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# Help needed with extended Euclidean algorithm (Bézout's identity) watch

1. I suppose this is rather a stupid question, but I'm currently looking at the working for a linear diophantine equation, and I can't understand where certain things come from.

The numbers in question are:

a = 75
b = 34
c = (a,b)

And the working is as follows:

1 = 7 - 1 * 6

= 7 - 1 * ( 34 - 4 * 7 )

= 5 * 7 - 1 * 34

= 5 * ( 75 - 2 * 34 ) - 1 * 34

= 5 * 75 - 11 * 34

It's probably something really simple that I'm missing, but I don't understand how you go from 7 - 1 * ( 34 - 4 * 7 ) to 5 * 7 - 1 * 34 , and from 5 * ( 75 - 2 * 34 ) - 1 * 34 to 5 * 75 - 11 * 34 .

I hope someone can help me.

Thanks
2. (Original post by und)
= 7 - 1 * ( 34 - 4 * 7 )

= 5 * 7 - 1 * 34
It's just a re-arrangement.

and your second query is similar.
3. (Original post by ghostwalker)
It's just a re-arrangement.

and your second query is similar.
Nevermind. I really am being stupid. I don't know why they didn't just write out the proof in letters. It would have been easier to understand.

Anyway, thanks for the help.

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Updated: March 19, 2011
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