Problem 1110

Solution 1. Let $A_{i, j}$ be the entry in the $i^{\text {th }}$ row and the $j^{\text {th }}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j}=A_{i, j}-1$ and $r_{i}=R_{i}-1$. We show that
$$\sum_{i=1}^{n} R_{i} \equiv(n-1)+P \quad\left(\bmod n^{4}\right)$$

Due to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1)+P$ modulo $n^{4}$, whence the conclusion.

By condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence
$$R_{i}=\prod_{j=1}^{n}\left(1+a_{i, j}\right)=1+\sum_{j=1}^{n} a_{i, j}+\sum_{1 \leqslant j_{1}<j_{2} \leqslant n} a_{i, j_{1}} a_{i, j_{2}}+\cdots \equiv 1+\sum_{j=1}^{n} a_{i, j} \equiv 1-n+\sum_{j=1}^{n} A_{i, j} \quad\left(\bmod n^{2}\right)$$
for every index $i$. Using condition (ii), we obtain $R_{i} \equiv 1\left(\bmod n^{2}\right)$, and so $n^{2} \mid r_{i}$.
Therefore, every product of at least two of the $r_{i}$ is divisible by $n^{4}$. Repeating the same argument, we obtain
$$P=\prod_{i=1}^{n} R_{i}=\prod_{i=1}^{n}\left(1+r_{i}\right) \equiv 1+\sum_{i=1}^{n} r_{i} \quad\left(\bmod n^{4}\right)$$
whence
$$\sum_{i=1}^{n} R_{i}=n+\sum_{i=1}^{n} r_{i} \equiv n+(P-1) \quad\left(\bmod n^{4}\right)$$
as desired.

Comment. The original version of the problem statement contained also the condition
(iii) The product of all the numbers in the table is congruent to 1 modulo $n^{4}$.

This condition appears to be superfluous, so it was omitted.
Solution 2. We present a more straightforward (though lengthier) way to establish (1). We also use the notation of $a_{i, j}$.
By condition $(i)$, all the $a_{i, j}$ are divisible by $n$. Therefore, we have
$$\begin{aligned} P=\prod_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv 1+\sum_{(i, j)} a_{i, j} & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\ & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \quad\left(\bmod n^{4}\right), \end{aligned}$$
where the last two sums are taken over all unordered pairs/triples of pairwise different pairs $(i, j)$; such conventions are applied throughout the solution.
Similarly,
$$\sum_{i=1}^{n} R_{i}=\sum_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv n+\sum_{i} \sum_{j} a_{i, j}+\sum_{i} \sum_{j_{1}, j_{2}} a_{i, j_{1}} a_{i, j_{2}}+\sum_{i} \sum_{j_{1}, j_{2}, j_{3}} a_{i, j_{1}} a_{i, j_{2}} a_{i, j_{3}} \quad\left(\bmod n^{4}\right)$$

Therefore,
$$\begin{array}{l} P+(n-1)-\sum_{i} R_{i} \equiv \sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right) \\ i_{1} \neq i_{2}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}}+\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2} \neq i_{3} \neq i_{1}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\ +\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2}=i_{3}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}\left(\bmod n^{4}\right) . \\ \end{array}$$

We show that in fact each of the three sums appearing in the right-hand part of this congruence is divisible by $n^{4}$; this yields (1). Denote those three sums by $\Sigma_{1}, \Sigma_{2}$, and $\Sigma_{3}$ in order of appearance. Recall that by condition (ii) we have
$$\sum_{j} a_{i, j} \equiv 0 \quad\left(\bmod n^{2}\right) \quad \text { for all indices } i \text {. }$$

For every two indices $i_{1}<i_{2}$ we have
$$\sum_{j_{1}} \sum_{j_{2}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \equiv 0 \quad\left(\bmod n^{4}\right),$$
since each of the two factors is divisible by $n^{2}$. Summing over all pairs $\left(i_{1}, i_{2}\right)$ we obtain $n^{4} \mid \Sigma_{1}$.
Similarly, for every three indices $i_{1}<i_{2}<i_{3}$ we have
$$\sum_{j_{1}} \sum_{j_{2}} \sum_{j_{3}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \cdot\left(\sum_{j_{3}} a_{i_{3}, j_{3}}\right)$$
which is divisible even by $n^{6}$. Hence $n^{4} \mid \Sigma_{2}$.
Finally, for every indices $i_{1} \neq i_{2}=i_{3}$ and $j_{2}<j_{3}$ we have
$$a_{i_{2}, j_{2}} \cdot a_{i_{2}, j_{3}} \cdot \sum_{j_{1}} a_{i_{1}, j_{1}} \equiv 0 \quad\left(\bmod n^{4}\right),$$
since the three factors are divisible by $n, n$, and $n^{2}$, respectively. Summing over all 4 -tuples of indices $\left(i_{1}, i_{2}, j_{2}, j_{3}\right)$ we get $n^{4} \mid \Sigma_{3}$.