Suppose you are allowed to make one mistake, that is, one pair of adjacent cells can have the same numeral. What is the minimum number of different numerals required to fill a 5×5 matrix?

For a $5 \times 5$ matrix with one allowed mistake (one pair of adjacent cells can have the same numeral), we want the minimum number of different numerals.

---

Step 1: Understanding the relaxed constraint

Allowing one mistake means exactly one adjacent pair can share the same numeral. In graph terms, this removes one edge from the adjacency graph.

---

Step 2: Original chromatic number

Without mistakes, we showed earlier that:

• Every $2 \times 2$ block forms a $4$-clique.
• Hence the chromatic number is at least $4$, and a $4$-coloring exists.

---

Step 3: Effect of removing one edge

If we pick a single $2 \times 2$ clique:

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix}$$

and allow $a$ and $b$ to share a numeral, that specific $4$-clique can now be $3$-colored.

But the grid contains many other $2 \times 2$ cliques, disjoint from this one. Those untouched cliques still require $4$ colors.

---

Step 4: Global analysis

Because the $5 \times 5$ grid is dense with overlapping $2 \times 2$ cliques, breaking just one edge cannot eliminate the $4$-clique requirement everywhere. At best, it helps locally, but globally there remain other $2 \times 2$ squares that still demand $4$ numerals.

---

Conclusion

The chromatic number of the adjacency graph does not drop with just one allowed mistake.

Thus, even with one mistake allowed, the minimum number of different numerals required for a $5 \times 5$ matrix is

$\boxed{4}$