INTEGRAL POINTS AND ARITHMETICPROGRESSIONS ON HUFF CURVESSZ. TENGELY
Abstract. In this paper we provide bounds for the size of theintegral points on generalized Huff curves
Ha,b :x(ay 2 − 1) = y(bx2 − 1)with a, b ∈ Z and
c,d
Ha,b
:

ax(y 2 − c) = by(x2 − d)with a, b, c, d ∈ Z. We also deal with integral points on these typesof curves with x-coordinates forming arithmetic progressions.

1. introduction
In 1948 Huff [21] studied a geometric problem and related to it a family of curves now called Huff curves. He considered rational distancesets. Given a, b ∈ Q∗ such that a2 6= b2 . Determine the set of points(x, 0) ∈ Q2 satisfying that d((0, ±a), (x, 0)) and d((0, ±b), (x, 0)) arerational numbers, where d denotes the usual Euclidean distance. Consider the Huff curve ax(y 2 − 1) = by(x2 −is a rational point 1). If there
2by(x, y) on the curve, then the point P = y2 −1 , 0 is in the distance set.

2000 Mathematics Subject Classification. Primary 11D61; Secondary 11Y50.
Key words and phrases. Diophantine equations, arithmetic progressions, elliptic
curves.
1

2

SZ. TENGELY

The point (2, 4) is on the curve 2x(y 2 − 1) = 5y(x2 − 1), hence

 

2·5·4
8
,0 =
,0
42 − 1
3
is in the distance set.
Elliptic curves were introduced in cryptography [23, 26]. Elliptic
curves can be represented in different forms having different arithmetic
properties. Many models have been studied recently: Edwards curves,
Huff curves, Montgomery curves, Weierstrass curves, Hessian curves,
Jacobi quartic curves and generalizations. In this paper we deal witharithmetic properties of two generalized Huff models introduced by Wu
and Feng [42] and by Ciss and Sow [15]. We provide bounds for the sizeof integral solutions using Runge’s method [31] combined with reduction method from [37]. In case of the family Ha,b all integral solutionsc,dare classified and in case of Ha,b
the obtained bound is polynomial ina, b, c, d and in case of many concrete equations the largest integral
point is very close to this bound.
Siegel [32] in 1926 proved that the equation y 2 = a0 xn + a1 xn−1 +. . . + an =: f (x) has only a finite number of integer solutions if f has
at least three simple roots. In 1929 Siegel [33] classified all irreducible
algebraic curves over Q on which there are infinitely many integral
points. These curves must be of genus 0 and have at most 2 infinite
valuations. These results are ineffective, that is, their proofs do notprovide any algorithm for finding the solutions. In the 1960’s Baker[4, 6] gave explicit lower bounds for linear forms in logarithms of the
form
n
X
Λ=
bi log αi 6= 0
i=1

where bi ∈ Z for i = 1, . . . , n and α1 , . . . , αn are algebraic numbers(6= 0, 1), and log αi , . . . , log αn denote fixed determinations of the logarithms. Baker [5] used his fundamental inequalities concerning linear
forms in logarithms to derive bounds for the solutions of the elliptic
equation y 2 = ax3 + bx2 + cx + d. This bound were improved by several
authors see e.g. [9, 20]. Baker and Coates [7] extended this result togeneral genus 1 curves. Lang proposed [24] proposed a different method
to prove the finiteness of integral points on genus 1 curves. This method
makes use of the group structure of the genus 1 curve. Stroeker andTzanakis [34] and independently Gebel, Pethő and Zimmer [17] worked
out an efficient algorithm based on this idea to determine all integral
points on elliptic curves. The elliptic logarithm method for determining all integer points on an elliptic curve has been applied to a variety

