ON INTEGRAL POINTS ON BIQUADRATIC CURVES AND NEAR MULTIPLESOF SQUARES IN LUCAS SEQUENCES
MAX A. ALEKSEYEV AND SZABOLCS TENGELY
Abstract. We describe algorithmic reduction of the search for integral points on a curvey2 = ax4 + bx2 + c with ac(b2 − 4ac) , 0 to solving a finite number of Thue equations. While
existence of such reduction is anticipated from arguments of algebraic number theory, ouralgorithm is elementary and to best of our knowledge is the first published algorithm ofthis kind. In combination with other methods and powered with existing software Thue
equations solvers, it allows one to efficiently compute integral points on biquadratic curves.
We illustrate this approach with a particular application of finding near multiples ofsquares in Lucas sequences.
As an example, we establish that among Fibonacci numbers only 2 and 34 are of the form
2m2 + 2; only 1, 13, and 1597 are of the form m2 − 3; and so on.
As an auxiliary result, we also give an algorithm for solving a Diophantine equationf (m,n)2
k = g(m,n) in integers m, n, k, where f and g are homogeneous quadratic polynomials.

1. Introduction
Siegel [23] proved that any equation y2 = f (x) with irreducible polynomial f ∈ Z[x]of degree at least 3 has finitely many integral points. With the method of Baker [2, 3, 4],
it became possible to bound the solutions and perform an exhaustive search. For thirddegree curves, Baker’s method was a subject to many practical improvements, and now
there exists a number of software implementations for finding integral points on elliptic
curves [6, 15, 24]. These procedures are based on a method developed by Stroeker andTzanakis [26] and independently by Gebel, Pethő and Zimmer [14].
Thue equations of the form g(x, y) = d where g is homogeneous irreducible polynomialof degree at least 3 were first studied by Thue [28], who proved that they have only a finite
number of solutions. In computer era, Thue equations became a subject to developmentsof computational methods, resulting in at least two implementations: in computer algebra systems MAGMA [6] and PARI/GP [27]. For our practical computations, we chose
SAGE [24], which adopts PARI/GP Thue equations solver based on Bilu and Hanrot’s
improvement [5] of Tzanakis and de Weger’s method [29].
In the current work, we show how to reduce a search for integral points on a biquadratic
curve
y2 = ax4 + bx2 + cwith integer (or, more generally, rational) coefficients a, b, c with ac(b2 − 4ac) , 0 firstly tof (m,n)a Diophantine equation k2 = g(m,n) in coprime integers m, n with homogeneous quadratic
polynomials f and g (Theorem 5), and then to a finite number of Thue equations (Theorem 4). While possibility of reduction to Thue equations was described by Mordell [17]based on arguments from algebraic number theory, to the best of our knowledge, there is1

