Quadratic number pyramids

Hans Humenberger, Norbert Hungerbühler, Berthold Schuppar

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Hans Humenberger, Norbert Hungerbühler, Berthold Schuppar. Quadratic number pyramids. 2025. hal-05273469

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<figure><img src="image_1.png" /></figure>

Distributed under a Creative Commons Attribution 4.0 International License
Quadratic number pyramids

Hans Humenberger

Faculty of Mathematics, University of Vienna, Oskar-Morgenstern-Platz 1, 1090 Vienna, Austria

hans.humenberger@univie.ac.at

Norbert Hungerbühler

Department of Mathematics, ETH Zentrum, Rämistrasse 101, 8092 Zürich, Switzerland

norbert.hungerbuehler@math.ethz.ch

Berthold Schuppar

Fakultät für Mathematik, TU Dortmund, Vogelpothsweg 87, 44221 Dortmund, Germany

berthold.schuppar@tu-dortmund.de

Abstract

We investigate quadratic number pyramids, i.e., rational values $(a, b, c)$ such that $a^2+b^2, b^2+c^2$, and $a^2+2b^2+c^2$ are rational squares. We show that every solution $(a, b, c)$ of this diophantine system corresponds to a rational point on an elliptic curve $\Gamma$. Vice versa, every rational point on $\Gamma$ yields a quadratic number pyramid. We provide explicit formulas that parametrize infinite families of quadratic number pyramids.

**Key words.** Quadratic number pyramid, elliptic curves, quadratic diophantine systems.

**Mathematics Subject Classification.** 11D09, 11G05

# 1 Introduction

Number pyramids are a simple and engaging mathematical play, often used in primary school education to help children develop number sense, logical thinking, and arithmetic skills. A number pyramid is made up of stacked bricks or boxes, where each brick is the sum of the two bricks directly beneath it. Sometimes additional conditions are imposed on the numbers in that only values from a certain given set are allowed. In this article we investigate quadratic number pyramids consisting of 6 bricks, each containing a square number. Here is a first example.

<figure><img src="image_2.png" /></figure>

A particular class of such quadratic number pyramids can be constructed from Pythagorean triples (see Section 2). We will call these examples trivial solutions. However the full class of solutions of quadratic number pyramids is a much richer family and is associated with elliptic curves (see Section 3).

Let us first formally describe the quadratic number pyramids to fix the notation.
**Definition 1.** A quadratic number pyramid is a diagram of the form

<figure><img src="image_3.png" /></figure>

such that

$$a^2 + b^2 = e^2, \quad b^2 + c^2 = f^2, \quad a^2 + 2b^2 + c^2 = g^2 \quad (1)$$

with $a, b, c, e, f, g \in \mathbb{Q}$. In this case we will say that $(a, b, c)$ is a solution of the quadratic number pyramid.

Note that the concept of quadratic number pyramids should not be confused with Euler bricks [2, 3]. There the condition is

$$a^2 + b^2 = e^2, \quad b^2 + c^2 = f^2, \quad c^2 + a^2 = g^2.$$

Hence an Euler brick is a rectangular cuboid whose edges and face diagonals all have rational lengths, as shown in Figure 1.

<figure><img src="image_4.png" /><figcaption>Figure 1: An Euler brick</figcaption></figure>

The geometric interpretation of a quadratic number pyramid looks slightly different. It corresponds to a tetrahedron with rational sides and three right angles as indicated in Figure 2.

Every solution $(a, b, c)$ of a quadratic number pyramid in $\mathbb{Q}$ can be transformed into a solution in $\mathbb{N}$ by expanding with the least common multiple of the denominators and changing the sign, if necessary. In the following, we will therefore concentrate on rational solutions.