INTEGRAL POINTS ON HUFF CURVES

3of elliptic equations (see e.g. [35, 36, 38, 39, 40]). The disadvantageof this approach is that there is no known algorithm to determine therank of the so-called Mordell-Weil group of an elliptic curve, which
is necessary to determine all integral points on the curve. There areother methods that can be used in certain cases to determine all integral solutions of genus 1 curves. Poulakis [30] provided an elementaryalgorithm to determine all integral solutions of equations of the form
y 2 = f (x), where f (x) is quartic monic polynomial with integer coefficients. Using the theory of Pellian equations, Kedlaya [22] described amethod to solve the system of equations(
x 2 − a1 y 2 = b 1 ,P (x, y) = z 2 ,where P is a given integer polynomial.
An arithmetic progression on a curve F (x, y) = 0, is an arithmetic
progression in either the x or y coordinates. One can pose the following
natural question. What is the longest arithmetic progression in the xcoordinates? In case of linear polynomials, Fermat claimed and Euler
proved that four distinct squares cannot form an arithmetic progression. Allison [2] found an infinite family of quadratics containing an
integral arithmetic progression of length eight. The curve is y 2 = 12 (k 2 −l2 )x2 − 52 (k 2 −l2 )x+(3k 2 −2l2 ), and the arithmetic progression is as follows (−1, 6k 2 − 5l2 ), (0, 3k 2 − 2l2 ), (1, k 2 ), (2, l2 ), (3, l2 ), (4, k 2 ), (5, 3k 2 −2l2 ), (6, 6k 2 − 5l2 ). Arithmetic progressions on Pellian equations x2 −dy 2 = m have been considered by many mathematicians. Dujella,
Pethő and Tadić [16] proved that for any four-term arithmetic progression, except {0, 1, 2, 3} and {−3, −2, −1, 0}, there exist infinitely
many pairs (d, m) such that the terms of the given progression are
y-components of solutions. Pethő and Ziegler [29] dealt with 5-term
progressions on Pellian equations. Aguirre, Dujella and Peral [1] constructed 6-term arithmetic progression on Pellian equations parametrized
by points on elliptic curve having positive rank. Pethő and Ziegler
posed several open problems. One of them is as follows: ”Can one
prove or disprove that there are d and m with d > 0 and not a perfect square such that y = 1, 3, 5, 7, 9 are in arithmetic progression on
the curve x2 − dy 2 = m?” Recenlty, González-Jiménez [18] answered
the question: there is not m and d not a perfect square such thaty = 1, 3, 5, 7, 9 are in arithmetic progression on the curve x2 − dy 2 = m.
He constructed the related diagonal genus 5 curve and he applied covering techniques and the so-called elliptic Chabauty’s method. Bremner[10] provided an infinite family of elliptic curve of Weierstrass form

4

SZ. TENGELY

with 8 points in arithmetic progression. González-Jiménez [18] showedthat these arithmetic progressions cannot be extended to 9 points arithmetic progressions. Bremner, Silverman and Tzanakis [12] dealt withthe congruent number curve y 2 = x3 − n2 x, they considered integral
arithmetic progressions. If F is a cubic polynomial, then the problem
is to determine arithmetic progressions on elliptic curves. Bremnerand Campbell [13] found distinct infinite families of elliptic curves,
with arithmetic progression of length eight. Campbell [13] produced
infinite families of quartic curves containing an arithmetic progressionof length 9. Ulas [41] constructed an infinite family of quartics containing a progression of length 12. Restricting to quartics possessing
central symmetry MacLeod [25] discovered four examples of length 14
progressions (e.g. y 2 = −17x4 + 3130x2 + 8551, x = −13, −11, . . . , 13.)
Alvarado [3] extended MacLeod’s list by determining 11 more examplesof length 14 progressions (e.g. y 2 = 627x4 −87870x2 +3312859) Moody[27] proved that there are infinitely many Edwards curves with 9 points
in arithmetic progression. Bremner [11] and independently GonzálezJiménez [18, 19] proved using elliptic Chabauty’s method that Moody’s
examples cannot be extended to longer arithmetic progressions. Moody[28] produced six infinite families of Huff curves having the property
that each has rational points with x-coordinate x = −4, −3, . . . , 3, 4.
That is he obtained arithmetic progressions of length 9. Choudhry [14]improved the result of Moody, he found infinitely many parametrizedfamilies of Huff curves on which there are arithmetic progressions oflength 9, as well as several Huff curves on which there are arithmetic
progressions of length 11.
In this article we characterize the arithmetic progressions in case ofc,dthe curve Ha,b and we provide infinite families of curves Ha,bcontaining
arithmetic progressions of length 9. It is important to note that we only
consider arithmetic progressions related to integral points.