no published algorithm applicable for the general case. Furthermore, in contrast to traditional treatment of this kind of problems with algebraic number theory [17, 25, 11, 18], ourreduction method is elementary. It may be viewed as a generalization of the one of Steiner
and Tzanakis [25] who reduced Ljunggren equation y2 = 2x4 − 1 to two Thue equations(in particular, we obtain the same Thue equations).
There are other methods which can be used in certain cases to determine all integral
solutions of the equation y2 = ax4 + bx2 + c. Poulakis [19] provided an elementary algorithm to solve Diophantine equations of the form y2 = f (x), where f (x) is quartic monicpolynomial with integer coefficients, based on Runge’s method [21, 22, 30]. Here it is
crucial that the leading coefficient of f (x) is 1 (the idea also works if the leading coefficientis a perfect square). Using the theory of Pell equations, Kedlaya [16] described a method
to solve the system of equations( 2
x − a1 y2 = b1 ,P(x, y) = z2 ,where P is a given integer polynomial, and implemented his algorithm in Mathematica(http://math.ucsd.edu/∼kedlaya/papers/pell.tar). If we set P(x, y) = c1 x + d1 , thenwe obtain a quartic equation of the form (a1 c1 y)2 = a1 z4 − 2a1 d1 z2 − a1 b1 c21 . There is also a
simple reduction of the equation y2 = ax4 + bx2 + c to an elliptic equation: after multiplying
the equation by a2 x2 , one obtains:(xy)2 = (ax2 )3 + b(ax2 )2 + ac(ax2 ),which can be further written as
Y2 = X3 + bX2 + acX.
As we noted earlier to determine all integral points on a given elliptic curve one can follow
a method developed by Stroeker and Tzanakis [26] and independently by Gebel, Pethő
and Zimmer [14]. The disadvantage of this approach is that there is no known algorithm
to determine the rank of the so-called Mordell-Weil group of an elliptic curve, which isnecessary to determine all integral points on the curve.
For efficient computation of integral points on biquadratic curves, we implemented
in SAGE a combination of the elliptic curve and reduction to Thue equations methods. By default we employ the elliptic curve method and if it fails, we fallback
to our reduction to Thue equations. The SAGE code can be downloaded fromhttp://www.math.unideb.hu/∼tengely/biquadratic.sage.
Our approach also allows one to efficiently compute solutions to a system of Diophantine equations (Theorem 6):( 2
a1 x + c1 z = d1 ,b2 y2 + c2 z2 = d2 .
From this perspective, it continues earlier work [1], where the first author described analgorithm for computing solutions to a system of Diophantine equations:( 2
a1 x + b1 y2 + c1 z2 = d1 ,a2 x2 + b2 y2 + c2 z2 = d2 ,and demonstrated applications for finding common terms of distinct Lucas sequencesof the form U(P, ±1) or V(P, ±1), which include Fibonacci, Lucas, Pell, and Lucas-Pellnumbers. The current method also has applications for such Lucas sequences, allowing
one to find all terms of the form a · m2 + b for any fixed integers a, b. While the question offinding multiples of squares (i.e., with b = 0) in Lucas sequences has been widely studied,starting with the works of Cohn [9] and Wyler [31] (we refer to Bremner and Tzanakis [7]for an extensive review of the literature), finding near multiples of squares (i.e., with b , 0)
has got so far only a limited attention [12, 13, 20]. In the current work, we present an
unified computational approach for solving this problem. As an example, we establishthat among Fibonacci numbers only 2 and 34 are of the form 2m2 + 2; only 1, 13, and 1597
are of the form m2 − 3; and so on.
The paper is organized as follows. In Section 2, we develop our machinery for homogeneous quadratic polynomials with integer coefficients. In Section 3, we prove our method
for finding integral points on biquadratic curves and illustrate its workflow on Ljunggren
equation. In Section 4, we further demonstrate how our method can be used for finding
near multiples of squares in Lucas sequences and list some results of this kind.
2. Homogeneous quadratic polynomials
We start with studying properties of quadratic homogeneous polynomials with integer
coefficients in two and three variables. We do not distinguish between homogeneouspolynomials in two variables from their univariate counterparts (i.e., f (x, y) = ax2 +bxy+cy2
and f˜(z) = az2 + bz + c) that allows us to define resultant (Res) and discriminant (Disc) on
them.
Theorem 1 (Theorem 5 in [1]1). Let A, B, C be non-zero integers and let (x0 , y0 , z0 ) with z0 , 0
be a particular non-trivial integer solution to the Diophantine equation Ax2 + By2 + Cz2 = 0. Then
its general integer solution is given by(1)(x, y, z) =p
(Px (m, n), P y (m, n), Pz (m, n))qwhere m, n as well as p, q are coprime integers with q > 0 dividing 2 lcm(A, B)Cz20 , andPx (m, n) = x0 Am2 + 2y0 Bmn − x0 Bn2 ,P y (m, n) = −y0 Am2 + 2x0 Amn + y0 Bn2 ,Pz (m, n) = z0 Am2 + z0 Bn2 .
We refer to [8, 10] for general methods of finding a particular solution to a quadratic
homogeneous equation in three variables.2
Theorem 2. Let P1 (x, y) and P2 (x, y) be homogeneous quadratic polynomials with integer coefficients and R = Res(P1 , P2 ) , 0. Let G be the largest element in the Smith normal form of
the resultant matrix of P1 and P2 .3 Then for any coprime integers m, n, gcd(P1 (m, n), P2 (m, n))divides G.
1This theorem corrects an error in Corollary 6.3.8 of [8].
In PARI/GP, a particular solution can be computed with the function bnfisnorm.
3
While G = R would also satisfy the theorem statement, we want G as small as possible. The resultant R
is often much larger than G defined in the theorem.
2

