Pairs of Pythagorean triangles with given catheti ratios

TL;DR

This paper explores the existence of infinitely many non-similar Pythagorean triangle pairs with specified catheti ratios, under certain conditions, expanding understanding of Pythagorean triples and their proportional relationships.

Contribution

It proves the existence of infinitely many non-similar Pythagorean triangle pairs with prescribed ratios, given that the product of one triangle's hypotenuse and leg differs from the other.

Findings

Infinitely many non-similar Pythagorean pairs exist for given ratios.

Such pairs are characterized by the condition $Aa eq Bb$.

The result generalizes the understanding of Pythagorean triangle ratios.

Abstract

In this note we investigate the problem of finding pairs of Pythagorean triangles $(a, b, c), (A, B, C)$, with given catheti ratios $A/a, B/b$. In particular, we prove that there are infinitely many essentially different ("non-similar") pairs of Pythagorean triangles $(a, b, c), (A, B, C)$ satisfying given proportions, provided that $Aa\neq Bb$.

Equations

$$a^2+b^2=c^2\text{mbox}anda,b,c\in \mathbb{N}=\{1,2,\dots \}.$$

$$\frac{A^{\prime }}{a^{\prime }}=\frac{A}{a},\frac{B^{\prime }}{b^{\prime }}=\frac{B}{b}.$$

$$H:v^2=u(a^{2}u+b^2)(A^{2}u+B^2)$$

$$\left\{\begin{array}[]{ccccc}(as)^{2}&+&b^{2}&=&t^{2}\\ (As)^{2}&+&B^{2}&=&r^{2}\end{array}\right.$$

$$E:y^2=x(x+A^2{b}^2)(x+a^2{B}^2),$$

$$x(2Q^{\ast })=\frac{(a^2{A}^2-B^2{b}^2{)}^2}{4c^2{C}^2}$$

$$E_{a,b}:y^2=x(x+a^4)(x+b^4).$$

$$E_{a,b}:y^2=x(x+a^4)(x+b^4)$$

$$f(-a^{3}b)={\left(a^{3}b(b-a)\right)}^2(u^4+2u^{3}v+2u^2{v}^2-2u{v}^3+v^4).$$

$$C:W^2=U^4+2U^3+2U^2-2U+1.$$

$$E^{\prime }:Y^2=(X-24)(X-6)(X+30),$$

$$\varphi :C\ni (U,W)\mapsto (X,Y)\in E^{\prime },$$

$$(X,Y)=(6(3U^2+3U+1-3W),54(2U^3+3U^2+2U-1-(1+2U)W)),$$

$$(U,W)=\left(\frac{72+Y-3X}{6(X+3)},\frac{Y^2+162Y+6588-2X^3-9X^2}{36(X+3{)}^2}\right).$$

$$(x_n,y_n)=(-a_n^3{b}_n,4a_n{b}_n(b_n-a_n)w_n).$$

$$f_{A,B}(p^{2}x+q,p^{3}y)=C{f}_{a,b}(x,y),$$

$$\begin{cases}\begin{array}[]{lll}q(A^{4}+q)(B^{4}+q)&=&0\\ p^{2}(A^{4}B^{4}-a^{4}b^{4}p^{4}+2(A^{4}+B^{4})q+3q^{2})&=&0\\ p^{4}(A^{4}+B^{4}-(a^{4}+b^{4})p^{2}+3q)&=&0\end{array}\end{cases}.$$

