Solution to equation $4n+1 = (2k+1)^2-(2u)^2$ over the natural numbers(!) Let $n \in \mathbb{N}$ be given, then I want to know if there are $k ,u \in \mathbb{N}$ (with $k \neq n, u \neq n$) such that $4n+1 = (2k+1)^2-(2u)^2$ holds. 
What makes it difficult for me is the fact that I am looking for solutions over the natural numbers. Otherwise one would see that there are infinitely many solutions (for example by completing the square).

Given

$$
4 n + 1 = \Big( 2 k + 1 \Big)^2 - \Big( 2 u \Big)^2.
$$

Set

$$
k = u + a,
$$

then we get

$$
n = a^2 + a + \big( 2 a + 1 ) u.
$$

The case $a=0$ is the solution

$$
n = u = k,
$$

which is to be excluded.
For every $a > 0$ we have a linear relation between $n$ and $u$.

$$
\begin{array}{l|ll}
a & n\\
\hline
1 & 2 + 3u\\
2 & 6 + 5u\\
3 & 12 + 7u\\
4 & 20 + 9u & = 2 + 9(u + 2)\\
5 & 30 + 11u\\
6 & 42 + 13u\\
7 & 56 + 15u & = 2 + 3 ( 5 u + 18 ) = 6 + 5(3 u + 10)\\
\vdots & \vdots
\end{array}
$$

The equation is not general solvable for a given $n$.

In case $n$ is even, we can find solutions for:

$$
n = 8, 14, 16, 20, 26 (3\times),32, 36, 38 (2\times),40, 44, 46, 50, 52, 54, 56 (3\times),62,\\
66, 68 (2\times),74 (2\times),76, 82, 86 (2\times),92, 94, 96 (3\times), \cdots
$$

For example

$$
\begin{eqnarray}
4 \times 26 + 1 &=& \big( 2 \times 9 + 1 \big)^2 - \big( 2 \times 8 \big)^2\\
&=& \big( 2 \times 6 + 1 \big)^2 - \big( 2 \times 4 \big)^2\\
&=& \big( 2 \times 5 + 1 \big)^2 - \big( 2 \times 2 \big)^2
\end{eqnarray}
$$