2. main results
In the following theorem we characterize the integral points on the
curve Ha,b .
Theorem 1. The Diophantine equation
Ha,b :x(ay 2 − 1) = y(bx2 − 1)

INTEGRAL POINTS ON HUFF CURVES

5

with a, b, x, y ∈ Z has the following solutions(a, b, x, y)(a, b, x, y)(a, b, x, y)(a, b, x, y)(a, b, x, y)(a, b, x, y)(a, b, x, y)

=
=
=
=
=
=
=

(a, a, x, x) with a, x ∈ Z,
(1, 1, −1, 1),(1, 1, 1, −1),(−1, −1, −1, 1),(−1, −1, 1, −1),
(a, 2 − a, −1, 1) with a ∈ Z,(a, 2 − a, 1, −1) with a ∈ Z.

A direct consequence of the above theorem is as follows.
Corollary 1. Let (x1 , y1 ), (x2 , y2 ), (x3 , y3 ) be solutions of the equation
Ha,b for some a, b ∈ Z such that (x1 , x2 , x3 ) forms an arithmetic progression and at most one solution (xi , yi ) satisfies the condition xi = yi .
Then (x1 , x2 , x3 ) = (−3, −1, 1), (−1, 0, 1), (1, 0, −1) or (1, −1, −3).
c,dIn case of the second family Ha,b
we have the following result.

Theorem 2. Let a, b, c, d ∈ Z such that abcd(a2 c − b2 d) 6= 0. Define
L1 , L2 , U1 , U2 as follows
1√ 4 2L1 = −
a c − a2 b2 cd − 162 a2 c + 81 b2 d + 6561,
91√ 4 2−a c + a2 b2 cd − 162 a2 c + 81 b2 d − 6561,
L2 = −
91√ 4 2a c − a2 b2 cd − 162 a2 c + 81 b2 d + 6561,
U1 =9
1√ 4 2U2 =−a c + a2 b2 cd − 162 a2 c + 81 b2 d − 6561.
9
Let m0 = min({0} ∪ {Li : i = 1, 2, Li ∈ R}) and M0 = max({0} ∪ {Li :c,di = 1, 2, Li ∈ R}). If (x, y) is an integral point on Ha,b
, then we havethat eitherq
2 2
2 a4 c2− 2 a tb cd − 2 a2 c + b2 d + 21 ttx=±t ∈ {−161, . . . , 161}
b
or
m0
M0
≤x≤
if b > 0,
b
b
M0
m0
≤x≤
if b < 0.
bb−17,−6
Remark. In case of the curve H5,2
there is no solution coming fromthe formula for x, the bound is −29 ≤ x ≤ 29. The integral solutions are