$$F=$$

$$(A^2{b}^2+a^2{B}^2)(a^2{A}^2+b^2{B}^2)\times$$

$$(a^4{A}^4-A^4{b}^4-a^4{B}^4)(A^4{b}^4+a^4{B}^4-b^4{B}^4)\times$$

$$(a^4{A}^4-A^4{b}^4+b^4{B}^4)(a^4{A}^4-a^4{B}^4+b^4{B}^4).$$

$$2uv{a}_n=|u^2-v^2|b_n,w^2=u^4+2u^{3}v+2u^2{v}^2-2u{v}^3+v^4$$

$${\varphi }^{-1}(2P)=(1/4,-13/16),$$

$${\varphi }^{-1}(3P)=(69/35,-7274/1225)$$

$$E_{a,b}:y^2=x(x+a^4)(x+b^4)$$

$$\begin{array}[]{|l|l|l|}\hline\cr(a,b,c)&(a,b,c)&(a,b,c)\\ \hline\cr(3,4,5)&(581,3420,3469)&(2380,4611,5189)\\ (5,12,13)&(588,2365,2437)&(2436,2923,3805)\\ (12,35,37)&(612,1075,1237)&(2725,5628,6253)\\ (13,84,85)&(660,2989,3061)&(3267,6956,7685)\\ (19,180,181)&(685,9372,9397)&(3612,6955,7837)\\ (24,143,145)&(780,6059,6109)&(3751,6840,7801)\\ (33,56,65)&(913,3384,3505)&(4180,8541,9509)\\ (64,1023,1025)&(924,5893,5965)&(4251,5180,6701)\\ (69,2380,2381)&(949,2580,2749)&(4469,5100,6781)\\ (115,252,277)&(1403,1596,2125)&(4740,5341,7141)\\ (180,299,349)&(1507,9324,9445)&(5365,9828,11197)\\ (319,360,481)&(1820,8181,8381)&(5633,7656,9505)\\ (339,6380,6389)&(2059,2100,2941)&(6125,6612,9013)\\ (473,864,985)&(2147,6204,6565)&(6204,7747,9925)\\ (540,629,829)&(2299,7140,7501)&(6811,8460,10861)\\ \hline\cr\end{array}$$

$$\frac{A}{a}=\mu ,\frac{B}{b}=\nu$$

$$\frac{A}{a}=r_0,\frac{B}{b}=r_1,\frac{C}{c}=r_2.$$

$$a^2+b^2=c^2,a^2+r_1^2{b}^2=r_2^2{c}^2.$$

$$(u^2-v^2{)}^2{r}_1^2-(u^2+v^2{)}^2{r}_2^2=-(2uv{)}^2$$

$$r_1=\frac{((u^4-v^4)w+2uv)((v^4-u^4)w+2uv)}{2(u^2-v^2{)}^2(u^2+v^2)w},r_2=\frac{(u^4-v^4{)}^2{w}^2+4u^2{v}^2}{2(u^2-v^2)(u^2+v^2{)}^{2}w}.$$

$$E_{\nu }:y^2=x(x+1)(x+{\nu }^2)$$

Full text

1. Introduction

In [2] the author investigated the following problem stated by Leech (as was remarked by Smyth in [3]):

Find two rational right-angled triangles on the same base whose heights are in the ratio $n:1$ for $n$ an integer greater than 1.

By considering an equivalent elliptic curve problem, he find parametric solutions for certain values of $n$ and present results of a numerical search for solutions with $n\in\{1,\ldots,999\}$. Motivated by his findings we deal with the general problem of finding pairs of Pythagorean triangles with given catheti ratios. More precisely, throughout the paper $(a,b,c)$ will denote a Pythagorean triple, i.e. a triple of positive integers satisfying the condition

[TABLE]

Given two such triples $(a,b,c)$ and $(A,B,C)$ we are interested whether there exist essentially different pairs $(a^{\prime},b^{\prime},c^{\prime})$, $(A^{\prime},B^{\prime},C^{\prime})$ satisfying

[TABLE]

As we will see in the sequel under some mild conditions on the pair of triples $(a,b,c),(A,B,C)$, the problem has a positive answer. More precisely, in Section 2 we prove that there are infinitely many essentially different pairs $(a^{\prime},b^{\prime},c^{\prime}),(A^{\prime},B^{\prime},C^{\prime})$ solving (1) provided that $Aa\neq Bb$. In Section 3 we investigate the case $Aa=Bb$ and prove that there are infinitely many pairs $(a,b,c),(A,B,C)$ such that the system (1) has infinitely many solutions $(a^{\prime},b^{\prime},c^{\prime}),(A^{\prime},B^{\prime},C^{\prime})$. We also present results of our numerical search and observe that for certain pairs $(a,b,c),(A,B,C)$ there is no non-trivial solution. Finally, in the last section we investigate the general problem of finding pairs of Pythagorean triangles $(a,b,c),(A,B,C)$ such that $A/a=\mu,B/b=\nu$, where $\mu,\nu$ are given rational numbers. We reduce the problem to the case when $\mu=1$ and prove that the set of those $\nu$ is dense in the Euclidean topology in $\mathbb{Q}^{+}$.

