Let $n$ be a positive integer, and $x$ and $y$ be non-negative integers. We are to examine the solutions of the following equation in $x$ and $y$

$$x ^ { 2 } - y ^ { 2 } = n . \tag{1}$$

First of all, by transforming (1), we obtain

$$( x + y ) ( x - y ) = n . \tag{2}$$

(1) When we find the solutions $( x , y )$ of (1) in the cases where $n = 8$ and $n = 9$, we have that
if $n = 8$, then $( x , y ) = ( \mathbf { A } , \mathbf { B } )$,
and
if $n = 9$, then $( x , y ) = ( \mathbf { C } , \mathbf { D } ) , ( \mathbf { E } , \mathbf { F } )$.
Note that you should write the solutions in the order such that $\mathbf{C} \leq \mathbf{E}$.

(2) For each of $\mathbf { G }$ $\sim$ $\mathbf { R }$ in the following sentences, choose the correct answer from among (0) $\sim$ (9) given below.

The following is a proof that (3) given below is the necessary and sufficient condition for (1) to have a solution.

Proof: First, suppose that $( x , y )$ satisfies (1).
If $x$ and $y$ are both even or both odd, then both $x + y$ and $x - y$ are $\mathbf { G }$. Hence, by (2) we see that $n$ is a multiple of $\mathbf { H }$.

Next, if one of $x$ and $y$ is even and the other is odd, then both $x + y$ and $x - y$ are $\mathbf{I}$, and hence $n$ is $\mathbf{J}$.

Thus we see that

$$\text{``} n \text{ is a multiple of } \mathbf{H} \text{, or } n \text{ is } \mathbf{J} \text{''} \quad \ldots\ldots (3)$$

is a necessary condition for (1) to have a solution.

Conversely, suppose that $n$ satisfies the condition (3).
If $n$ is a multiple of $\mathbf{H}$, then $n$ can be represented as $n = \mathbf{H} \cdot k$, where $k$ is a positive integer. So, if for example we take $x + y = \mathbf { K } \cdot k$ and $x - y = 2$, then $( x , y ) = ( k + \mathbf { L } , k - \mathbf { M } )$, which shows that (1) has a solution.

On the other hand, if $n$ is $\mathbf{J}$, then $n$ can be represented as $n = \mathbf { N } \ell + \mathbf { O }$, where $\ell$ is a non-negative integer. So, if for example we take $x + y = \mathbf { P } \ell + \mathbf { Q }$ and $x - y = 1$, then $( x , y ) = ( \ell + \mathbf { R } , \ell )$, which shows that (1) has a solution.

From the above, we see that the necessary and sufficient condition for (1) to have a solution is (3).

(0) 0
(1) 1
(2) 2
(3) 3
(4) 4
(5) 5
(6) 6
(7) even
(8) odd
(9) prime