6

SZ. TENGELY

given by (x, y) ∈ {(−27, −9), (0, 0), (27, 9)}, that is the largest solution
is just 2 away from the bound. We determined all integral solutions onc,dwith −5 ≤ a, b, c, d ≤ 5 and abcd(a2 c−b2 d) 6= 0. In 1976
the curves Ha,bcases there exist solution different from (0, 0). Only in 36 cases there
exist solutions with |x| > 10, in the table below we provide complete
set of solutions for the corresponding curves.
(a, b, c, d)solutions(a, b, c, d)solutions(−5, −4, −3, 4)(−5, −4, 2, −3)(−5, −2, −5, 4)(−5, −2, −3, −4)(−5, −1, 2, 4)(−5, −1, 3, −4)(−5, 1, 2, 4)(−5, 1, 3, −4)(−5, 2, −5, 4)(−5, 2, −3, −4)(−5, 4, −3, 4)(−5, 4, 2, −3)(−4, −1, −2, 5)(−4, 1, −2, 5)(−3, −2, −5, 4)(−3, −1, −5, 4)(−3, 1, −5, 4)(−3, 2, −5, 4)[(0, 0) , (12, 9) , (−12, −9)][(0, 0) , (−12, −10) , (12, 10)][(0, 0) , (12, 3) , (−12, −3)][(0, 0) , (16, 6) , (−16, −6)][(0, 0) , (12, 3) , (−12, −3)][(0, 0) , (−16, −4) , (16, 4)][(0, 0) , (12, −3) , (−12, 3)][(0, 0) , (−16, 4) , (16, −4)][(0, 0) , (12, −3) , (−12, 3)][(0, 0) , (16, −6) , (−16, 6)][(0, 0) , (12, −9) , (−12, 9)][(0, 0) , (12, −10) , (−12, 10)][(0, 0) , (15, 3) , (−15, −3)][(0, 0) , (15, −3) , (−15, 3)][(0, 0) , (16, 10) , (−16, −10)][(0, 0) , (16, 4) , (−16, −4)][(0, 0) , (16, −4) , (−16, 4)][(0, 0) , (16, −10) , (−16, 10)](3, −2, −5, 4)(3, −1, −5, 4)(3, 1, −5, 4)(3, 2, −5, 4)(4, −1, −2, 5)(4, 1, −2, 5)(5, −4, −3, 4)(5, −4, 2, −3)(5, −2, −5, 4)(5, −2, −3, −4)(5, −1, 2, 4)(5, −1, 3, −4)(5, 1, 2, 4)(5, 1, 3, −4)(5, 2, −5, 4)(5, 2, −3, −4)(5, 4, −3, 4)(5, 4, 2, −3)[(0, 0) , (16, −10) , (−16, 10)][(0, 0) , (16, −4) , (−16, 4)][(0, 0) , (16, 4) , (−16, −4)][(0, 0) , (16, 10) , (−16, −10)][(0, 0) , (15, −3) , (−15, 3)][(0, 0) , (15, 3) , (−15, −3)][(0, 0) , (12, −9) , (−12, 9)][(0, 0) , (−12, 10) , (12, −10)][(0, 0) , (12, −3) , (−12, 3)][(0, 0) , (16, −6) , (−16, 6)][(0, 0) , (12, −3) , (−12, 3)][(0, 0) , (−16, 4) , (16, −4)][(0, 0) , (12, 3) , (−12, −3)][(0, 0) , (−16, −4) , (16, 4)][(0, 0) , (12, 3) , (−12, −3)][(0, 0) , (16, 6) , (−16, −6)][(0, 0) , (12, 9) , (−12, −9)][(0, 0) , (12, 10) , (−12, −10)]c,dOn the curves Ha,b
we consider the question of long arithmetic progressions, we have the following statement.

Theorem 3. There exist infinitely many tuples (a, b, c, d), a, b, c, d ∈ Zc,dsuch that on the curve Ha,bthere is a length 9 arithmetic progression
formed by x-coordinates of integral points of the curve.
3. proof of the results
Proof of Theorem 1. Consider the case a = b. We obtain thataxy(y − x) = x − y.
Therefore x = y is a solution for all x ∈ Z. Assume that x 6= y. We getthat axy = −1. Hence (a, b, x, y) ∈ {(−1, −1, ∓1, ±1), (1, 1, ∓1, ±1)}
are the possible solutions of the equation, and one can check that these
are in fact solutions.
We may assume that |a| > |b|. We rewrite the equation in the form
byx2 + (1 − ay 2 )x − y = 0.
Thus there exists an integer t such that(1)

F (y) := a2 y 4 + (4b − 2a)y 2 + 1 = t2 .