2. The case $Aa\neq Bb$

In this section we are interested in finding the solutions of system (1) under the assumption $Aa\neq Bb$. More precisely, together with given triples $(a,b,c)$ and $(A,B,C)$ we will consider the following elliptic curve

[TABLE]

The basic observation is that each point $Q=(u,v)\in H(\mathbb{Q})$ of the form $2P$ ($P\in H(\mathbb{Q})$) gives the desired pair of triangles $(a^{\prime},b^{\prime},c^{\prime})$, $(A^{\prime},B^{\prime},C^{\prime})$. Namely, if $(u,v)=P+P$ then $u=s^{2}$, $a^{2}u+b^{2}=t^{2}$ and $A^{2}u+B^{2}=r^{2}$ with some $s,t,r\in\mathbb{Q}$. Hence

[TABLE]

and $(a^{\prime},b^{\prime},c^{\prime})=(as,b,t)$, $(A^{\prime},B^{\prime},C^{\prime})=(As,B,r)$ do work. The point $Q^{\prime}=(1,cC)$ corresponds to the initial situation $(a^{\prime},b^{\prime},c^{\prime})=(a,b,c)$, $(A^{\prime},B^{\prime},C^{\prime})=(A,B,C)$. Hence the natural question in this context is whether the point $Q^{\prime}$ is of infinite order or (at least) is the rank of $H$ positive? We will investigate both problems in the sequel.

After change of variables $x=a^{2}A^{2}u$, $y=a^{2}A^{2}v$ we get a standard form

[TABLE]

and the image of the point $Q^{\prime}$ takes the form $Q^{\ast}=(a^{2}A^{2},a^{2}A^{2}cC)$. The group $E(\mathbb{Q})$ contains exactly 3 points of order 2, hence the torsion part $\operatorname{Tor}(E(\mathbb{Q}))$ cannot be cyclic. By the Mazur theorem $\operatorname{Tor}(E(\mathbb{Q}))=\mathbb{Z}_{2}\times\mathbb{Z}_{2n}$ for some $n\in\{1,2,3,4\}$. If $P+P=(0,0)$ then $|P|=4$, hence $n=2$ or $n=4$. It follows that if $Q^{\ast}$ has a finite order then $|Q^{\ast}|=4$ and $|2Q^{\ast}|=2$. By duplication formula

[TABLE]

we infer that $|2Q^{\ast}|=2$ if and only if $Aa=Bb$. Geometrically this means that $(a,b,c)$ and $(A,B,C)$ are " skew-similar". In other cases the point $Q^{\ast}$ has infinite order. In this way we have proved

Theorem 2.1**.**

If $Aa\neq Bb$ then there exist infinitely many essentially different ("non-similar") pairs of Pythagorean triangles $(a^{\prime},b^{\prime},c^{\prime})$, $(A^{\prime},B^{\prime},C^{\prime})$ satisfying given proportions (1).

3. The case $Aa=Bb$

In previous section we have proved that provided $Aa\neq Bb$, then there are infinitely many Pythagorean triangles $(a^{\prime},b^{\prime},c^{\prime})$, $(A^{\prime},B^{\prime},C^{\prime})$ satisfying given proportions (1).

What is going on in the case $Aa=Bb$? First of all, let us note that primitivity of triangles $(a,b,c),(A,B,C)$ immediately implies that $A=b,B=a$. Then $C=c$ and we deal with the case of cross-similarity of pairs of Pythagorean triangles. In other words, we are interested in the existence of rational points on the curve

[TABLE]

We prove

Theorem 3.1**.**

There are infinitely many primitive Pythagorean triples $(a,b,c)$ such that the elliptic curve

[TABLE]

has positive rank.

Proof.

