## Task 6/82

Given the inequality $|x|+|y| \leq n$ with $n \in N, x ; y \in G$ (where $G$ denotes the set of integers). Determine the number of ordered solution pairs $(x ; y)$.
We replace the inequalities $|x|+|y| \leq n$ with $n \in \mathbb{N}, x ; y \in G$ by the set of all equations $|x|+|y|=k$ with $k \in \mathbb{N}, 0 \leq k \leq n, x ; y \in G$.

Obviously, a pair $(x ; y)$ is a solution to the given inequality if and only if it is a solution to one of these equations. The problem is thus reduced to determining the number of solutions each of these equations has.

If $k=0$, then the pair $(x=0 ; y=0)$ is the only solution. If $0<k \leq n$, we distinguish two cases:

1. One of the pair elements is zero. Then the other pair element is $\pm k$. In this case, there are thus the four solutions $(x=0 ; y= \pm k),(x= \pm k ; y=0)$.
2. Neither of the pair elements is zero. Then one pair element can take the values $\pm 1 ; \pm 2 ; \pm 3 ; \ldots ; \pm(k-1)$, and the other the values $\pm(k-1) ; \pm(k-2) ; \ldots ; \pm 1$. Since there are four possible combinations of signs, this results in $4(k-1)$ solutions.

For $0<k \leq 11$, there are thus $4+4(k-1)=4 k$ solutions for each $k$. By summing, the total number of solutions is

$$
1+\sum_{k=1}^{n} 4 k=1+4 \sum_{k=1}^{n} k=1+4 \frac{(n+1) n}{2}=n^{2}+(n+1)^{2}
$$