Proof. Let P1 (x, y) = a1 x2 +b1 xy+c1 y2 and P2 (x, y) = a2 x2 +b2 xy+c2 y2 where a1 , b1 , c1 , a2 , b2 , c2
are integer. Consider polynomials x · P1 (x, y), y · P1 (x, y), x · P2 (x, y), y · P2 (x, y) as linear
combinations with integer coefficients of the basis terms x3 , x2 y, xy2 , y3 . Our goal is to find
two linear combinations of these polynomials: one equal an integer multiple of x3 and theother equal an integer multiple of y3 . This corresponds to the following two systems oflinear equations with the same resultant matrix:
 
 
  

x3 :
0
1
a1 0 a2 0  t1 






2
0






x y:
0
b1 a1 b2 a2  t2 (2) ·   =   or  

2
xy :
0
0
c1 b1 c2 b2  t3 
 
 
  

3
1
0
t4
0 c1 0 c2
y :with respect to rational numbers t1 , t2 , t3 , t4 . Since the matrix determinant equals theresultant R , 0, the systems have unique solutions of the form:1
1 ∆1 ∆2 ∆3 ∆4(t1 , t2 , t3 , t4 ) = (∆1 , ∆2 , ∆3 , ∆4 ) =, , ,RG d3 d3 d3 d3
where ∆i are determinant of certain 3 × 3 minors of the resultant matrix and d3 is its third
determinant divisor. Here we used the fact that dR3 = G is the fourth elementary divisor of
the resultant matrix (and the largest element of its Smith normal form). It is important to
notice that all vector components ∆d3i are integer.
So we have two linear combinations of x · P1 (x, y), y · P1 (x, y), x · P2 (x, y), y · P2 (x, y) withinteger coefficients ∆d31 , ∆d32 , ∆d33 , ∆d34 equal G·x3 and G· y3 , respectively. For (x, y) = (m, n) wherem, n are coprime integers, these linear combinations imply that gcd(P1 (m, n), P2 (m, n))divides both G · m3 and G · n3 . Therefore, gcd(P1 (m, n), P2 (m, n)) divides gcd(G · m3 , G · n3 ) =G · gcd(m3 , n3 ) = G.
We remark that in practice to compute G, we do not need to compute Smith normal
form of the resultant matrix. Instead, we simply solve the two linear systems (2) anddefine G as the least common multiple of all denominators in both solutions (t1 , t2 , t3 , t4 ).
Theorem 3. Any homogeneous quadratic polynomial with integer coefficients and non-zero discriminant can be represented as a linear combination with non-zero rational coefficients of squaresof two homogeneous linear polynomials. Moreover, these polynomials are linearly independent.
Proof. Let ax2 + bxy + cy2 be a homogeneous quadratic polynomial with integer coefficientsand non-zero discriminant, i.e., b2 − 4ac , 0. If b = 0 then ac , 0 and the statement istrivial.
Suppose that b , 0. If a , 0, then we have
1 
ax2 + bxy + cy2 =· (2ax + by)2 + (4ac − b2 )y2 .
4aSimilarly, if c , 0, then we have
1 
ax2 + bxy + cy2 =· (4ac − b2 )x2 + (bx + 2cy)2 .
4cFinally, if a = c = 0, thenb 
bxy = · (x + y)2 − (x − y)2 .
4
It is easy to see that in all cases, the linear polynomials are linearly independent.
Theorem 4. Let P1 (x, y) and P2 (x, y) be homogeneous quadratic polynomials with integer coefficients such that Disc(P1 ) , 0, Disc(P2 ) , 0, and Res(P1 , P2 ) , 0. Then the equation
z2 =