Let $a=2uv,b=u^{2}-v^{2},c=u^{2}+v^{2}$ be a Pythagorean triple and write $f(x)=x(x+a^{4})(x+b^{4})$. We note the crucial identity

[TABLE]

In other words $f(-a^{3}b)$ is a square if and only if there is a rational point on the quartic curve

[TABLE]

The curve $C$ is birationally equivalent with the elliptic curve

[TABLE]

via the mapping

[TABLE]

where

[TABLE]

and the inverse $\phi^{-1}:\;E^{\prime}\ni(X,Y)\mapsto(U,W)\in C$ is given by

[TABLE]

A quick computation with Magma [1], reveals that rank of $E^{\prime}(\mathbb{Q})$ is equal to 1. One can check that the generator of the infinite part is $P=(42,-216)$.

Now we compute multiplies $nP=(X_{n},Y_{n})$ for $n=1,2,\ldots$ and then consider corresponding points $(U_{n},W_{n})=\phi^{-1}(nP)$. By writing $U_{n}=u_{n}/v_{n},W_{n}=w_{n}/v_{n}^{2}$ with $\gcd(u_{n}w_{n},v_{n})=1$ we get the triple $(a_{n},b_{n},c_{n})=(2u_{n}v_{n},|u_{n}^{2}-v_{n}^{2}|,u_{n}^{2}+v_{n}^{2})$ which leads to the curve $E_{a_{n},b_{n}}$ with point of infinite order

[TABLE]

In order to finish the proof we need to know that for a given pair $(a_{n},b_{n})$ there are only finitely many pairs $(a_{m},b_{m})$ with $m>n$ such that the curves $E_{a_{n},b_{n}}$, $E_{a_{m},b_{m}}$ are isomorphic. This can be seen as follows: by standard properties of elliptic curves we know that the curves $E_{a,b},E_{A,B}$ are isomorphic if and only there is a linear isomorphism $\psi:\;E_{a,b}:\;(x,y)\mapsto(p^{2}x+q,p^{3}y)\in E_{A,B}$ for some $p,q\in\mathbb{Q}$. Let $f_{a,b}(x,y)$ be the polynomial defining the curve $E_{a,b}$. We see that $E_{a,b}\simeq E_{A,B}$ if and only if

[TABLE]

for some $C\in\mathbb{Q}$. By comparing the coefficients on both sides we get that $C=p^{6}$ and need to consider the following system of equations:

[TABLE]

Let $G$ be the Gröbner basis of the polynomials defining the above system in the ring $\mathbb{Q}[a,b,A,B][p,q]$. In other words we treat $a,b,A,B$ as constants. We have $|G|=14$ and note that $G\cap\mathbb{Q}[a,b,A,B]=\{F\}$, where

[TABLE]

Assuming that $a,b,A,B$ are positive we see that the only possibility for vanishing of $F(a,b,A,B)$ is the condition $(Ab-aB)(aA-bB)=0$. In other words, if $a_{n},b_{n}$ and $a_{m},b_{m}$ with $n\neq m$ come from our construction we need to have $a_{n}a_{m}=b_{n}b_{m}$ or $a_{n}b_{m}=a_{m}b_{n}$. However, let us note that for given $n\in\mathbb{N}$ the system of equations

[TABLE]

has only finitely many solutions in integers $u,v$. This means that there are only finitely many values of $m>n$ such that $a_{n}a_{m}=b_{n}b_{m}$. Similar reasoning works in the case of the equality $a_{n}b_{m}=a_{m}b_{n}$. Thus, in both cases, for only finitely many values of $m>n$ we have that the curves $E_{a_{n},b_{n}},E_{a_{m},b_{m}}$ are isomorphic. In consequence we can construct an infinite set $\mathcal{A}\subset\mathbb{N}$ such that for each $s_{1},s_{2}\in\mathcal{A},s_{1}\neq s_{2}$ we have non-isomorphic curves corresponding to $s_{1},s_{2}$.

∎

Example 3.2**.**

Let $P=(42,-216)$ be the generator of the free part of $E^{\prime}(\mathbb{Q})$, where $E^{\prime}$ is constructed in the proof of the theorem above. We have $2P=(105/4,405/8)$ and

[TABLE]

i.e., $u_{2}=1,v_{2}=4$, with $2u_{2}v_{2}=8,u_{2}^{2}-v_{2}^{2}=-15$. The corresponding triple is $(a_{2},b_{2},c_{2})=(8,15,17)$. On the curve $E_{8,15}$ we get the point $(8^{3}\cdot 15,2296320)$ of infinite order.

