-n'=\frac{132}{\gcd(132,34)}k$$
Hence all the solutions satisfy
$$m'=m+\frac{34}{\gcd(132,34)}k, n'=n-k\frac{132}{\gcd(34,132)}$$
Siong Thye Goh
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5Perhaps it is helpful to the reader to remark why these are all the solutions. – Improve Nov 29 '16 at 23:49
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Hint $\ $ Since $\,ax+by\,$ is linear in $\,x,y,\,$ the general solution of $\ ax+by=c\ $ is the sum of any particular solution $\,(x_0,y_0)\,$ plus the general solution solution of the associated homogeneous equation $\,ax+by = 0,\,$ which is $\,(x,y)= n(-\bar b,\bar a),\,$ where $\,\bar a/\bar b\,$ is the lowest terms rep of $\,a/b,\,$ i.e. where $\,\bar a,\bar b\,$ are coprime. Summing these two components gives the general solution
$$\,(x,y)\, =\, (x_0,y_0)+n(-\bar b,\bar a)\, =\, (x_0-n\bar b,\,y_0 + n\bar a)$$
Bill Dubuque
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