Let $f(N,x)$ is $N$ repeated $x$ times. So $f(123,2)=123123$ and
$f(123,3)=123123123$
In the problem we are given $a$ and $b$ and $N$ and we need to calculate $\gcd(f(N,a),f(N,b))$.
How do we prove that $\gcd(f(N,a),f(N,b))=f(N,\gcd(a,b))$?
It has been proved that $\gcd(f(N,a),f(N,b))=\gcd(f(N,a-b),f(N,b))$.
Well, the 2nd one gives you the solution, I think.
$\gcd(f(N,a),f(N,b))=\gcd(f(N,a-b),f(N,b))$
If you continue writing/applying it (i.e. if you keep descending), you will end up with $\gcd(f(N,d),f(N,d))$ where this $d$ is the $\gcd(a,b)$. So this 2nd equation basically repeats the algorithm for finding the $\gcd(a,b)$.
Example:
$\gcd(f(N,24),f(N,44))=\gcd(f(N,20),f(N,24))=\gcd(f(N,4),f(N,20))=\gcd(f(N,4),f(N,16))=\gcd(f(N,4),f(N,12))=\gcd(f(N,4),f(N,8))=\gcd(f(N,4),f(N,4)) = f(N,4) = f(N,gcd(24,44))$
It's not a surprise that $4=gcd(24,44)$.