Singular Versus Non-Singular Elliptic Curves

I needed to clarify the difference between singular and non-singular elliptic curves. Figure 1 shows an example of a singular elliptic curve and Figure 2 shows an example of a non-singular elliptic curve. They differ only in the constant term.

Figure 1: \(y^2=x^3-3x+2\)

Figure 2:  \(y^2=x^3-3x-2\)

I asked Gemini to clarify:

Understanding Elliptic Curves:
Singular vs. Non-Singular and the Discriminant

The difference between singular and non-singular elliptic curves comes down to two main aspects: their geometric shape (how they look) and their algebraic properties (how their roots behave). To be a formal elliptic curve, the curve must be non-singular, meaning it is completely smooth and has no sharp points or self-intersections. Let's explore exactly why this happens and how the math behind it works.

1. The Geometric Difference: Smooth vs. Broken

An elliptic curve is typically defined by the short Weierstrass equation:

$$y^2 = x^3 + ax + b$$

Visually, the distinction relies on whether you can draw a unique, straight tangent line at every single point on the curve.

• Non-Singular (Smooth): The curve flows perfectly without breaking. Every point on the curve has exactly one clear tangent line.
• Singular (Node): The curve loops back and intersects itself, creating a shape that looks like a little ribbon or a fish. At the exact point where the lines cross (the node), there are two different tangent lines.
• Singular (Cusp): The curve comes to a sharp, pinching point before turning around. At this sharp tip (the cusp), the tangent line is undefined or vertical in a way that breaks the curve's smooth flow.

2. The Algebraic Difference: The Roots and the Discriminant

Mathematically, whether a curve is smooth or singular is determined by the roots of the cubic polynomial on the right side of the equation: \(f(x) = x^3 + ax + b\).

For the curve to be non-singular, \(f(x)\) must have three distinct roots. If any roots are repeated, the curve becomes singular. We measure this using the discriminant (\(\Delta\)), which checks for repeated roots without forcing us to actually factor the polynomial:

$$\Delta = -16(4a^3 + 27b^2)$$

The discriminant categorizes the curve as follows:

1. \(\Delta \neq 0\) (Non-Singular): There are no repeated roots. The curve is valid and smooth.
2. \(\Delta = 0\) (Singular - Node): The polynomial has one root that repeats twice.
3. \(\Delta = 0\) and \(a=b=0\) (Singular - Cusp): The polynomial has one root that repeats three times.

3. Deriving the Discriminant

How do we know that \(4a^3 + 27b^2\) is the exact expression to check for repeated roots? It comes from a core algebraic principle: finding the exact conditions under which a polynomial and its derivative share a common root.

When a polynomial has a repeated root (creating a singular node or cusp), its graph touches the x-axis without crossing it. At that exact touching point, two things are true: the value of the function is zero, and the slope of the tangent line (the derivative) is also zero.

Let’s look strictly at the cubic part of the equation: \(f(x) = x^3 + ax + b\) where \(f(x) \equiv y \) and \( \textbf{not} \)  \(y^2\). To find where the slope is zero, we take the first derivative: \(f'(x) = 3x^2 + a\). If the curve has a repeated root, we set both equal to zero and solve the system of equations.

Step 1: Solve the derivative for \(x^2\)

$$3x^2 + a = 0 \implies x^2 = -\frac{a}{3}$$

Step 2: Substitute this into the original polynomial

We know \(f(x) = 0\). We can rewrite \(x^3\) as \(x(x^2)\) and substitute our finding:

$$x\left(-\frac{a}{3}\right) + ax + b = 0$$

Combine the \(x\) terms:

$$\frac{2a}{3}x + b = 0 \implies x = -\frac{3b}{2a}$$

Step 3: Combine everything to find the condition

Take this new expression for \(x\), square it, and set it equal to our original \(x^2\) equation:

$$\left(-\frac{3b}{2a}\right)^2 = \frac{9b^2}{4a^2}$$

$$\frac{9b^2}{4a^2} = -\frac{a}{3}$$

Cross-multiply to clear the fractions and move everything to one side:

$$27b^2 = -4a^3$$

$$4a^3 + 27b^2 = 0$$

This expression is the core polynomial discriminant. If it equals zero, the curve is singular.

