Let un be the nth Fibonacci number. Prove that the Euclidean algorithm takes precisely n steps to prove that gcd(un+1, un) = 1.

There are a few ways you can modify thestatement to make it true. We could prove that it takesn-1steps for alln≥2. Or we couldreindex the Fibonacci sequence sou0= 1,u1= 1, and prove the result for alln≥1. I’ll do thelatter.Proceed by induction onn. For the base case,u1= 1,u2= 2. So the Euclidean algorithm1 = 1·2 + 0takes exactly 1 step to prove gcd(u2,u1) = 1.Now suppose the Euclidean algorithm takesksteps to prove gcd(uk+1,uk) = 1 for somek≥1.Nowuk+2=uk+1+ukand 0<uk<uk+1(sinceuk+1=uk+uk-1). So performing the Euclideanalgorithm to find gcd(uk+2,uk+1), the first step of the algorithm will beuk+2=uk+1+uk.But now this reduces to finding gcd(uk+1,uk), which will take an additionalksteps by the inductionhypothesis. Hence, the Eucliean algorithm will take exactlyk+1 steps to find gcd(uk+2,uk+1).squaresolid

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