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I need to prove that $\gcd(a,a+b) = \gcd(a,b)$ in order to prove that two successive fibonacci numbers are coprime. I want to use Bezout's Identity, would this be correct?

Let $ \gcd(a, a+b) = x \ ,$ by Bezout's identity we have: $$ \ ma \ + n(a+b) = x$$ $$(m+n) a + bn = x$$ Now we can say that, at least, $\ x$ divides $a$ and $b$, so $\gcd(a,b) \ge x = \gcd(a,a+b).$

Suppose now that $d = \gcd(a,b) > x = \gcd(a,a+b) $, then we have: $$a = d\lambda_1,\quad b= d\lambda_2$$ Hence: $$\gcd(a,a+b) = \gcd (d\lambda_1, d\lambda_1 +d\lambda_2 ) = d \cdot \gcd(\lambda_1, \lambda_1+ \lambda_2)$$

And because $\gcd(\lambda_1, \lambda_1+\lambda_2) \ge 1$ we get that $d = \gcd(a,b) \gt d \cdot \gcd(\lambda_1,\lambda_1+ \lambda_2)$, a contradiction.

So $\gcd(a,a+b) = \gcd(a,b)$.

Bill Dubuque
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Samir El Karrat Moreno
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    Compare with the proofs at this duplicate. And yes, two successive Fibonacci terms are coprime, see this duplicate. Note that there are several different answers, so you can choose the one you like most. – Dietrich Burde Jan 19 '23 at 22:13
  • As here in the linked dupe. your equalities imply the set equality $,a\Bbb Z + (a+b)\Bbb Z = a\Bbb Z + b\Bbb Z,,$ so the sets have the same least positive elements, i.e. $\gcd(a,a+b) = \gcd(a,b),$ by Bezout. This method will be clearer when you learn about ideals. – Bill Dubuque Jan 19 '23 at 23:08

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