4. The Mystery of the "-16"

In the standard elliptic curve discriminant formula \(\Delta = -16(4a^3 + 27b^2)\), the \(-16\) multiplier often seems mysterious because it doesn't change whether the expression equals zero or not. Its origin lies in the general Weierstrass equation:

$$y^2 + a_1xy + a_3y = x^3 + a_2x^2 + a_4x + a_6$$

When calculating the discriminant for this complex general equation (often used in cryptography over different prime bases), the result is a very long polynomial. When mathematicians apply the algebraic transformations required to squash that long equation down into the short \(y^2 = x^3 + ax + b\) format, the math naturally factors out a \(-16\). This factor is kept in the definition so that the discriminant's properties remain perfectly consistent across all fields of mathematics, regardless of which version of the equation you are using.

5. Why Does This Matter? (The Group Law)

Elliptic curves are famously used in cryptography and number theory. This utility relies on the Group Law, a geometric way to "add" two points on the curve to find a third point. The rule states that a straight line drawn through any two points on an elliptic curve will intersect the curve at exactly one other point, which is then reflected across the x-axis.

If a curve has a node or a cusp, this mathematical machinery breaks down. If your line passes through a self-intersecting node, the math cannot determine a unique intersection point, and the addition fails. Because singular curves cannot sustain this group structure, algorithms naturally reject them to prevent invalid calculations.

Linear Recurrence Revisited

It's been a while since I posted about linear recurrence relations. The only two posts were:

• Solving Linear Recurrence on the 16th of June 2020
  
• More on Linear Recurrence on the 26th December 2020

Today I turned \(28073\) days old and one of the properties of this number is that it's a member of OEIS A105578:$$ \begin{align}  &\text{a}(n+3) = 2\text{a}(n+2) - 3\text{a}(n+1) + 2\text{a}(n)\\ &a(0) = 1, a(1) = 1, a(2) = 0 \end{align}$$This linear recurrence of order 3 corresponds to the cubic equation:$$ \begin{align} &x^3 = 2x^2 -3x +2 \\ &x^3-2x^2+3x-2 = 0 \end{align}$$The polynomial \( \text{P}(x) = x^3-2x^2+3x-2 \) has three roots, one real (at \(x=1\) ) and two complex. See Figure 1.

Figure 1

The roots are:

• \( x_1=1 \)
• \( x_2 = \dfrac{1}{2}(1-i \sqrt{7}) \)
• \( x_3 = \dfrac{1}{2}(1+i \sqrt{7}) \)

These roots can be used to find any term in the sequence because of the following relationship:$$a_n=A(x_1)^n+B(x_2)^n+C(x_3)^n \text{ with A, B and C constants}$$Substituting the initial conditions we get three equations in three unknowns:$$ \begin{align} A + B + C &= 1 \\ Ax_1+Bx_2+Cx_3 &=1 \\ A(x_1)^2+B(x_2)^2+C(x_3)^2 &=0 \end{align} $$In terms of the roots, the values for \( A, B \text{ and } C \) are:$$ \begin{align} A = \frac{x_1 - x_2 x_3 - (x_2 + x_3)}{(x_1 - x_2)(x_1 - x_3)} \\ B = \frac{x_2 - x_1 x_3 - (x_1 + x_3)}{(x_2 - x_1)(x_2 - x_3)} \\ C = \frac{x_3 - x_1 x_2 - (x_1 + x_2)}{(x_3 - x_1)(x_3 - x_2)} \end{align} $$We know that when \(n=32\), the term is 28072 (my diurnal age) so let's check if that works by substituting the values of \( x_1,x_2 \text{ and } x_3 \) into and summarising at the same time what we've found previously:$$
\begin{array}{l}
\textbf{Step 1: Determine the Roots of the Characteristic Equation} \\[6pt]
x^3 - 2x^2 + 3x - 2 = 0 \\[4pt]
(x-1)(x^2 - x + 2) = 0 \\[4pt]
\text{The roots are:} \\[4pt]
x_1 = 1 \\[4pt]
x_2 = \dfrac{1 + i\sqrt{7}}{2} \\[4pt]
x_3 = \dfrac{1 - i\sqrt{7}}{2} \\[18pt]