In case of $3P=(10698/49,1097928/343)$ we get

[TABLE]

and the corresponding triple is $(a_{3},b_{3},c_{3})=(4830,3536,5986)$. The point of infinite order on $E_{a_{3},b_{3}}$ is $(-4830^{3}\cdot 3536,3750258651849283392000)$. **

We believe that the following is true.

Conjecture 3.3**.**

There are infinitely many primitive Pythagorean triples $(a,b,c)$ such that the elliptic curve

[TABLE]

has rank zero.

Remark 3.4**.**

We performed a small numerical search for primitive triples $(a,b,c)$ such that the rank of $E_{a,b}$ is zero. We used the MAGMA computational package and checked that in the range $a<b<10^{4}$ (in this range there are exactly 890 primitive Pythagorean triples) there are at least 45 triples $(a,b,c)$ for which the rank of $E_{a,b}$ is equal to zero. In order to get the data we were interested in, first we generated all triples $(a,b,c)$ in the considered range and dealt with the elliptic curve $E_{a,b}$. Next, we used the procedure RankBound(E), implemented in Magma, which gives an upper bound for the rank of the elliptic curve $E$. If the computed bound were equal to 0, we printed the triple $(a,b,c)$. In the table below we collect the data we obtained. However, it is very likely that in the range $a<b<10^{4}$ there are many more triples $(a,b,c)$ such that the rank of $E_{a,b}$ is 0.

[TABLE]

Table. Primitive Pythagorean triples $(a,b,c)$ such that the curve $E_{a,b}$ has rank 0.

4. A general question

Now we ask a more difficult question: Does there exist a pair of Pythagoran triangles $(a,b,c)$ and $(A,B,C)$ satisfying

[TABLE]

where $\mu,\nu$ are given positive rationals. If at least one Pythagorean triple can be non-primitive it is easy to see that the answer depends only on the ratio $\nu/\mu$. Indeed, if triples $(a,b,c),(A,B,C)$ solve the equation (4) then the triples $(\mu a,\mu b,\mu c),(A,B,C)$ solve the equation (4) with $\mu=1$ and $\nu$ replaced by $\nu/\mu$. Therefore in the sequel we confine ourselves to the pairs with $\mu=1$.

However, before we concentrate on the system (4) we state an easier question concerning characterization of all possible pairs $r_{1},r_{2}$ of positive rational numbers such that there are Pythagorean triangles $(a,b,c)$ and $(A,B,C)$ satisfying

[TABLE]

As we already noted, without loss of generality, we can assume that $r_{0}=1$. Thus, our problem is equivalent with characterization of solutions of the system

[TABLE]

By writing $a=2uv,b=u^{2}-v^{2},c=u^{2}+v^{2}$ we are interested in solutions of the Diophantine equation

[TABLE]

with respect to $r_{1},r_{2}$. However, this is simple because our equation is quadratic in $r_{1},r_{2}$ and we have the rational point at infinity $(u^{2}+v^{2}:u^{2}-v^{2}:0)$. In consequence, we obtain the parametrization in the following form

[TABLE]

A quick computation reveals that our ratios $r_{1},r_{2}$ are positive (with fixed values of $u,v$) if and only if the condition $0<w<\frac{2uv}{u^{4}-v^{4}}$ is satisfied.

Let us return now to our initial question. Of course finding the solutions of the system (4) is more difficult. Indeed, for a concrete $\nu$ we are confronted with the problem of positivity of the rank of the elliptic curve

[TABLE]

(compare the paper [2] for the case $\nu\in\mathbb{N}$). We prove a general but weaker theorem.

Theorem 4.1**.**

For each $\bar{\nu}\in\mathbb{Q}^{+}$ there exist $s\in\mathbb{Q}^{+}$ such that for $\nu=\bar{\nu}s^{2}$ there exist $(a,b,c)$ and $(A,B,C)$ satisfying $A=a$, $B=\nu b$.

Proof.

