one can express the gcd of two natural numbers using Pick´s theorem via
$$ gcd(a,b) = a - b - ab + 2 \sum_{k=1}^{a} \lfloor \frac{b}{a} k \rfloor $$
I wonder how to proof the associativity. It becomes a mess.
$$ gcd(gcd(a,b),c) = gcd(a,b) - c - gcd(a,b)c + 2 \sum_{k=1}^{c} \lfloor \frac{gcd(a,b)}{c} k \rfloor $$
$$ gcd(gcd(a,b),c) = [a - b - ab + 2 \sum_{k=1}^{a} \lfloor \frac{b}{a} k \rfloor] - c - [a - b - ab + 2 \sum_{k=1}^{a} \lfloor \frac{b}{a} k \rfloor]c + 2 \sum_{k=1}^{c} \lfloor \frac{a - b - ab + 2 \sum_{j=1}^{a} \lfloor \frac{b}{a} j \rfloor}{c} k \rfloor $$
(some magic happens)
$$ gcd(gcd(a,b),c) = a - gcd(b,c) - a gcd(b,c) + 2 \sum_{k=1}^{gcd(b,c)} \lfloor \frac{a}{gcd(b,c)} k \rfloor $$
$$ gcd(gcd(a,b),c) = gcd(a,gcd(b,c)) $$
Can someone help me out?
What might help are following relations:
$$ a + \sum_{k=1}^{a} \lfloor \frac{b}{a} k \rfloor = b + \sum_{k=1}^{b} \lfloor \frac{a}{b} k \rfloor $$
For natural t: $$ \sum_{k=1}^{ta} \lfloor \frac{b}{a} k \rfloor = t \sum_{k=1}^{a} \lfloor \frac{b}{a} k \rfloor + \frac{abt}{2} (t-1) $$