INTEGRAL POINTS ON HUFF CURVES

7

This equation satisfies Runge’s condition so we apply Runge’s method
to determine all the integral solutions. Define P (y) = ay 2 + 2b−a
. Weahave that2
4b 2 4b2 4b1
1= 2y 2 +− − 2 + 2 − 2,
F (y) − P (y) −
a
a
a
a
a
a
2

1
4b 2 4b2 4b1+ − 2 − 2 − 2.
F (y) − P (y) += −2y 2 +
a
a
a
a
a
a
These two quadratic polynomials have opposite signs if |y| ≥ 3, since|a| > |b|. Therefore one has that
2

2

1
1
2
< F (y) = t < P (y) +P (y) −aa
if |y| ≥ 3. It yields that t = ay 2 + 2b−a
. Equation (1) implies that b = 0.
a
In this case
(
)
r
−1
1
1
y∈
±
+
2ax
4a2 x2 aand we obtain that |y| ≤ 1. It remains to check the cases y ∈ {0, ±1, ±2}.
One gets that (x, y) = (∓1, ±1) and b = 2 − a.
c,dProof of Theorem 2. Rewrite the equation of Ha,b
as follows

axy 2 − b(x2 − d)y − acx = 0.
Hence there exists an integer u for which
G(X) := X 4 + (4a2 c − 2b2 d)X + b4 d2 = u2 ,where X = bx. Let R(X) = X 2 +2a2 c−b2 d. Let R(X) = X 2 +2a2 c−b2 d.
We obtain that
G(X) − (R(X) − 162)2 = 324 x2 − 4 a4 c2 + 4 a2 b2 cd +648 a2 c − 324 b2 d − 26244,
G(X) − (R(X) + 162)2 = −324 x2 − 4 a4 c2 + 4 a2 b2 cd −648 a2 c + 324 b2 d − 26244.
The roots of the above polynomials are defined in Theorem 2 as L1 , U1
and L2 , U2 . If x is not an element of the interval [min(L1 , L2 ), max(U1 , U2 )],then
G(X) > (R(X) − 162)2

and G(X) < (R(X) + 162)2 .

8SZ. TENGELY

Since G(X) = u2 we get that u = ±(R(X) + t) for some integer |t| <162. It follows thatq
2 2
2 a4 c2− 2 a tb cd − 2 a2 c + b2 d + 12 ttx=±t ∈ {−161, . . . , 161}.
b
It remains to bound the ”small” solutions, that is to compute min(L1 , L2 )and max(U1 , U2 ), these are roots of the above defined polynomials. 
Proof of Theorem 3. Based on numerical experience we fix b = ma andd = a + 1 for some integer m. The integral point (0, 0) is on the curvec,dfor any integral tuple (a, b, c, d). If we have an integral solution
Ha,bwith x = 1, thenn2 − m2 a2c=4
for some integer n. In a similar way x = 2 corresponds to an integral
2
2 a2
solution if 2y 2 − m(3 − a)y − n −m
= 0. Hence 4n2 − 3m2 (a2 + 2a − 3)2
is a square. We look for solutions of the form n = ua + v for someu, v ∈ Z. We get that(v − u)2 − 4u2 + 3m2 = 0.
Parametric solution of the above equation is given by−2p2 + 6q 2,
Gp,qp2 + 3q 2
u =
,
Gp,q
4pq
,
m =
Gp,q

v−u =

for some integers p, q, where Gp,q = gcd(−2p2 + 6q 2 , p2 + 3q 2 , 4pq). We
deal with the case given by Gp,q = 1. To obtain an integral solution
with x = 3 the polynomial
9(a2 − 2a + 1)p4 − 2(37a2 + 74a − 431)p2 q 2 + 81(a2 + 6a + 9)q 4
has to be a square. Computing discriminant we have that 2560(a +7)(a + 4)(a − 2)(a − 5) = 0, therefore a ∈ {−7, −4, 2, 5}. Using theabove formulas we obtain that if a = −7, then(x, y) = (3, 4p2 + 10pq − 6q 2 )is a point on the curve, if a = −4, then5
3(x, y) = (3, p2 + 8 pq + q 2 )22