In order to get the result we consider the following Pythagorean triples

[TABLE]

Thus, we consider the equation

[TABLE]

Equivalently, putting $S=\frac{x^{2}-1}{x}s$, we are interested in the surface

[TABLE]

defined over the rational function field $\mathbb{Q}(\bar{\nu})$. First we construct a rational curve lying on $H$. We are looking for a rational curve of the form

[TABLE]

where $p,T$ need to be determined. A quick computation with the form of our substitution and the equation defining $H$, reveals that

[TABLE]

does the job. In consequence, we get that the curve

[TABLE]

lies on the surface $H$. We show that there are infinitely many rational curves lying on $H$. In order to do this we treat the surface $H$ as a genus one curve, say $\mathcal{E}$, defined over the rational function field $\mathbb{Q}(\bar{\nu},t)$ in the $(y,S)$ plane. Here we use the base change $x=f(t)$. The curve $\mathcal{E}$ is written in the Weierstrass form

[TABLE]

where the map $\psi:\;H\ni(y,S)\mapsto(X,Y)\in\mathcal{E}$ is given by

[TABLE]

On the curve $\mathcal{E}$ we have the point $P$ coming from the rational curve $L$ lying on $H$. More precisely, $P=(X,Y)$, where

[TABLE]

Because for any given $\bar{\nu}$ positive rational we have $X,Y\in\mathbb{Q}(t)\setminus\mathbb{Q}[t]$ we immediately get (via the analogue of the Nagell-Lutz theorem for a rational function field) that the point $P$ is of infinite order on $\mathcal{E}$. This implies the existence of infinitely many rational curves which lie on $H$ (coming from the points $\psi^{-1}(mP)$, where $m=1,2,\ldots$), with fixed $x=f(t)$. Finally, we see that, for any given $\bar{\nu}$ there are infinitely many parametric families of pairs of Pythagorean triples $(a,b,c),(A,B,C)$ satisfying $a=A,B=\nu b=\bar{\nu}s^{2}b$ and hence the result.

∎

Corollary 4.2**.**

The set

[TABLE]

is dense in the Euclidean topology in the set $\mathbb{R}_{>0}$.

Proof.

In order to get the result we could use the parametric curve constructed in the proof of Theorem 4.1. However, instead we will treat the equation $y^{2}=x(x+1)(x+\nu^{2})$ defining the curve $E_{\nu}$ as an equation in the $(y,\nu)$ plane, i.e.,

[TABLE]

which is a genus 0 curve defined over rational function field $\mathbb{Q}(x)$. After rational base change $x=t^{2}-1$ we see that the curve

[TABLE]

has $\mathbb{Q}(t)$-rational point $(\nu,y)=(0,t(t^{2}-1))$. Thus we obtain the parametric solution in the following form

[TABLE]

We thus see that, for given $t\in\mathbb{Q}$, on the elliptic curve $E_{\nu(u)}$, treated over rational function field $\mathbb{Q}(u)$, we have the point $P=(t^{2}-1,y(u))$. It is clear that the point $P$ is of infinite order on $E_{\nu(u)}$.

To finish the proof, note that for any given positive $t\in\mathbb{Q}$ satisfying the property $t^{2}(t^{2}-1)>0$, the rational map $\nu:\mathbb{R}\ni u\mapsto\frac{2t\left(t^{2}-1\right)u}{t^{2}\left(t^{2}-1\right)-u^{2}}\in\mathbb{R}$ is continuous and has the obvious property

[TABLE]

where $u_{0}=\sqrt{t^{2}(t^{2}-1)}$. The density of $\mathbb{Q}$ in $\mathbb{R}$ together with the above property of the map $\nu=\nu(u)$, immediately implies that the set $\nu(\mathbb{Q})\cap\mathbb{R}_{>0}$ is dense in $\mathbb{R}_{>}$ in the Euclidean topology. The theorem follows.

∎

Remark 4.3**.**

Parameters $\nu$ for which the rank of $E_{\nu}$ is positive arise from a geometric problem, similar to that of congruent numbers. But our Theorem 4.1 illustrates a difference. Whereas beeing congruent depends only on the square-class of a considered number the situation with numbers $\nu$ is completely different: by our theorem each square-class contains relevant numbers $\nu$! **

Acknowledgments. We are grateful to the anonymous referee for careful reading of the manuscript and useful remarks which have improved the presentation.