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'Grid Points on Hyperbolas' printed from https://nrich.maths.org/
Harry solved this problem:
I think that a condition is that $k$ is either odd or a multiple of
4. Here's why.
We need to be able to write $k=(y-x)(y+x)$ where $x$ and $y$ are
integers. But $y-x$ and $y+x$ are either both even or both odd, so
either their product is odd or it is a multiple of 4. So it is
certainly necessary for $k$ to be odd or a multiple of 4.
It is also sufficient. If $k$ is odd, then $x=\frac{k-1}{2}$ and
$y=\frac{k+1}{2}$ are integers, and $y^2-x^2=k$. If $k$ is a
multiple of 4, then $x=\frac{k}{4}-1$ and $y=\frac{k}{4}+1$ are
integers, and $y^2-x^2=k$.
Gabriel extended this further by noting
that there are multiple solutions for various values of
$k$
If $k$ is a multiple of $2n-1$ but not of $4n-2$, then
$$
x=\frac{k}{4n-2}-\frac{2n-1}{2}\quad\quad
y=\frac{k}{4n-2}+\frac{2n-1}{2}
$$
If k is a multiple of $4n$, then
$$
x=\frac{k}{4n}-n \quad\quad y=\frac{k}{4n}+n
$$
So, for example, if $k=21=3\times 5\times
7$ then there are solutions
$$
(x, y ) = \left(\frac{105}{6}-\frac{3}{2},
\frac{105}{6}+\frac{3}{2}\right)\mbox{ or
}\left(\frac{105}{10}-\frac{5}{2},
\frac{105}{10}+\frac{5}{2}\right)\mbox{ or
}\left(\frac{105}{14}-\frac{7}{2},
\frac{105}{14}+\frac{7}{2}\right)
$$
We can see the patterns in these fractions.
Simplifying gives us three solutions
$$
(x, y) = (16, 19), (8, 13), (4, 11)
$$
In this case, we can see how we might take
the analysis further by writing $k$ as its prime
factorisation
$$
(y-x)(x+y) = 3\times 5\times 7
$$
From this we can see, for example, $(x+y)$ is either $3,5,7,15,21$
or $105$.