Find two simple solutions to the equation: $(x-1)^{2}+(x+1)^{2}=y^{2}+1$, where $x$ and $y$ are non-negative integers.

Find three integers $a, b$, and $c$ such that if $(x, y)$ is a solution to the equation, then $(a x+b y, c x+a y)$ is also a solution.

Deduce that the equation has infinitely many solutions.
The simplest solutions are those satisfying: $(x-1)^{2}=1, y=x+1$, but there are only two of them: $(0,1)$ and $(2,3)$. It is from these that one can construct an infinity of other solutions to the Diophantine equation.

If we expand, the equation reads: $y^{2}-2 x^{2}=1$. For $(a x+b y, c x+a y)$ to be a solution when $(x, y)$ is a solution, it is sufficient that for all $(x, y),(c x+a y)^{2}-2(a x+b y)^{2}=y^{2}-2 x^{2}$. By studying the term in $x^{2}$, the term in $x y$, and the term in $y^{2}$ separately, we notice that this is true if we simultaneously have: $c^{2}-2 a 2=-2, 2 a c-4 a b=0, a^{2}-2 b^{2}=1$. The middle condition implies $c=2 b$, in which case the other two conditions are equivalent. Now, $a^{2}-2 b^{2}=1$ is precisely the equation we are studying, for which we know a root other than $(1,0):(3,2)$. If $(x, y)$ is a solution, $(3 x+2 y, 4 x+3 y)$ is also a solution, strictly greater than $(3 x+2 y > x, 4 x+3 y > y)$, which allows us to construct an infinity of solutions: starting from $(2,3)$, we get $(12,17)$, then $(70,99)$, etc... because $12=(3 \times 2)+(2 \times 3), 17=(4 \times 2)+(3 \times 3)$, and we indeed have $17^{2}-\left(2 \times 12^{2}\right)=1$, and so on... We could show that we thus reach all the solutions of the Diophantine equation, but this is not required.