(3)

P1 (x, y)
P2 (x, y)has a finite number of integer solutions (x, y, z) = (m, n, k) with gcd(m, n) = 1.
Proof. Suppose that (x, y, z) = (m, n, k) with gcd(m, n) = 1 satisfies the equation (3). Thenby Theorem 2, gcd(P1 (m, n), P2 (m, n)) = P2 (m, n) divides certain integer G. Then for somedivisor4 g of G, we have P2 (m, n) = g and P1 (m, n) = gk2 . So (x, y, z) = (m, n, k) represents asolution to the following system of equations:(
P1 (x, y) = g · z2 ,(4)
P2 (x, y) = g.
Therefore, to find all solutions to (3), we need to solve the systems (4) for g ranging overthe divisors of G. Each such system is solved as follows.
We start with using Theorem 3 to represent P1 (x, y) as a linear combination with rationalcoefficients of squares of two linearly independent polynomials, say, P1 (x, y) = a·Q1 (x, y)2 +b · Q2 (x, y)2 . Substituting this representation into the first equation of (4), we get thefollowing equation:a · Q1 (x, y)2 + b · Q2 (x, y)2 − g · z2 = 0.

(5)p

We solve this equation using Theorem 1 to obtain Q1 (x, y) = q · R1 (m, n) and Q2 (x, y) =p
· R2 (m, n), where q ranges over the positive divisors of a certain integer and integer p isqcoprime to q. We solve these linear equations with respect to x, y to obtainp (x, y) = · Sx (m, n), S y (m, n) ,qwhere Sx (m, n) and S y (m, n) are linear homogeneous polynomials with rational coefficients(which depend only on g but not p, q). Plugging this into the second equation of (4), we
get the following quartic equations with integer coefficients w.r.t. m, n:` g · P2 (Sx (m, n), S y (m, n)) = q2 ·(6)

g · `g.
p2

where ` g is the least common multiple of the coefficients denominators of
P2 (Sx (m, n), S y (m, n)) (notice that ` g depends only on g but not p, q). Here p2 must divide g` and thus there is a finite number of such equations. By Theorem 3 in [1], eachsuch equation has a finite number of solutions, unless P2 (Sx (m, n), S y (m, n)) = c · T(m, n)2
for some polynomial T(x, y), which is not the case since Disc(P2 ) , 0.
From practical perspective, we remark that different choices of values for g, p, q may
result in the same equation (6). In particular, if g0 = g · d2 for some integer d, then we canrepresent equation (5) for g0 in the form a · Q1 (x, y)2 + b · Q2 (x, y)2 − g · (d · z)2 = 0 so thatit has the same solutions w.r.t. Q1 (x, y) and Q2 (x, y). Then the equation (6) for g0 has thesame left hand side as the one for g (with ` g0 = ` g ), while the former has an extra factor
d2 in the right hand side. Therefore, to reduce the number of equations, we can restrict g
4Unless specified otherwise, the divisors of an integer include both positive and negative divisors.

to the square-free divisors of G: for each such g, we compute the left hand side of (6) anditerate over all distinct integer right hand sides of the form q2 ·g·` g ·d2
, where d2 divides Gg .
p2

3. Finding integral points on biquadratic curves
Now we are ready to prove our main result:
Theorem 5. Finding integral points on a biquadratic curve(7)

y2 = a · x4 + b · x2 + cwith integer coefficients a, b, c, ac(b2 − 4ac) , 0, reduces to solving a finite number of quartic Thue
equations.
Proof. Multiplying the equation (7) by 4c, we can rewrite it as a linear combination of threesquares with non-zero integer coefficients:5(b2 − 4ac)(x2 )2 + 4cy2 − (bx2 + 2c)2 = 0.
Denoting X = x2 , Y = y, Z = bx2 + 2c, A = b2 − 4ac, B = 4c, C = −1, we get a Diophantine
equation :(8)

A · X2 + B · Y2 + C · Z2 = 0.

