Blogrolling On IV

Now the function $J(1,y)$ really likes to be a complex number. Since it is always of norm 1 when the discriminant is non-positive, it is only real when $J=\pm 1$. Note that $J(1,y)$ may be generalised yet again using the rule

$xn{J}^2+xyJ+n=J$

which is also designed to give a norm 1 complex number at $x=1$. In this case, the condition for $J(1,y)=\pm 1$ is

$y=1\pm Q$

for $n^2=Q^2-N^2$ where $Q$ and $N$ are whole numbers. When $n=1$ this reduces to the previous case of $N=0$, but a general Pythagorean triple gives a solution for $y$. Note that each such choice of $y$ leads to a new Abel series with coefficients somehow depending on $n$ and all of these series sum to $\pm 1$ although the substitution rule is different for each choice of $n$. Other choices for $y$ give roots of unity. For example, when $y=6$ and $n=3$ we find that $J=\frac{1}{6}[-5\pm \sqrt{11}i]$.

Update: Here is the original post on the Everything Seminar, with nice examples of series summing to -1.