Let us briefly sort out some special cases. The following solutions $(a, b, c)$ are not interesting

$$(0, 0, 0), (a, 0, 0), (0, 0, c).$$

The triple $(0, b, 0)$ cannot be a quadratic number pyramid for $b \neq 0$ since $2b^2 \neq 0$ is not a rational square. The triple $(a, 0, c)$ is also not interesting, as it is simply
<figure><img src="image_5.png" /><figcaption>Figure 2: A tetrahedron with sides given by a quadratic number pyramid</figcaption></figure>

a Pythagorean triple $a^2 + c^2 = g^2$. More interesting is the case $(0, b, c)$ or the symmetric situation $(a, b, 0)$. In the latter case we have

$$a^2 + b^2 = e^2, \quad a^2 + 2b^2 = g^2.$$

This means that we have three squares $a^2, a^2 + b^2, a^2 + 2b^2$ in an arithmetic progression. This is related to the famous congruum problem: A congruum is the difference between successive square numbers in an arithmetic progression of three squares (see, e.g., [7, §8.4]). In our case the congruum is $b^2$, and we have

$$e^2 + b^2 = g^2, \quad e^2 - b^2 = a^2,$$

and hence

$$2b^2 = g^2 - a^2 = (g - a)(g + a)$$

which can be rewritten as

$$\frac{b^2}{g-a} = \frac{g+a}{2}.$$

Taking the square on both sides and adding $(\frac{g-a}{2})^2$ leads to

$$(\frac{b^2}{g-a})^2 + (\frac{g-a}{2})^2 = (\frac{g+a}{2})^2 + (\frac{g-a}{2})^2 = e^2.$$

Hence, the three rational numbers $\frac{b^2}{g-a}$, $\frac{g-a}{2}$, $e$ are a Pythagorean triple. Therefore we have

$$\frac{b^2}{g-a} = k(m^2 - n^2), \quad \frac{g-a}{2} = 2kmn$$

for some rational numbers $k, m, n$. The product yields for the congruum

$$b^2 = 4k^2mn(m^2 - n^2).$$

In particular, $mn(m^2 - n^2) = y^2$ must be the square of a rational number $y$. Substituting $y = n^2Y$ and $m = nX$ yields $Y^2 = X^3 - X = X(X - 1)(X + 1)$. However, this elliptic curve has rank 0 and torsion group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ (see Section 3). Therefore, the only rational points on this curve are $(0, 0)$, $(1, 0)$, $(-1, 0)$. In other words, there is no quadratic number pyramid solution of the form $(0, b, c)$ or $(a, b, 0)$ with $(a, b, c) \neq (0, 0, 0)$. Summarizing we have the following.

**Proposition 1.** Let $(a, b, c) \neq (0, 0, 0)$. Then we have:
(a) The triple $(0, b, 0)$ is not a solution of the quadratic number pyramid.

(b) The triples $(0, b, c)$ and $(a, b, 0)$ are not solutions of the quadratic number pyramid.

## 2 Quadratic number pyramids from Pythagorean triples

It is easy to see that one can create quadratic number pyramids with Pythagorean triples.

**Proposition 2.** Let $(u, v, w)$ be a Pythagorean triple, i.e. $u^2 + v^2 = w^2$. Then, a triple $(a, b, c)$ with $|a| = u^2$, $|b| = |uv|$, $|c| = v^2$ yields a quadratic number pyramid.

*Proof.* The triple $(a, b, c)$ generates the quadratic number pyramid

<figure><img src="image_6.png" /></figure>

□

Notice that the first example in the introduction is of this form with $(u, v, w) = (3, 4, 5)$. We call solutions of this form trivial. More interesting are the non-trivial solutions. Here is a list of the non-trivial solutions $(a, b, c)$ below 10,000 which are primitive in the sense that $\gcd(a, b, c) = 1$.

