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I'm having trouble determining u and v using gcd algorithm, stuck on combining reverse of 308 and 273. I'm lost on what to do after laying out d=rk=... Is there a fixed order to what lines to use?

Work:

273u+308v=GCD(308,273)

a=273

b=308

d=7

273=35x7+28

308=273x1+35

35=308+(-273)x1

28=273+(-35)x7

7=35+(-28)x1

7=28x1+7

=28x1+(35+(-28)x1)

=??

JanusP
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3 Answers3

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Once you have $d=7$ (which is the GCD) you're done, as it divides both 308 and 273. Once you have $d$ you can compute $u$ and $v$.

David G. Stork
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The way I would obtain the GCD would be through the Euclidean algorithm therefore y would write: $$308=273(1)+35 \tag{1.}$$ $$\Rightarrow 273=35(7)+28 \tag{2.}$$ $$\Rightarrow 35=28(1)+7 \tag{3.}$$

Once you have the GDC, you use the operations you already have to rewrite the GDC. Therefore: $$7=35-28$$ Then susbsitute 28 with the identity of (2) and repeat the process for 35. That should in the end give you u and v.

Steven Alexis Gregory
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Miguel Angel Andrade Velázquez
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  • repeat substituting identity of (2) or which step;for 35? How did you simplify the statement so as to get individually u and v? – JanusP Apr 04 '20 at 01:55
  • @AdmiralPyro No, susbstitute 35 from (1). You should maybe read up on the euclidean algorithm, this would solve your problem and any other problems where you are asked to give u and v such that x(u)+y(v)=GCD(x,y) https://en.wikipedia.org/wiki/Euclidean_algorithm – Miguel Angel Andrade Velázquez Apr 04 '20 at 01:57
  • Thank you for the link, however I only understood it in terms of "use the 2nd last line from GCD steps as 1st line in euclidean algorithm" and am not clear on rk,rk-1 etc and how to interpret the theory. – JanusP Apr 04 '20 at 02:01
  • @AdmiralPyro When you're trying to obtain the linear combination, you are trying to go "backwards" from the method you used to obtain the GDC. So you start from (3), then subsititute 28 from (2). Then you rearrange and then susbstitute 35 from (1) and so on. So in general you want to go back the way you obtain the GDC to get finally to 308 and 273 – Miguel Angel Andrade Velázquez Apr 04 '20 at 02:23
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I like the method of continued fractions to find the Bezout coefficients

$$ \gcd( 308, 273 ) = ??? $$

$$ \frac{ 308 }{ 273 } = 1 + \frac{ 35 }{ 273 } $$ $$ \frac{ 273 }{ 35 } = 7 + \frac{ 28 }{ 35 } $$ $$ \frac{ 35 }{ 28 } = 1 + \frac{ 7 }{ 28 } $$ $$ \frac{ 28 }{ 7 } = 4 + \frac{ 0 }{ 7 } $$ Simple continued fraction tableau:
$$ \begin{array}{cccccccccc} & & 1 & & 7 & & 1 & & 4 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 8 }{ 7 } & & \frac{ 9 }{ 8 } & & \frac{ 44 }{ 39 } \end{array} $$ $$ $$ $$ 44 \cdot 8 - 39 \cdot 9 = 1 $$

$$ \gcd( 308, 273 ) = 7 $$
$$ 308 \cdot 8 - 273 \cdot 9 = 7 $$

Will Jagy
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    I'm lost on how bezout results in 8 and 9 for u and v respectively, How does that lead to simplified fraction tableau then result in "44⋅8−39⋅9=1"? – JanusP Apr 04 '20 at 01:57
  • @AdmiralPyro In a continued fraction, take two consecutive fractions, called "convergents." The little cross product is either $1$ or $-1.$ For example, $8 \cdot 8 - 9 \cdot 7 = 1.$ Then $9 \cdot 39 - 44 \cdot 8 = -1$ – Will Jagy Apr 04 '20 at 02:01