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July 11th, 2009, 02:59 PM   #1
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Bezout's identity and odd co primes

Good evening,

Let a and b two odd positive integers >1 and e = +1 or -1, is it true that...

...these propositions are equivalent :

1) a and b are co primes
2) there exist two odd positive integers, u<b, v<a such au-bv = 2e
3) there exist two odd positive integers, x<b, y<a such by-ax = 4e

For example

3 and 7 :
7-3 = 4
3.3-7 = 2

3 and 5 :
5-3 =2
3.3-5 = 4

5 and 7 :
7-5 = 2
5.5 - 7.3 = 4

11 and 15 :
15 - 11 = 4
11.7 - 15.5 = 2

7 and 17 :
7.3 - 17 = 4
17.3 - 7.7 = 2

Euzenius
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July 12th, 2009, 04:02 AM   #2
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Re: Bezout's identity and odd co primes

And what about :

2') there exists one and only one pair (u,v)...
3') ... one and only one pair (x,y)...

instead of (2) and (3)

Euzenius
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July 13th, 2009, 07:39 AM   #3
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Re: Bezout's identity and odd co primes

But,

9x3 - 5x5 = 2
9 - 5 = 4

No crossing, I was surely tired...

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July 16th, 2009, 10:41 AM   #4
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Re: Bezout's identity and odd co primes

Hello,

But in fact if p,q are odd positive integers >1, these propositions are equivalent :

1) p and q are co primes
2) there exists only one pair ( s, t), q>s>0, p>t>0 such sp - tq = 1
3) there exist only one pair (u, v), q>u>0, p>v>0 such vq - up = 1

As you can see s+u = q and t+v = p (but the linear system 4x4 in s,t,u,v is linked, determinant of the matrix is zero)

But could you prove that sv - tu = +1 or -1 ? (Or find a counter example ?)

Sorry I don't get any proof...


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