A2. For every integer $n \geqslant 1$ consider the $n \times n$ table with entry $\left\lfloor\frac{i j}{n+1}\right\rfloor$ at the intersection of row $i$ and column $j$, for every $i=1, \ldots, n$ and $j=1, \ldots, n$. Determine all integers $n \geqslant 1$ for which the sum of the $n^{2}$ entries in the table is equal to $\frac{1}{4} n^{2}(n-1)$.
Answer: All integers $n$ for which $n+1$ is a prime.
Solution 1. First, observe that every pair $x, y$ of real numbers for which the sum $x+y$ is integer satisfies
$$
\lfloor x\rfloor+\lfloor y\rfloor \geqslant x+y-1 .
$$

The inequality is strict if $x$ and $y$ are integers, and it holds with equality otherwise.
We estimate the sum $S$ as follows.
$$
\begin{array}{r}
2 S=\sum_{1 \leqslant i, j \leqslant n}\left(\left\lfloor\frac{i j}{n+1}\right\rfloor+\left\lfloor\frac{i j}{n+1}\right\rfloor\right)=\sum_{1 \leqslant i, j \leqslant n}\left(\left\lfloor\frac{i j}{n+1}\right\rfloor+\left\lfloor\frac{(n+1-i) j}{n+1}\right\rfloor\right) \\
\geqslant \sum_{1<i, j \leqslant n}(j-1)=\frac{(n-1) n^{2}}{2} .
\end{array}
$$

The inequality in the last line follows from (1) by setting $x=i j /(n+1)$ and $y=(n+1-$ i) $j /(n+1)$, so that $x+y=j$ is integral.

Now $S=\frac{1}{4} n^{2}(n-1)$ if and only if the inequality in the last line holds with equality, which means that none of the values $i j /(n+1)$ with $1 \leqslant i, j \leqslant n$ may be integral.

Hence, if $n+1$ is composite with factorisation $n+1=a b$ for $2 \leqslant a, b \leqslant n$, one gets a strict inequality for $i=a$ and $j=b$. If $n+1$ is a prime, then $i j /(n+1)$ is never integral and $S=\frac{1}{4} n^{2}(n-1)$

Solution 2. To simplify the calculation with indices, extend the table by adding a phantom column of index 0 with zero entries (which will not change the sum of the table). Fix a row $i$ with $1 \leqslant i \leqslant n$, and let $d:=\operatorname{gcd}(i, n+1)$ and $k:=(n+1) / d$. For columns $j=0, \ldots, n$, define the remainder $r_{j}:=i j \bmod (n+1)$. We first prove the following
Claim. For every integer $g$ with $1 \leqslant g \leqslant d$, the remainders $r_{j}$ with indices $j$ in the range
$$
(g-1) k \leqslant j \leqslant g k-1
$$
form a permutation of the $k$ numbers $0 \cdot d, 1 \cdot d, 2 \cdot d, \ldots,(k-1) \cdot d$.
Proof. If $r_{f^{\prime}}=r_{j}$ holds for two indices $j^{\prime}$ and $j$ in (2), then $i\left(j^{\prime}-j\right)=0 \bmod (n+1)$, so that $j^{\prime}-j$ is a multiple of $k$; since $\left|j^{\prime}-j\right| \leqslant k-1$, this implies $j^{\prime}=j$. Hence, the $k$ remainders are pairwise distinct. Moreover, each remainder $r_{j}=i j \bmod (n+1)$ is a multiple of $d=\operatorname{god}(i, n+1)$. This proves the claim.
We then have
$$
\sum_{j=0}^{n} r_{j}=\sum_{g=1}^{d} \sum_{\ell=0}^{(n+1) / d-1} \ell d=d^{2} \cdot \frac{1}{2}\left(\frac{n+1}{d}-1\right) \frac{n+1}{d}=\frac{(n+1-d)(n+1)}{2} .
$$

By using (3), compute the sum $S_{i}$ of row $i$ as follows:
$$
\begin{aligned}
S_{i}=\sum_{j=0}^{n}\left\lfloor\frac{i j}{n+1}\right\rfloor= & \sum_{j=0}^{n} \frac{i j-r_{j}}{n+1}=\frac{i}{n+1} \sum_{j=0}^{n} j-\frac{1}{n+1} \sum_{j=0}^{n} r_{j} \\
& =\frac{i}{n+1} \cdot \frac{n(n+1)}{2}-\frac{1}{n+1} \cdot \frac{(n+1-d)(n+1)}{2}=\frac{(i n-n-1+d)}{2} .
\end{aligned}
$$

Equation (4) yields the following lower bound on the row sum $S_{1}$, which holds with equality if and only if $d=\operatorname{gcd}(i, n+1)=1$ :
$$
S_{i} \geqslant \frac{(i n-n-1+1)}{2}=\frac{n(i-1)}{2} .
$$

By summing up the bounds (5) for the rows $i=1, \ldots, n$, we get the following lower bound for the sum of all entries in the table
$$
\sum_{i=1}^{n} S_{i} \geqslant \sum_{i=1}^{n} \frac{n}{2}(i-1)=\frac{n^{2}(n-1)}{4} .
$$

In (6) we have equality if and only if equality holds in (5) for each $i=1, \ldots, n$, which happens if and only if $\operatorname{god}(i, n+1)=1$ for each $i=1, \ldots, n$, which is equivalent to the fact that $n+1$ is a prime. Thus the sum of the table entries is $\frac{1}{4} n^{2}(n-1)$ if and only if $n+1$ is a prime.

Comment. To simplify the answer, in the problem statement one can make a change of variables by introducing $m:=n+1$ and writing everything in terms of $m$. The drawback is that the expression for the sum will then be $\frac{1}{4}(m-1)^{2}(m-2)$ which seems more artificial.