Switzerland — Second Round, 2022

Solution:
Assume that the fraction is an integer, write
$$\frac{d^{2}+d+1}{n+1}=m$$
Obviously $m$ is going to be positive, as both $n$ and $d$ are also positive. Since $d$ divides $n$, we can write $n=k d$. Plugging it into the above equation we get
$$d^{2}+d+1=(k d+1) m \Leftrightarrow d^{2}+d-k d m=m-1$$
So $d$ also divides $m-1$, we can write $m=l d+1$ for $l \in \mathbb{Z}, l \geq 0$ ($l \geq 1$ if and only if $m-1>0$). Plugging that in again, gives us
$$d^{2}+d+1=(k d+1)(l d+1)=k l d^{2}+d(k+l)+1 \Leftrightarrow d+1=k l d+k+l$$
As $k \in \mathbb{N}$, we have $k \geq 1$. Now assume $m \neq 1$. From this it would follow that $m-1$ is strictly positive, leading to $l \geq 1$. Under this assumption we would get $k l d \geq d$, leading to
$$k d l+k+l \geq d+k+l \geq d+1+1>d+1=k d l+k+l$$
which is impossible. Contradiction! So we must have $l<1 \Rightarrow l=0$ leading to $m=0 \cdot d+1=1$ as wanted.

Solution:
Write $n=k d$ as before, and we obtain that $(k d+1) \mid\left(d^{2}+d+1\right)$. As $a \mid b$ implies $a \mid b-a$ we get
$$k d+1 \mid d^{2}+d-k d=d \cdot(d+1-k)$$
As $\gcd(k d+1, d)=1$ by Euclid's Algorithm, we also get
$$k d+1 \mid d+1-k$$
This follows as $a \mid b c$ and $\operatorname{gcd}(a, b)=1$ implies $a \mid c$. We do a case distinction on the size of $k$
- If $k<d+1$, we have $d+1-k>0$ which implies
$$k d+1 \leq d+1-k \leq d$$
This is impossible
- If $k>d+1$, we instead have $k-d-1>0$. As $a \mid b$ implies $a \mid-b$ we also have $k d+1 \mid k-d-1$. We get $k d+1 \leq k-d-1 \leq k$ which is also impossible.
- The final case $k=d+1$ gives us $n=d(d+1)$. Then $n+1=d(d+1)+1=d^{2}+d+1$, which means that the fraction is equal to one

Note that in the first two cases we used $a \mid b$ implies $a \leq b$, if both $a, b>0$. This finishes the proof.