Prove that the values differ at least by $n+1$

Question

The integers $1, 2, 3, \ldots, n^2$ are placed without duplication onto an $n \times n$ chessboard, with one integer per square. Show that there exist two (horizontally, vertically, or diagonally) adjacent squares whose values differ by at least $n + 1$.
(We assume $n > 1$.)

Answer 1

We assume $n > 1$. If $(a,b)\in \Bbb [1, n]^2 \cap \Bbb N^2$, $C_{(a, b)}$ denotes the integer placed on the square $(a, b)$ of the $n \times n$ chessboard.

Assume $|C_x - C_y| \leq n$ for all squares $x$, $y$ that are horizontally, vertically, or diagonally adjacent. Pick the square $z$ where $C_z =1$.

A King can move to a square that is horizontally, vertically, or diagonally adjacent to his current square. Therefore a King needs at most $n-1$ moves to move from square $z$ to an arbitrary square $x.$ So $$|C_x - C_z| \leq n(n-1) = n^2 - n$$ obtain $$C_x \leq n^2 - n + 1 \tag {1}$$

Because the largest value placed on the chessboard is $n^2$, we can choose $C_x = n^2$, which contradicts $(1)$.

Answer 2

A possible answer...........

This proof will be a kind of quasi-use of mathematical induction...

Let $x_{i,j}$ represent a point on the board (with i and j both in the range of 1 to n). Notice that the max difference possible between $x_{i,j}$ and n is given by the formula $n^{2}-x_{i,j}$, which can be factored into $$a)\; [(n+\sqrt(x_{i,j}))(n-\sqrt(x_{i,j}))]$$. Given that 1 will always be the smallest value, we can further change this into $$a_i)\;[(n+1)(n-1)]$$. Now, assume n=2 for a base case. Plugging this into the formula given, we have the claim is true. Let m be some arbitrary natural. Assume that the claim is true for m. Now consider $r=m+1$. After plugging this into formula $a_i$, we get that the distance between m+1 and 1 is $(m)(m+2)$. But notice that m is always greater than 1. So we have the proof of the claim.