Largest value of greatest common divisor

Let $M$ be the largest value of $U_n=\gcd\left(n!+101, (n+1)!+101\right)$ .

It is known that $\gcd(a,b)=\gcd(a,b+xa)$ for some integer $x$ . Using this property repeatedly, we have $U_n=\gcd\left(n!+101, 101n\right)$ .

When $n=101$ , $101n=101^2$ and $n!+101=101!+101=101(100!+1)$ . Since 101 is a prime number, $100! \equiv -1 \pmod{101}$ by Wilson's Theorem. So $100!=101m-1$ for some positive integer $m$ . This means that $101!+101$ is a multiple of $101^2$ . Thus $U_{101}=101^2=10201$ .

We claim that $M=10201$ .

If $n<101$ , then $101n<101^2$ and hence $U_n<101^2$ .

If $n>101$ , then $\frac{n!}{101}=kn$ for some positive integer $k$ . Hence $U_n=\gcd\left(101\left(\frac{n!}{101}+1\right),101n\right)=\gcd(101(kn+1),101n)=101$ .

This completes the proof.