How many solutions in the positive integers does the equation $x^2-y^2=4k$ have? How many solutions in the positive integers does the equation $x^2-y^2=4k$ have, given that $k$ is an odd integer? 
I want to say the answer is none, because $4(2k+1)$ is an even times an odd, and the different parity means we cannot find an integer solution to the difference of squares. Is this the right thinking? Thank you! 
 A: $(4w)^2-2^2=16w^2-4=4(4w^2-1)$ So there are clearly an infinite number of solutions.
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$\ds{x^{2} - y^{2} = 4k.\quad}$ $\ds{k}$ is an $\underline{odd\ integer}$.
  $\ds{\quad x, y\ \mbox{are integers}.\quad x, y:\ ?}$.

\begin{align}
\mbox{Lets}\quad &
\left.\begin{array}{rcl}
\ds{x} & \ds{=} & \ds{u + v}
\\
\ds{y} & \ds{=} & \ds{u - v}
\end{array}\right\}
\quad\iff\quad
\left\{\begin{array}{rcl}
\ds{u} & \ds{=} & \ds{x + y \over 2}
\\[2mm]
\ds{v} & \ds{=} & \ds{x - y \over 2}
\end{array}\right.\,,\qquad
\mbox{Then,}\quad \bbx{\ds{uv = k}}
\end{align}

 1. $\ds{\large k\ \underline{is\ prime}}$.

In that case we can take $\ds{u = k}$ and $\ds{v = 1 \implies x = k + 1\ \mbox{and}\ y = k - 1}$. For example: with $\ds{k = 17}$,
$$
18^{2} - 16^{2} = 324 - 256 = 68 = 4\times 17
$$

 2. $\ds{\large k\ \underline{is\ NOT\ prime}}$.

There is, at least, some integer $\ds{\pars{~n\ \mid\ 1 < n < k~}}$ where $\ds{k/n}$ is an integer. We can take $\ds{u = \max\braces{k/n,n}}$ and
$\ds{v = k/u}$. For example, with $\ds{k = 81}$ we can take $\ds{u = 27}$ and $\ds{v = 3}$ such that $\ds{x = 30}$ and $\ds{y = 24}$:
$$
30^{2} - 24^{2} = 900 - 576 = 324 = 4 \times 81
$$

$$\bbx{\quad%
\mbox{The above discussion shows that there is always, at least, one solution for the original}\quad
\\
\quad\mbox{question. In addition, our answer shows a systematic procedure to find a pair}\ \ds{\pars{x,y}}.}
$$