If this equation is solvable with a particular solution (X, Y, Z) = (X0 , Y0 , Z0 ), Z0 , 0, thenby Theorem 1 its general solution is given by:(X, Y, Z) = r · Px (m, n), P y (m, n), Pz (m, n) ,where Pi (m, n) are polynomials defined by (2), m, n are coprime integers, and r is a rational
number. In our case, this solution should additionally satisfy the relation:
2c = Z − bX = r · (Pz (m, n) − b · Px (m, n)) ,implying that2c.
Pz (m, n) − b · Px (m, n)
So we get a constraining Diophantine equation:r=

2c · Px (m, n),Pz (m, n) − b · Px (m, n)which reduces to a finite number of Thue equations by Theorem 4, unless Res(Px , Pz − b ·Px ) = 0 or Disc(Pz − b · Px ) = 0.
The case of Res(Px , Pz − b · Px ) = Res(Px , Pz ) = 0 is impossible since by direct computation
we have Res(Px , Pz ) = −4AB2 CZ40 , 0.
In the case of Disc(Pz − b · Px ) = 0, we have Disc(Pz − b · Px ) = 4b2 B2 Y02 − 4AB(Z20 −b2 X02 ) = 0 and thus b2 BY02 = A(Z20 − b2 X02 ). Since BY02 = −AX02 − CZ20 , we further getb2 (−AX02 − CZ20 ) = A(Z20 − b2 X02 ), which reduces to A + b2 C = 0. However this is impossible
since A + b2 C = −4ac , 0.
Therefore, Theorem 4 is applicable.

(9)

5

x2 =

Alternatively, we can multiply (7) by 4a and obtain another linear combination of three squares: (2ax2 +b) + (4ac − b2 ) · 12 − 4ay2 = 0, which has smaller coefficients and thus may be preferrable when c  a.
2As a corollary we get:
Theorem 6. The following system of Diophantine equations:(z = ax2 + d1 ,z2 = by2 + d2 ,where a, b, d1 , d2 are rational numbers with abd2 (d21 − d2 ) , 0 reduces to a finite number of Thue
equations.
Proof. The system implies that (ax2 + d1 )2 = by2 + d2 or (by)2 = a2 bx4 + 2abd1 x2 + b(d21 − d2 ).
Since (2d1 ab)2 − 4a2 b2 (d21 − d2 ) = 4a2 b2 d2 , 0, Theorem 5 applies.

Example: Ljunggren equation. We illustrate our method on Ljunggren equation y2 =2x4 − 1 whose reduction to Thue equations was first obtained in [25]. Here we have(a, b, c) = (2, 0, −1). First we compute A = b2 − 4ac = 8, B = 4c = −4, and C = −1 andconsider equation (8). Its particular solution is (1, 1, 2) yields by Theorem 1 a generalsolution:p(X, Y, Z) = (Px (m, n), P y (m, n), Pz (m, n))q
withPx (m, n) = 8m2 − 8mn + 4n2 ,P y (m, n) = −8m2 + 16mn − 4n2 ,Pz (m, n) = 16m2 − 8n2 .
Now we consider equation (9):x2 =2cPx (m, n)−2(8m2 − 8mn + 4n2 ) −2m2 + 2mn − n2==
Pz (m, n) − b · Px (m, n)
16m2 − 8n2
2m2 − n2and use Theorem 4 to solve it. We take the resultant matrix of P1 (x, y) = −2x2 +2xy− y2 andP2 (x, y) = 2x2 − y2 and solve the two linear systems (2) to obtain (t1 , t2 , t3 , t4 ) = 41 (−2, −1, 0, 1)
and (t1 , t2 , t3 , t4 ) = (−1, −1, −1, 0). So we get G = 4. Let g range over the divisors of G, i.e.,g ∈ {−4, −2, −1, 1, 2, 4}.
We use Theorem 3 to represent P1 (x, y) as a linear combination of squares:


