find integers n m such that

Question

find integers m and n such that gcd(258,180 ) = 258m + 180n

so far I have

258 = (1) 258 + (0) 180

180 = (0) 258 + (1) 180

78 = (1) 258 - (1) 180

-54 = (1)180 - 3(78)

however, im abit stuck once I get to this point

Answer 1

You don't have to but it's a good idea to avoid negative values.

$258 = 1*180 + 78$

$180 = 2*78 + 24$

$78 = 3*24 + 6$

$24 = 4*6 + NOTHING$

So that is as far we can go. $\gcd(180,258) = 6$-- as $6|180$ and $6|258$ but nothing larger than $6$ divides both.

So we want $258m + 180n = 6$

And we have $6 = 78 -3*24$.

But $24 = 180 - 2*78$ so $6 = 78 -3(180 - 2*78)$.

And $78 = 258 - 180$.

So $6= (258-180) - 3(180 - 2*(258-180))$.

And that's it.

$6 = (258 - 180) - (3*180 -6(258 - 180))=$

$258 - 180 - 3*180 + 6(258 -180) =$

$258 - 180 -3*180 + 6*258 - 6*180 =$

$7*258 - 10*180$

So $m=7$ and $n=-10$.

.......

Oh I forget to mention that $m=7$ and $n=-10$ are not that only such integers of course.

$180 = 6*30$ and $258 = 6*43$

And $6=258*7 +180*(-10) =$

$258*7 + 180*(-10) + 0 =$

$258*7 + 180*(-10) + (M - M)=$ (for any $M$)

$258*7 + 180*(-10) + (k*6*30*43 - k*6*30*43)=$ (for any integer $k$)

$[258*7 + k*6*30*43] + [180*(-10) - k*6*30*43] = $

$[258*7+258*30k] + [-180*10 - 180*43k] = $

$258(7+30k) + 180(-10 - 43k)$

So $m = 7+30k$ and $n= -10 -43k$ for any integer $k$ will be a solution so there are infinitely many solutions.

Including $m = -23$ and $n=33$

But $m=7$ and $n=-10$ are the "smallest" solution (where $|m-n|$ is least)

Answer 2

Euclidean algorithm:

$258=180+78$

$180=2\times78+24$

$78=3\times24+6$

$24=4\times6$

Therefore,

$78-3\times24=6$

$78-3\times(180-2\times78)=6$

$7\times78-3\times180=6$

$7\times(258-180)-3\times180=6$

$7\times258-10\times180=6$

Answer 3

You started out correctly applying the forward extended Euclidean algorithm but went astray at the 4th line by not using the (positive) remainder. Instead, it should proceed from there as

$$\begin{align} r_4 = 24\, &= \ \ {-}2(258)\ +\ 3(180)\ \ \,[=\, r_2 - 2\, r_3]\\[.2em] {\rm Bezout\ Identity}\ \rightarrow\ r_5\, =\ 6\, &= \ \ \ \ \ 7 (258) -10 (180)\ \ \,[=\, r_3 - 3\, r_4] \\[.2em] r_6\, =\ 0\, &= -30(258)+43(180)\ \ \,[=\, r_4 - 4\, r_5] \end{align}\qquad\qquad\qquad\qquad$$

Remark $ $ You could get back on track by adding your final two equations, but it is usually more efficient to choose $\,q_i\,$ to obtain the least (positive) value $\, r_{i+1} = r_{i-1} - q_i r_i,\,$ which occurs when $\,q_i$ is the quotient $r_{i-1}\div r_i,\,$ so $\,r_{i+1} = r_{i-1}\bmod r_i\,$ as in the Euclidean algorithm. Any value of $\,q_i\,$ will preserve the gcd since $\,(r_i,r_{i+1}) = (r_i,\, r_{i-1} - q_i r_i)= (r_i,r_{i-1})\,$ but the goal is to generate a minimal decreasing (remainder) sequence $\,r_i\,$ in order to optimize the search for the gcd (= least positive linear combination of gcd arguments). In fact generally we can eliminate half of the computations by using least magnitude (signed) remainders, e.g. here.