<table><thead><tr><th>a</th><th>b</th><th>c</th></tr></thead><tbody><tr><td>105</td><td>252</td><td>64</td></tr><tr><td>220</td><td>528</td><td>171</td></tr><tr><td>360</td><td>480</td><td>31</td></tr><tr><td>480</td><td>360</td><td>319</td></tr><tr><td>704</td><td>2772</td><td>129</td></tr><tr><td>891</td><td>912</td><td>380</td></tr><tr><td>1080</td><td>576</td><td>943</td></tr><tr><td>1197</td><td>3696</td><td>172</td></tr><tr><td>1276</td><td>1440</td><td>93</td></tr><tr><td>1408</td><td>456</td><td>855</td></tr><tr><td>3959</td><td>5040</td><td>2688</td></tr><tr><td>4209</td><td>4788</td><td>1216</td></tr><tr><td>5005</td><td>4368</td><td>324</td></tr><tr><td>5184</td><td>7980</td><td>1463</td></tr><tr><td>6175</td><td>1716</td><td>288</td></tr></tbody></table>
<table><tr><td>6237</td><td>6384</td><td>5612</td></tr><tr><td>7353</td><td>6696</td><td>928</td></tr><tr><td>7695</td><td>2760</td><td>1472</td></tr><tr><td>9715</td><td>9120</td><td>2052</td></tr></table>

The question now is where these non-trivial solutions come from.

### 3 Quadratic number pyramids and elliptic curves

We will now examine quadratic number pyramids in general. In a first step, we restrict ourselves to the rational solutions of $a^2 + 2b^2 + c^2 = g^2$ and keep the remaining conditions in (1) for later.

**Lemma 1.** The rational solutions of the equation $a^2 + 2b^2 + c^2 = g^2$ are given by

$$a = g \cdot \frac{u^2 - 2v^2 - w^2}{u^2 + 2v^2 + w^2} \qquad (2)$$

$$b = g \cdot \frac{2uv}{u^2 + 2v^2 + w^2} \qquad (3)$$

$$c = g \cdot \frac{2uw}{u^2 + 2v^2 + w^2} \qquad (4)$$

for $g, u, v, w \in \mathbb{Q}$ with $(u, v, w) \neq (0, 0, 0)$.

*Proof.* We leave aside the trivial case $g = 0$ and continue with $g \neq 0$. Consider the ellipsoid

$$E := \{(x, y, z) \in \mathbb{R}^3 : x^2 + 2y^2 + z^2 = g^2\} \subset \mathbb{R}^3.$$

It contains the rational point $P = (g, 0, 0)$. If $Q$ is another rational point on $E$, then the vector $Q - P$ has rational components. Vice versa, if $(u, v, w)$ is a non-zero rational vector, then the line $P + \lambda(u, v, w)$ intersects $E$ in $P$ and a rational point $Q$, $Q \neq P$ if $u \neq 0$. The coordinates $(a, b, c)$ of the intersection point $Q$ are given by (2)–(4). $\square$

Let us now turn our attention to the other two conditions in (1) for a quadratic number pyramid. Using the expressions for $a, b, c$ from Lemma 1, these conditions are

$$a^2 + b^2 = g^2 \cdot \frac{u^4 - 2u^2w^2 + (2v^2 + w^2)^2}{(u^2 + 2v^2 + w^2)^2} = e^2 \qquad (5)$$

$$b^2 + c^2 = 4g^2 \cdot \frac{u^2(v^2 + w^2)}{(u^2 + 2v^2 + w^2)^2} = f^2. \qquad (6)$$

Condition (6) is satisfied for

$$v = k(m^2 - n^2), \quad w = 2kmn \qquad (7)$$
with $k, m, n \in \mathbb{Q}$, and no other rational solutions of (6) exist. Observe that $|m| = |n|$ would result in $v = 0$ in (7) and hence $b = 0$. As we have discussed this case in the introduction we will assume $|m| \neq |n|$ in the sequel.

Using (7), the condition (5) reduces to the problem of finding the rational points $(u, y)$ on the quartic curve

$$ \gamma : u^4 - 8k^2m^2n^2u^2 + 4k^4(m^4 + n^4)^2 = y^2 $$

in the variables $u$ and $y$. We immediately identify the following rational points on $\gamma$:

$$ \begin{aligned} p_0 &= (0, 2k^2(m^4 + n^4)) & p_1 &= (2kmn, 2k^2(m^4 - n^4)) \\ q_0 &= (0, -2k^2(m^4 + n^4)) & q_1 &= (2kmn, -2k^2(m^4 - n^4)) \end{aligned} $$

Hence we may use the $u$-coordinate of these points to satisfy (5). Namely using $u = 0$ and (7) in (2)-(4) yields the solution $(a, b, c) = (g, 0, 0)$ which is not interesting. Using $u = 2kmn$ and (7) in (2)-(4) yields the formula

$$ (a, b, c) = (g(m^2 - n^2)^2, 2gmn(m^2 - n^2), 4gm^2n^2) $$

which are (multiples of) trivial solutions. So, the points $p_0, p_1, q_0, q_1$ do not give new solutions, but they will be useful in the next step.