1
1
P1 (x, y) = −2x2 + 2xy − y2 = − (−4x + 2y)2 + 4y2 = − (−2x + y)2 + y2
82
and obtain the equation (5) (multiplied by −2):(−2x + y)2 + y2 + 2gz2 = 0.
Clearly, it may have non-trivial solutions only for g < 0 and only when −1 is a squaremodulo 2g, which leaves us the only suitable value g = −1. The corresponding equation
has a particular solution (1, 1, 1) and by Theorem 1 its general solution is:p (−2x + y, y, z) = · m2 + 2mn − n2 , −m2 + 2mn + n2 , m2 + n2q
where (p, q) = 1 and q > 0 divides 4.

pp

From this solution we express x = q · (−m2 + n2 ) and y = q · (−m2 + 2mn + n2 ) and plugthem into the equation P2 (x, y) = g to obtain the following quartic equations:m4 + 4m3 n − 6m2 n2 − 4mn3 + n4 = q2 ·−1
p2

and conclude that p2 = 1. Since the left hand side is irreducible, these are Thue equations.
We used PARI/GP to solve the resulting three Thue equations (for q = 1, 2, 4) and foundthat only the one with q = 2 has solutions, which are (m, n) = ±(5, −1), ±(1, 5), (±1, ±1). The
P (m,n)corresponding solutions to P21 (m,n) = x2 are (m, n) = ±(12, 17) and ±(0, 1), giving respectively
x = ±13 and ±1.
4. Near multiples of squares in Lucas sequences
The pair of Lucas sequences U(P, Q) and V(P, Q) are defined by the same linear recurrent
relation with the coefficient P, Q ∈ Z but different initial terms:
U0 (P, Q) = 0, U1 (P, Q) = 1, Un+1 (P, Q) = P · Un (P, Q) − Q · Un−1 (P, Q), n ≥ 1;
V0 (P, Q) = 2, V1 (P, Q) = P, Vn+1 (P, Q) = P · Vn (P, Q) − Q · Vn−1 (P, Q), n ≥ 1.
Some Lucas sequences have their own names:
Sequence
U(1, −1)V(1, −1)U(2, −1)
V(2, −1)

Name
Fibonacci numbers
Lucas numbers
Pell numbers
Pell-Lucas numbers

Initial terms
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . .
2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, . . .
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, . . .
2, 2, 6, 14, 34, 82, 198, 478, 1154, . . .

Other examples include Jacobsthal numbers U(1, −2), Mersenne numbers U(3, 2) etc.
The characteristic polynomial of Lucas sequences U(P, Q) and V(P, Q) is λ2 − Pλ + Q
with the discriminant D = P2 − 4Q. For non-degenerate sequences, the discriminant D isa positive non-square integer.
Terms of Lucas sequences satisfy the following identity:(10)

Vn (P, Q)2 − D · Un (P, Q)2 = 4Qn .

In the current paper, we focus on the case of Q = 1 or Q = −1, implying that the pairs(Vn (P, Q), Un (P, Q)) satisfy the equation:6(11)x2 − Dy2 = ±4.

The converse statement can be used to prove that given positive integers belong to V(P, Q)
or U(P, Q) respectively:
Theorem 7 (Theorem 1 in [1]). Let P, Q be integers such that P > 0, |Q| = 1, (P, Q) , (3, 1),
and D = P2 − 4Q > 0. If positive integers u and v are such thatv2 − Du2 = ±4,
then u = Un (P, Q) and v = Vn (P, Q) for some integer n ≥ 0.
6

Here and everywhere below ± in the r.h.s. of an equation means that we accept both signs as solutions.

Theorem 8. For fixed integers a , 0 and b, finding terms of the form am2 + b in nondegenerate
Lucas sequences U(P, Q) or V(P, Q) with Q = ±1 reduces to a finite number of Thue equations,
unless this sequence is V(P, Q) and b = ±2. Thus there are only a finite number of such terms inthese Lucas sequences.
Proof. Let ax2 + b be a term of U(P, ±1). By Theorem 7 we have y2 = D(ax2 + b)2 ± 4 =Da2 x4 + 2Dabx2 + (Db2 ± 4). Theorem 5 applies, since Db2 ± 4 , 0 (notice that D , ±1) and(2Dab)2 − 4Da2 (Db2 ± 4) = ∓16Da2 , 0.
Now let ax2 + b be a term of V(P, ±1). By Theorem 7 we have (Dy)2 = D((ax2 + b)2 ∓ 4) =Da2 x4 + 2Dabx2 + D(b2 ∓ 4). We have (2Dab)2 − 4D2 a2 (b2 ∓ 4) = ±16D2 a2 , 0. Theorem 5
applies here as soon as b2 ∓ 4 , 0 (i.e., b , ±2).