INTEGRAL POINTS ON HUFF CURVES

9

is a point, if a = 2, then


1 2
15 2
(x, y) = 3, p + 4 pq +q2
2
is a point on the curve and if a = 5, then(x, y) = (3, 2p2 + 2pq − 12q 2 )is a solution. We handle the latter two cases, the first two can betreated in a similar way. If p4 + 135p2 q 2 + 225q 4 is a square, then we
get a rational point on the curve with a = 2, x = 4. Hence we have agenus 1 curve which can be transformed to the elliptic curve
E1 :Y 2 = X 3 − 60750X 2 + 877078125X.

We use the computer algebra software Magma [8] to compute the maps
between curves and ranks of elliptic curves. The rank of the MordellWeil group is 2. The points (0, 0), (37125, 0) generate the torsion subgroup and the free part is generated by(18900, 1275750), (39375, −1181250).
These points map to +∞, (0, −15) and (20/3, −815), (−15, 285). It remains to provide infinitely many rational points on E1 which map torational points having an x-coordinate with odd numerator and denominator. The points (2k + 1)(39375, −1181250) have this property
for k ∈ Z. To avoid cases with abcd(a2 c − b2 d) = 0 we need points withx-coordinate different from ±1, ±3, ±5, ±15. As examples we computethe cases with k = 3, 5.
(p, q)(−182745, 68681)(0, 0)(±1, ±102280403100)
points (±2, ±6129121350)(±3, ±1871528340)(±4, ±1164860400)(−2384753104425, 47115188177959)(0, 0)(±1, ±17101083066610011832819307220)(±2, ±16533605562876168234812883720)(±3, ±16202218336426365555936358620)(±4, ±15931270770270530449534665405)

We note that instead of odd multiples of the point (39375, −1181250)
one may use points of the form
(2k + 1)(18900, 1275750) + (2l + 1)(39375, −1181250)
to obtain appropriate rational points.
If 4p4 − 27p2 q 2 + 144q 4 is a square, then we get an integral point on
the curve with a = 5, x = 4. Again we obtain an elliptic curve. It is
given by
E2 :Y 2 = X 3 + 1990656X 2 − 2140353331200X.

10

SZ. TENGELY

The rank of the Mordell-Weil group of E2 is 1. The torsion subgroupis generated by the points (0, 0), (774144, 0) and the free part by the
point (−552960, 1274019840). Any point of the form
k1 (−552960, 1274019840) + k2 (0, 0) + k3 (774144, 0),
where k1 ∈ Z, k2 , k3 ∈ {0, 1} maps to appropriate rational point on
the quartic model. We note that in this case abcd(a2 c − b2 d) 6= 0
if pq 6= ±1, ±2, ±3, ±6. Let us consider the cases with (k1 , k2 , k3 ) =(3, 0, 0) and (4, 0, 0). We have that (p, q) = (16, 3) and in the lattercase (−66, 35).
(p, q)(16, 3)(−66, 35)(0, 0)(0, 0)(±1, ±140) (±1, ±46512)
points (±2, ±308) (±2, ±10608)(±3, ∓500) (±3, ±1368)(±4, ∓700) (±4, ±612)
We note that if a = −7, −4, then the corresponding elliptic curves havepositive rank as well (1 and 2).
Acknowledgement. Szabolcs Tengely was supported by the European Union and the State of Hungary, co-financed by the European
Social Fund in the framework of TÁMOP 4.2.4. A/2-11-1-2012-0001”National Excellence Program”.
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Mathematical Institute
University of Debrecen
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Hungary
E-mail address: tengely@science.unideb.hu