Now we employ the theory of elliptic curves to find the rational points on $\gamma$ which are related to non-trivial solutions. Following Mordell in [6, §10, Theorem 2] we use the birational transformation

$$ u = -\frac{3Y}{4k^2m^2n^2 - 3X} \qquad (8) $$

$$ y = \frac{64k^6m^6n^6 - 108k^2m^2n^2X^2 + 54X^3 - 27Y^2}{3(4k^2m^2n^2 - 3X)^2} \qquad (9) $$

which maps the quartic curve $\gamma$ to the elliptic curve

$$ \Gamma : Y^2 = X^3 + \alpha X + \beta \qquad (10) $$

where

$$ \alpha = -\frac{k^4}{3}(3m^4 + n^4)(m^4 + 3n^4), \quad \beta = \frac{4}{27}k^6m^2n^2(9m^8 + 14m^4n^4 + 9n^8). $$

The discriminant of $\Gamma$ is $\Delta = 4\alpha^3 + 27\beta^2 = -4k^{12}(m^4 - n^4)^4(m^4 + n^4)^2$ and hence the curve is regular for $k \neq 0, |m| \neq |n|$. The transformation which maps $\Gamma$ to $\gamma$ is given by

$$ X = \frac{1}{6}(3(u^2 + y) - 4k^2m^2n^2) \qquad (11) $$

$$ Y = \frac{u}{2}(u^2 + y - 4k^2m^2n^2). \qquad (12) $$

If we transform the points $p_0, p_1, q_0, q_1$ from $\gamma$ to $\Gamma$, we obtain the following
rational points on $\Gamma$:

$$P_0 = \left( \frac{k^2}{3} (3m^4 - 2m^2n^2 + 3n^4), 0 \right)$$

$$Q_0 = \left( -\frac{k^2}{3} (3m^4 + 2m^2n^2 + 3n^4), 0 \right)$$

$$P_1 = \left( \frac{k^2}{3} (3m^4 + 4m^2n^2 - 3n^4), 2k^3mn(m^4 - n^4) \right)$$

$$Q_1 = \left( -\frac{k^2}{3} (3m^4 - 4m^2n^2 - 3n^4), -2k^3mn(m^4 - n^4) \right).$$

In addition

$$R = \left( \frac{4}{3} k^2 m^2 n^2, 0 \right)$$

is another rational point on $\Gamma$. In particular, we see three rational points on the X-axis which allows us to factor $\Gamma$ as follows:

$$\Gamma : Y^2 = \left( X - \frac{4}{3} k^2 m^2 n^2 \right) \left( X - k^2 \left( m^4 - \frac{2}{3} m^2 n^2 + n^4 \right) \right) \left( X + k^2 \left( m^4 + \frac{2}{3} m^2 n^2 + n^4 \right) \right).$$

Now, we can use the group structure of the elliptic curve $\Gamma$ to find new rational points generated by $P_0, Q_0, P_1, Q_1, R$. These points can then be transformed to rational points on $\gamma$, and each such point then corresponds to a quadratic number pyramid. But before we do this, let us briefly recall the group operation on $\Gamma$. Adding two different points $P$ and $Q$ on $\Gamma$ is done with the geometric construction in Figure 3 on the left. To double a point, i.e. to calculate $P+P =: 2P$, replace the line through $PQ$ by the tangent in $P$, as indicated in Figure 3 on the right. The neutral element is the point at infinity in the direction of the Y-axis.

