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A friend and I are in an intro to number theory class at UK and were struggling to prove the theorem that states that for two relatively prime integers $a$ and $b$ there exist integers x and y which satisfy the equation $ax+by=1$. We have now proven the theorem but while attempting to prove it began generating lists of relatively prime pairs with the same difference i.e. $(2,7)$, $(3,8)$, $(4,9)$..., and noticed patterns emerging in the solutions such as for differences of $5$:

(a,b)  (x,y)
(2,7)  (-3,1)
(3,8)  (3,-1)
(4,9)  (-2,1)
(6,10) (2,-1)

We are attempting to find a rule which will quickly generate a solution for any given pair, even of large numbers, not only for the differences of $5$ but for every given difference n. Any suggestions? Have you heard of this before?

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  • $\begingroup$ sorry the part in the middle looks a little confusing let me clarify it should read as for (2,7), (x,y)=(-3,1), for (3,8), (x,y)=(3,-1) so on and so forth $\endgroup$ – Carson Sep 23 '14 at 19:22
  • $\begingroup$ Thank you. That's much better. $\endgroup$ – Carson Sep 24 '14 at 15:21
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Hint: Try extended Euclide algorithm and Bézout coefficients.

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  • $\begingroup$ right I understand this but i do not understand how this goes into predicting based upon a given difference n. I understand the idea of getting a single pair from the extended algorithm but I am personally more interested in the side about the patterns occur in these differences $\endgroup$ – Carson Sep 23 '14 at 19:41

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