Large Fibonacci Numbers Part 1

-GCD from two fibonacci numbers is a fibonacci number too. source $gcd({ F_{m}, F_{n} }) = F_{ gcd (m, n)}$ - where $F_{n }$ is $n_{th}$ fibonacci number

-From the question above, m = 100, n = 110, so: $gcd( 100, 110 ) = 10$ $F_{10} = 55$ $gcd({ F_{100}, F_{110} }) = 55$ $1 + 2 + 3 + ... + n = 55$ $\frac{ n \times (n+1) }{2} = 55$ $n = 10$

sorry for bad formatting

All python :)

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from fractions import gcd
import math
gcdd=gcd(354224848179261915075,43566776258854844738105)

def quad(a,b,c):
    array=[]
    d = b**2-4*a*c
    if d < 0:
        return 0
    elif d == 0:
        x1 = -b / (2*a)
        return x1
    else:
        array.append((-b + math.sqrt(d)) / (2*a))
        array.append((-b - math.sqrt(d)) / (2*a))
        for i in array:
            if i>0:
                return int(i)

print quad(1,1,-2*gcdd)

Credit for quad function goes to Organis here whose code I modified because I was to lazy to reuse mine. Inside it, I returned first positive value if there were two solutions. What I did is got gcd (cheap way) and solved equation (n n+1/2=gcdd which is equal to n^2+n-2 gcdd=0) this is why I did 'print quad(1,1,-2*gcdd)'