<figure><img src="image_7.png" /><figcaption>Figure 3: The group operation on the elliptic curve $\Gamma$</figcaption></figure>

For $P = (x_P, y_P) \neq Q = (x_Q, y_Q)$ the coordinates of $R = P + Q = (x_R, y_R)$ are given by the formulas

$$\lambda = \frac{y_P - y_Q}{x_P - y_Q}$$

$$x_R = \lambda^2 - x_P - x_Q$$

$$y_R = \lambda(x_P - x_R) - y_P.$$
The formula for $R = P + P$ is

$$ 
\begin{aligned}
\lambda &= \frac{3x_P + \alpha}{2y_P} \\
x_R &= \lambda^2 - 2x_P \\
y_R &= \lambda(x_P - x_R) - y_P
\end{aligned}
 $$

where $\alpha$ is the parameter in (10). At this point it is necessary to briefly consider the structure of the group $\Gamma(Q)$ of the rational points on $\Gamma$. By the theorems of Mordell-Weil [5, 8] and of Mazur [4], this group is of the form

$$ \Gamma(Q) = \Gamma(Q)_{\text{tors}} \times \mathbb{Z}^r $$

where the torsion group $\Gamma(Q)_{\text{tors}}$ is one of the following

$$ \mathbb{Z}/t\mathbb{Z} \text{ with } 1 \le t \le 10 \text{ or } t = 12, \text{ or } \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/t\mathbb{Z} \text{ with } 1 \le t \le 4 $$

and $r$ is the rank of the curve.

## 3.1 The torsion group and the rank of $\Gamma$

The points $P_0, Q_0$, and $R$ are points of order 2. Hence, according to Mazur's Theorem, the torsion group of $\Gamma$ is one of the following:

$$ \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/t\mathbb{Z} \text{ with } 1 \le t \le 4. $$

We can exclude the case $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/8\mathbb{Z}$ as follows. Assume there is a rational point $P$ with coordinates $(x_P, y_P)$ on $\Gamma$ of order 4, $y_P \ne 0$. Hence, we have $4P = 0$ or equivalently $3P = -P$. The coordinates $(x_{3P}, y_{3P})$ of $3P$ can be computed by the above formulas, and we have $(x_{3P}, y_{3P}) = -(x_P, y_P) = (x_P, -y_P)$. The explicit expression for $x_{3P}$ is rather unpleasant, but still we have the system

$$ x_{3P} = x_P \quad (13) $$

$$ y_P^2 = x_P^3 + \alpha x_P + \beta \quad (14) $$

which has the following solutions for $x_P$:

$$ x_P = \frac{k^2}{3} \left( 3m^4 - 2m^2n^2 + 3n^4 \pm 3(m-n)(m+n)\sqrt{2(m^4+n^4)} \right) $$

However, $2(m^4 + n^4)$ is never a square by a result of Euler [1, §XXII].

The case $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ occurs exactly if the inflection point of $\Gamma$ is rational. To compute the inflection point we proceed as follows. First, by (10), we have

$$ 2YY' = 3X^2 + \alpha $$

and

$$ 2Y'^2 + 2YY'' = 6X. $$

Using $Y'' = 0$ in an inflection point and eliminating $Y'$ we find that the X-value of an inflection point satisfies the quartic equation

$$ (3X + \alpha)^2 - 12X(X^3 + \alpha X + \beta) = 0 \quad (15) $$
Observe that $P_1$ cannot be the inflection point. Indeed, if we replace $X$ in (15) by the $x$-coordinate of $P_1$, we get

$$-36k^8(m^4 - n^4)^3(m^4 - 4m^2n^2 - n^4)$$

which is not zero for rational values of $|m| \neq |n|, k \neq 0$. Hence $P_1$ cannot be a point of order 3. In the same way, one can see that

$$2P_1 = \left( \frac{k^2(3m^8 + 10m^4n^4 + 3n^8)}{12m^2n^2}, \frac{k^3(n^4 - m^4)^3}{8m^3n^3} \right) \quad (16)$$

cannot be the inflection point. Hence $P_1$ is not a point of order 6. Therefore, $P_1$ cannot be a torsion point. Summarising we have the following result.

**Proposition 3.** The elliptic curve $\Gamma$ has torsion group $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/6\mathbb{Z}$ if the inflection point is rational, and $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ otherwise. The rank of $\Gamma$ is at least 1, and $P_1$ is a point of infinite order.

