How to generate another solution $\in \mathbb N$ of $x^2=2y^2+7$ knowing one. This is from a new book I started reading; it asks whether I can construct another solution for $x^2=2y^2+7$, given that $(x_0,y_0)$ is one. It gives the following diagram as a hint:

However, this diagram only "tells me" that there are no integer solutions ($\neq 0$) for $x^2=2y^2$, and I really don't see how I can use that information. I wrote a little program to find the solutions up to 1000:
x | 3  5  13  27  75  157  437  915
y | 1  3   9  19  53  111  309  647

but cannot work out any patterns (except for all the numbers being odd). Any further hints would be appreciated.
EDIT: The table above is supposedly superfluous. I am really curious as to how the diagram constitutes a hint.
Here's the German excerpt:

IV . Die Basis eines gleichschenkeligen rechtwinkligen Dreiecks sei $x$ und die Schenkellänge sei $y$.
a) [...]
b) Entwickle aus der Zeichnung ein Verfahren, aus einer Lösung $(x,y)$ der Gleichung $x^2-2y^2=7$ eine weitere zu erhalten.
c) [...]

What might be relevant, is that in one of the previous examples it was shown, that if $(x_0,y_0)$ a soltion to $x^2+xy-y^2=1$, others can be obtained with $(y,x+y)$ and $(y-x,x)$, and that the same equation with a $0$ instead of a $1$ has no solutions.
 A: Suppose you have a solution $(x_0, y_0)$ for $x^2 = 2y^2 + 7$.  Now suppose you have another solution $(a_0, b_0)$ for $a^2 = 2b^2 + 1$.  Then because $$\begin{align*} 7 = (7)(1) &= (x^2-2y^2)(a^2-2b^2) \\ &= (ax)^2 - 2(bx)^2 - 2(ay)^2 + 4(by)^2 \\ &= (ax)^2 + 4(ax)(by) + 4(by)^2 - 2(bx)^2 - 2(ay)^2 - 4(bx)(ay) \\ &= (ax+2by)^2 - 2(ay+bx)^2, \end{align*}$$ we see that $$(x_1, y_1) = (a_0 x_0 + 2b_0 y_0, a_0 y_0 + b_0 x_0)$$ is also a solution to $x^2 = 2y^2 + 7$.  Now since $(a_0, b_0) = (3,2)$ is a fairly easy solution of the second identity, we get $$(x_1, y_1) = (3x_0 + 4y_0, 2x_0 + 3y_0)$$ is a solution if $(x_0, y_0)$ is a solution.

What is curious about this result is that it actually also finds additional solutions to the auxiliary equation $a^2 = 2b^2 + 1$, since our derivation did not rely on the fact that we had a $7$.  If that $7$ had been $1$, it still would have held, meaning that we could generate other solutions to $a^2 = 2b^2 + 1$ using the same recursion formula.  Indeed, we find $(a_1, b_1) = (17, 12)$ also works.
Furthermore, we can also see how to generalize the above for equations of the form $$x^2 = Dy^2 + N,$$ for square-free $D$ and positive integers $N$.  I leave it as an exercise for the reader to determine if such a method can be used to generate all integer solutions.
A: given a solution $(x,y)$ of $$ x^2 - 2 y^2 = k, $$ you get a new solution with
$$ (3x+4y, 2x+3y) $$
The matrix
$$
\left(
\begin{array}{rr}
3 & 4 \\
2 & 3
\end{array}
\right)
$$
 and its inverse generate the (oriented) automorphism group of the quadratic form $x^2 - 2 y^2$
It should be noted that you must use two seed solutions to get your entire list, $(3,1)$ and $(3,-1).$ The two infinite sequences of pairs are interlaced in your list
By Cayley-Hamilton, we get a one-variable recurrence, however it applies to every other number;
$$ x_{2n+4} = 6 x_{2n+2} - x_{2n}  $$
$$ x_{2n+5} = 6 x_{2n+3} - x_{2n +1}  $$
$$ y_{2n+4} = 6 y_{2n+2} - y_{2n}  $$
$$ y_{2n+5} = 6 y_{2n+3} - y_{2n +1}  $$
For example, $6 \cdot 13 - 3 = 75,$  $6 \cdot 75 - 13 = 437.$ 