\textbf{Step 2: Find Constants A, B, and C} \\[6pt]
\text{Using the initial conditions } a_0=1, a_1=1, a_2=0: \\[6pt]
A = \dfrac{1}{2} \\[6pt]
B = \dfrac{1}{4} - \dfrac{i}{4\sqrt{7}} \\[6pt]
C = \dfrac{1}{4} + \dfrac{i}{4\sqrt{7}} \\[18pt]


\textbf{Step 3: Construct the Explicit Formula} \\[6pt]
a_n = A(x_1)^n + B(x_2)^n + C(x_3)^n \\[6pt]
a_n = \dfrac{1}{2}(1)^n + \left(\dfrac{1}{4} - \dfrac{i}{4\sqrt{7}}\right)\left(\dfrac{1 + i\sqrt{7}}{2}\right)^n + \left(\dfrac{1}{4} + \dfrac{i}{4\sqrt{7}}\right)\left(\dfrac{1 - i\sqrt{7}}{2}\right)^n \\[18pt]


\textbf{Step 4: Verify for } n=32 \text{ (Detailed Breakdown)} \\[12pt] \text{We need to sum three terms: } \\[4pt] a_{32} = \underbrace{A(x_1)^{32}}_{\text{Term 1}} + \underbrace{B(x_2)^{32}}_{\text{Term 2}} + \underbrace{C(x_3)^{32}}_{\text{Term 3}} \\[18pt] \textbf{1. Calculate Term 1 (The Real Root)} \\[6pt] \text{Since } x_1 = 1 \text{ and } A = 0.5: \\[6pt] \text{Term 1} = 0.5 \cdot (1)^{32} = \mathbf{0.5} \\[18pt] \textbf{2. Calculate Terms 2 and 3 (The Complex Roots)} \\[6pt] \text{Notice that } C \text{ is the complex conjugate of } B, \text{ and } x_3 \text{ is the conjugate of } x_2. \\[6pt] \text{This implies that Term 3 is the complex conjugate of Term 2.} \\[6pt] \text{Mathematical Rule: } Z + \bar{Z} = 2 \cdot \text{RealPart}(Z). \\[6pt] \text{Therefore, we only need to calculate Term 2 and double its real component.} \\[12pt] \text{Let } x_2 = \sqrt{2}e^{i\theta} \text{ (Polar form, where } \sqrt{2} \text{ is the magnitude)}. \\[6pt] (x_2)^{32} = (\sqrt{2})^{32} e^{i32\theta} = 2^{16} (\cos(32\theta) + i\sin(32\theta)) \\[6pt] (x_2)^{32} = 65536 (\cos(32\theta) + i\sin(32\theta)) \\[12pt] \text{When we multiply this by } B \text{ and take } 2 \times \text{Real Part, the result is exactly:} \\[6pt] \text{Term 2} + \text{Term 3} = \mathbf{28071.5} \\[18pt] \textbf{3. Final Total} \\[6pt] a_{32} = \text{Term 1} + (\text{Terms 2 \& 3}) \\[6pt] a_{32} = 0.5 + 28071.5 \\[6pt] a_{32} = \mathbf{28072} \end{array} $$

Getting back to OEIS A105578, a plot of the first 52 members is show in Figure 2:

1, 1, 0, -1, 0, 3, 4, -1, -8, -5, 12, 23, 0, -45, -44, 47, 136, 43, -228, -313, 144, 771, 484, -1057, -2024, 91, 4140, 3959, -4320, -12237, -3596, 20879, 28072, -13685, -69828, -42457, 97200, 182115, -12284, -376513, -351944, 401083, 1104972, 302807, -1907136, -2512749, 1301524, 6327023, 3723976, -8930069, -16378020, 1482119, 34238160, 31273923, -37202396

Figure 2: permalink

It can be seen that after a while the values between to fluctuate wildly between larger and larger positive and negative values.

There is a database for a great many linear recurrences of order 3 listed here.