We conjecture that the inflection point of the curve $\Gamma$ is never a rational point.

## 3.2 Generating quadratic number pyramids

We use (8) and (9) to transform the point $2P_1$ in (16) from the curve $\Gamma$ to the curve $\gamma$. The resulting point is

$$2p_1 = \left( \frac{k(n^4 - m^4)}{2mn}, \frac{k^2(m^8 + 14m^4n^4 + n^8)}{4m^2n^2} \right).$$

Taking $u = \frac{k(n^4 - m^4)}{2mn}$, the first coordinate of $2p_1$, and using (7) in (2)–(4) with $g = 1$ we get the following solutions for quadratic number pyramids.

**Theorem 1.** Let $m, n, g \in \mathbb{Q}$ be arbitrary rational numbers. Then

$$a = g \cdot (m^8 - 8m^6n^2 - 2m^4n^4 - 8m^2n^6 + n^8) \quad (17)$$

$$b = g \cdot (4mn(m^2 - n^2)^2(m^2 + n^2)) \quad (18)$$

$$c = g \cdot (8m^2n^2(m^4 - n^4)) \quad (19)$$

are solutions of the quadratic number pyramid. Apart from the cases $|m| = |n|$ and $mng = 0$ these solutions are nontrivial.

*Proof.* The fact that the formulas (17)–(19) form a quadratic number pyramid follows from our construction. Observe that for trivial solutions we have $|ac| = b^2$ (compare Proposition 2). For the values given by (17)–(19) and $g = 1$ we find

$$ac - b^2 = -8m^2n^2(m^{12} + 4m^{10}n^2 + m^8n^4 + 8m^6n^6 - 5m^4n^8 - 12m^2n^{10} + 3n^{12})$$

and

$$ac + b^2 = 8m^2n^2(3m^{12} - 12m^{10}n^2 - 5m^8n^4 + 8m^6n^6 + m^4n^8 + 4m^2n^{10} + n^{12}).$$

Using the rational root theorem it is easy to check that these values cannot be zero for rational $m, n$ unless $|m| = |n|$ or $mn = 0$. $\square$
If we consider only integer values for $m$ and $n$, and $g = 1$, we find that $\gcd(a, b, c) = 1$ if and only if $\gcd(m, n) = 1$ and $m$ is even and $n$ is odd, or vice versa.

For example, with $m = 2, n = 1, g = 1$ we obtain the pyramid

<figure><img src="image_8.png" /></figure>

which occurs on the fourth line in the table in Section 1.

For $m = 3, n = 1, g = 1/16$ we get the solution $(31, 480, 360)$ which occurs on the third line in the table in Section 1. In general, if the solution $(a, b, c)$ comes from the values $m, n, g$, then the values $m+n, m-n, g/16$ generate the solution $(a', c, b)$.

Notice that we can start with any other multiple $tP_1$, $2 \le t \in \mathbb{N}$, to produce other formulas for $a, b, c$ in the same way. For example, the formulas for the starting point $3P_1$ yield a solution $(a, b, c) = (6917751, 346332, 461776)$ which is not reproduced by the formulas (17)–(19). Also notice that in general there are $r$ generators $P_1, P_2, \dots, P_r$ for curves $\Gamma$ of rank $r$, and each of these generators or their linear combinations can be used to find quadratic number pyramids. For example, the curve $\Gamma$ has rank 4 for $m = 2/13, n = 9/10$.

# 4 Conclusion

We have shown that Pythagorean triples lead to solutions $(a, b, c)$ of the quadratic number pyramid. However, these solutions only form a small subset of the solutions with the additional property that $|ac| = b^2$. Our main result is, that every solution $(a, b, c)$ of the quadratic number pyramid corresponds to a rational point on the elliptic curve $\Gamma$. Vice versa, every rational point on $\Gamma$ yields a quadratic number pyramid. It would be interesting to investigate larger quadratic number pyramids with $n > 3$ bricks at the base. Solutions can also be constructed for these pyramids using Pythagorean triples. But what does the general solution set look like?

# References

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