Is there an integer $m$ for which $100+m \cdot 101$ and $101-m \cdot 100$ are not coprime?
Solution. Let $a=100+101 m$ and $b=101-100 m$, and let the greatest common divisor of $a$ and $b$ be $d$. If $d \mid a$ and $d \mid b$, then $d \mid 100 a+101 b$.

$$
100 a+101 b=100^{2}+100 \cdot 101 m+101^{2}-101 \cdot 100 m=100^{2}+101^{2}=20201
$$

20201 is a prime number, so if $a$ and $b$ have a proper divisor greater than 1, it can only be 20201.

From the divisibility $d \mid a$ and $d \mid b$, it also follows that $d \mid a+b$, so $d \mid 100+101 m+101-100 m=201+m$. This, however, implies $m=k \cdot d-201$ (where $k$ is an integer). If $d=20201$, then $m=k \cdot 20201-201$.

For integers of the form $m=k \cdot 20201-201$,

$$
\begin{aligned}
& a=100+101(20201 k-201)=100+101 \cdot 20201 k-101 \cdot 201=20201(101 k-1) \\
& b=101-100(20201 k-201)=101+100 \cdot 20201 k+100 \cdot 201=20201(1-100 k)
\end{aligned}
$$

In this case, both $a$ and $b$ are multiples of 20201, so this is their greatest common divisor, meaning they are not relatively prime.

Note. Let's observe that we have determined all $m$ for which the greatest common divisor of $100+101 m$ and $101-100 m$ is 20201. (If $d=-20201$, we also get the same values for $m$.) For all other $m$, (100+ $101 m ; 101-100 m)=1$.