Other interesting consequences of $d=163$?

Solution 1:

Noam Elkies kindly pointed out that in his preprint "Three Lectures on elliptic surfaces and curves of high rank" (page 9), there is a discriminant $-163$ elliptic surface with torsion group $\mathbb{Z}/4\mathbb{Z}$. Explicitly, this is,

$$y^2 + a x y + a b y = x^3 + b x^2\tag{1}$$

where,

$$a = (8t - 1)(32t + 7),\;\;b = 8(t + 1)(15t - 8)(31t - 7)$$

Elkies gave several parametric solutions to (1). After some fiddling around, I found it can also be solved as,

$$x = \frac{-a^2}{8},\;\;y = \frac{a(a^2-8b)}{16},\;\;t = \frac{3(3-2v)}{4(2+v)}$$

and where $v$ (among others) are the roots of the discriminant $-163$ cubic,

$$v^3-6v^2+4v-2=0$$

mentioned in the post. (Though I don't know why it works.)

P.S: Another version of Elkies' K3 surface can be found at http://www.math.rice.edu/~hassett/conferences/Clay2006/Elkies/CMIPelkies.pdf

Solution 2:

$\left(\frac{-163}{n}\right) = \mu(n)$ for squarefree $n \leq 40$, where the LHS is the Kronecker symbol and the RHS is the Mobius function. (Iwaniec & Kowalski pg. 520)

Solution 3:

As pointed out by Greg Martin in the context of the abc conjecture,

$$3^37^211^219^2127^2 163- 2^{12}5^323^329^3=1\tag{1}$$

Note this equality has the form,

$$pr^2-qs^2=1$$

which solves the Pell equation,

$$(p r^2 + q s^2)^2 - p q(2r s)^2 = 1$$

In fact, the fundamental solution to,

$$x^2-(163)(640320)y^2=1$$

is given by the $p,q,r,s$ derivable from $(1)$. See also the related post https://mathoverflow.net/questions/154655/.