We remark that for b = ±2, the sequence V(P, Q) may have an infinite number ofterms of the form am2 + b. In particular, Lucas sequence V(1, −1) (Lucas numbers) has
infinitely many terms of the forms m2 + 2 and m2 − 2 since V4n (1, −1) = V2n (1, −1)2 − 2 andV4n+2 (1, −1) = V2n+1 (1, −1)2 + 2. The following theorem allows to find all terms solutionsof the form am2 ± 2 in V(P, ±1).
Theorem 9. For fixed integers a , 0 and b = ±2, finding terms of the form am2 +b in nondegenerate
Lucas sequence V(P, Q) with Q = ±1 reduces to a finite number of Thue equations and a PellFermat equation, which may or may not have infinitely many solutions.
Proof. Let ax2 + b with b = ±2 be a term of V(P, ±1). By Theorem 7, we have the following
equations:(Dy)2 = D((ax2 + b)2 + 4) = Da2 x4 + 2Dabx2 + 8D
and(Dy)2 = D((ax2 + b)2 − 4) = Da2 x4 + 2Dabx2 .
The former equation is addressed by Theorem 5, while the latter equation always has
solution x = 0 (corresponding to the term V0 = 2) and for x , 0 is equivalent to thePell-Fermat equation:
 Dy 2x− Da2 x2 = 2Dab.

For solution of this equation, we refer to Section 6.3.5 in [8].



In the table below we list all the terms of the form am2 + b for 1 ≤ a ≤ 3 and −3 ≤ b ≤ 3
in Fibonacci, Lucas, Pell, and Pell-Lucas numbers:

Form
m2
m2 + 1m2 − 1m2 + 2m2 − 2
m2 + 3m2 − 32m2
2m2 + 12m2 − 12m2 + 22m2 − 22m2 + 32m2 − 33m2
3m2 + 13m2 − 13m2 + 23m2 − 23m2 + 33m2 − 3

Fibonacci numbers
0, 1, 144
1, 2, 50, 3, 8
2, 3
2, 34
3
1, 13, 1597
0, 2, 8
1, 3
1
2, 34
0
3, 5, 21
5
0, 3
1, 13
2
2, 5
1
3
0, 144

Lucas numbers
1, 4
1, 2
3
2, 11, and V4n+2
V4n3, 4, 7, 199
1
2, 18
1, 3
1, 7, 199
2, 4
none
3, 11
29, 47, 64079
3
1, 4, 76
2, 11, 47
2, 29
1
3
none

Pell numbers
0, 1, 1691, 2, 5
0
2
2
12
1
0, 2
1
1
2
0, 70
5
5, 29
0, 12
1
2
2, 5, 291none
0Pell-Lucas numbers
none
2, 82
none
2 and V4n+2
14 and V4n
none
6
2
none
none
2 and V4n
V4n+2
none
none
none
none
2
2, 14
none
6
none

Acknowledgements
We thank Dr. John Cremona, Dr. Nikos Tzanakis, and Dr. Noam Elkies for a numberof helpful suggestions and discussions.
The first author was supported by the National Science Foundation under Grant No.
IIS-1253614. The second author was partially supported by the European Union and theEuropean Social Fund through project Supercomputer, the national virtual lab (grant no.:TÁMOP-4.2.2.C-11/1/KONV-2012-0010).
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Max A. Alekseyev
University of South Carolina, Columbia, SC, USA
E-mail address: maxal@cse.sc.edu
Szabolcs Tengely
Mathematical Institute, University of Debrecen, Hungary
E-mail address: tengely@science.unideb.hu