Pseudo-Sphenic Number Sequences

A cubic polynomial with three real roots and rational coefficients can be written in the following form:

(a\(x\) + b) \(\times \)  (c\(x\) + d) \( \times \) (e\(x\) + f) 
where a, b, c, d, e and f are rational numbers

Let's modify the conditions so that a, b, c, d, e and f are integers (positive or negative) and \(x\) can only take integer values greater than 1. Let's change the \(x\) to an \(n\) so that we have:

(a\(n\) + b) \( \times \) (c\(n\) + d) \( \times \) (e\(n\) + f)

Let's take a specific example where a=1, b=0, c=1, d=1, e=2 and f=3. This gives us:$$n \times (n+1) \times (2n+3)$$As we plug in different values for \(n\), starting with \(n=2\), a series of terms arises. In this case, the terms begin:

42, 108, 220, 390, 630, 952, 1368, 1890, 2530, 3300, 4212, 5278, 6510, 7920, 9520, 11322, 13338, 15580, 18060, 20790, 23782, 27048, 30600, 34450, 38610, 43092, 47908, 53070, 58590, 64480, 70752, 77418, 84490, 91980, 99900, 108262, 117078, 126360, 136120

These terms constitute OEIS A163815 (although the terms for \(n=0\) and \(n=1\) are included). These sorts of sequences involve the multiplication of triple linear combinations of \(n\), in this case \(n\), \(n+1\) and \(2n+3\). This permalink leads to a SageMath algorithm that will generate a sequence of terms for varying values of a, b, c, d, e and f. 

If we impose the condition that each of the linear factors must be distinct (and this is the case for the example just shown), then we have a sequence where each member is a sort of pseudo-sphenic number that can be written as a product of three of its divisors but each divisor is a linear combination of an underlying integer. For example, take the number 27048 (my diurnal age yesterday). It can be written as:$$27048 = 23 \times 24 \times 49\\ \text{where } 23=n, 24=n+1 \text{ and } 49=2n+13$$The associated polynomial will cut the \(x\) axis in three locations. Figure 1 shows the situation for \(y=x \times (x+1) \times (2x+3) \) where \(x\)=-1.5, -1 and 0.

Figure 1: Geogebra link

Let's try another example where a, b, c, d, e, f = 1, -1, 1, 1, 1 , 2. The initial terms generated will be of the form \( (n-1) \times (n+1) \times (n+2) \) and are as follows (starting with \(n=3\) because we want to avoid getting a 1 as a factor):

40, 90, 168, 280, 432, 630, 880, 1188, 1560, 2002, 2520, 3120, 3808, 4590, 5472, 6460, 7560, 8778, 10120, 11592, 13200, 14950, 16848, 18900, 21112, 23490, 26040, 28768, 31680, 34782

This sequence does not appear in the OEIS but it is recognised as being generated from a polynomial (I included the terms for \(n=1\) and \(n=2\) when searching in the OEIS). See Figure 2.

Figure 2

For a given number to be a pseudo-sphenic number, it must have more than three, not necessarily distinct, prime factors. For example, my earlier example of 27048:$$27048=2^3 \times 3 \times 7^2 \times 23=23 \times 24 \times 49$$Sometimes the members of a polynomial sequence will be genuinely sphenic as with the case of:$$n \times (n+6) \times (n+12)$$Figure 3 shows the presence of triplets of so-called "sexy" primes: (5, 11, 17), (11, 17, 23), (17, 23, 29) and (31, 37, 43).

Figure 3: permalink

Any sphenic number will be a member of some polynomial sequence. Let's arbitrarily choose the sphenic number formed by the prime factors 79, 101 and 197. We have:$$1571863=79 \times 101 \times 197$$Let's subtract 77 from each number so that they become smaller. This gives us 2, 24 and 120 so that we consider the polynomial:$$ (n+2) \times (n+24) \times (n+ 120)$$Figure 4 shows the result when we generate the sequence up to 1571863.

Figure 4: permalink

Looking at the results we can see that:

• 106723 = 19 * 41 * 137
• 244807 = 31 * 53 * 149
• 906277 = 61 * 83 * 179
• 1571863 = 79 * 101 * 197

Thus the seemingly unrelated sphenic numbers 106723, 244807, 906277 and 1571863 are in fact very closely related because they all arise when different values of \(n\) are assigned to the polynomial \( (n+2) \times (n+24) \times (n+120) \), specifically \(n=17, 29, 59\) and \(77\). Thus a hidden link between sphenic numbers is revealed.