PORTUGALIAE MATHEMATICA
Vol. 55 Fasc. 3 – 1998
THE DIOPHANTINE
EQUATIONS
µ 2
¶
b
±2
x2 − k = 2 · T n
2
Gheorghe Udrea
It is the object of this note to demonstrate that the two equations of the
title have only finitely many solutions in positive integers x and n for any given
integers b and k, k 6= ±2. In these equations (Tn (x))n≥0 is the sequence of the
Chebyshev polynomials of the first kind.
1 – Chebyshev polynomials of the first kind, (Tn (x))n≥0 , are defined by the
recurrence relation
(1.1)
Tn+1 (x) = 2x · Tn (x) − Tn−1 (x) ,
∀ x ∈ C, ∀ n ∈ N ∗ ,
where T0 (x) = 1 and T1 (x) = x, C being the set of complex numbers.
Also, we have the sequence (Ten (x))n≥0 of polynomials “associated” of the
Chebyshev polynomials (Tn (x))n≥0 , defined as it follows:
(1.2)
Ten+1 (x) = 2x · Ten (x) + Ten−1 (x) ,
∀ x ∈ C, ∀ n ∈ N ∗ ,
with Te0 (x) = 1 and Te1 (x) = x.
The connection between the sequence (Ten )n≥0 and the sequence (Tn )n≥0 is
given by the simple relations:
(1.3)
where i2 = −1.

T (i · x)

 Tek (x) = k
,
k


i
Tk (x) =
Tek (i·x)
ik
,
k ∈ N, x ∈ C ,
Received : November 12, 1996; Revised : April 29, 1997.
AMS Subject Classification: 11B37, 11B39, 11D25.
256
G. UDREA
Two important properties of the polynomials (Tn (x))n≥0 are given by the
formulas:
(1.4)
ϕ ∈ C, n ∈ N ,
Tn (cos ϕ) = cos n ϕ ,
and
(1.5)
∀ m, k ∈ N, ∀ x ∈ C .
Tm (Tk (x)) = Tmk (x) ,
2 – We are going to prove the following lemmas:
Lemma 1. If (Tn (x))n≥0 is the sequence of Chebyshev polynomials of the
first kind, then one has
(2.1)
µ
µ
a
Tn (a2 − 1) = 2 · Tn √
2
¶¶2
∀ n ∈ N, ∀ a ∈ C .
− 1,
Proof: Indeed, we have
µ
¶
¶
µ
µ
µ
¶¶
a
a 2
− 1 = T n T2 √
Tn (a − 1) = Tn 2 · √
2
2
µ µ
¶¶
µ µ
¶¶2
a
a
− 1,
= 2 · Tn √
= T 2 Tn √
2
2
2
Now, if we put in (2.1) a =
√b ,
2
q.e.d. .
we obtain
2 · Tn
µ
b2 − 2
2
¶
= zk2 − 2 ,
where zn = 2 · Tn ( 2b ) ∈ Z, ∀ n ∈ N, ∀ b ∈ Z.
Thus, we have
i.e.,
(2.2)
x2 − k = zn2 − 2 ,
k 6= 2 ,
x2 − zn2 = k − 2 ,
k 6= 2 .
Lemma 2. If (Ten (x))n≥0 is the sequence of polynomials “associated” of the
Chebyshev polynomials (Tn (x))n≥0 , then one has:
a)
(2.3)
b)
¶
µ 2
µ ¶
b +2
b
e
T2n
= Tn
,
2
2
b ∈ C, n ∈ N ;
µ ¶
µ ¶
b
2 b
e
e
T2n
= 2 · Tn
− (−1)n ,
2
2
b ∈ C, n ∈ N .
THE DIOPHANTINE EQUATIONS x2 − k = 2 · Tn ( b
2
257
±2
)
2
Proof: We have:
a)
b)
Te2n
µ ¶
b
2
T2n
=
µ
b
i·
2
i2n
µ
b
Te2n
2
µ
b
2
µ
b
i·
2
= (−1)n · T2n i ·
µ
¶¶
¶
µ
µ
¶
¶
b
i·b 2
= (−1)n · Tn 2 ·
= (−1) · Tn T2 i ·
−1
2
2
µ 2
¶¶
¶
µ µ 2
b +2
b
+1
= (−1)2n · Tn
, q.e.d.
= (−1)n · Tn −
2
2
n
µ ¶
¶
=
µ
T2n i ·
b
2
i2n
n
¶
µ
n
= (−1) · T2n
= (−1) · T2 Tn
Ã
µ
µ
b
i·
2
¶¶
= (−1)n · 2 · in · Ten
µ
= 2 · Ten2
µ ¶
b
2
n
b
2
µ ¶
− (−1)n ,
µ
= (−1) · 2 ·
µ ¶¶2
= (−1)n · 2 · (−1)n · Ten2
¶
b
2
−1
−1
!
Tn2
µ
b
i·
2
¶
−1
¶
¶
q.e.d. .
Now, from Lemma 2 we obtain:
2 · Tn
µ
b2 + 2
2
¶
µ
µ ¶
¶
µ ¶
b
2 b
n
e
e
= 2 · T2m
= 2 · 2 · Tn
− (−1)
2
2
b
− 2(−1)n
= 2 · Ten
2
= zen2 − (−1)n · 2 ,
µ
µ ¶¶2
where zen = 2 · Ten ( 2b ) ∈ Z, ∀ b ∈ Z, ∀ n ∈ N.
Thus, we have
(2.4)
or, equivalently,
(2.5)
x2 − k = zen2 − 2(−1)n ,
x2 − zen2 = k ± 2 ,
k 6= ±2 .
It will be observed that for given k ∈ Z, k 6= ±2, the set of values of x
satisfying equations (2.2) and (2.5) is finite and, accordingly, there are finitely
2
many values of n satisfying the equations x2 − k = 2 · Tn ( b 2±2 ), n ∈ N, b ∈ Z.
258
G. UDREA
Thus, for each given k ∈ Z, k 6= ±2, there are finitely many possible values
2
of x, n ∈ N, satisfying the equation x2 − k = 2 · Tn ( b 2+2 ), b ∈ Z. This concludes
the proof of the result of this paper.
Remarks.
α) For b = 1 in x2 − k = 2 · Tn ( b
2 +2
2
), we obtain the equation
x2 − k = L2n ,
(2.6)
n∈N,
where (Ln )n≥0 is the sequence of the Lucas numbers, defined as it follows:
Ln+1 = Ln + Ln−1 ,
L0 = 2, L1 = 1 .
Clearly, in (2.6) we utilized the identity
(2.7)
Tn
µ ¶
3
2
=
1
· L2n ,
2
∀n ∈ N .
β) If we put k = 0 in (2.6) one obtains (see (2.5)) that the numbers L2n ,
n ∈ N, are not perfect squares.
γ) For b = 4 in x2 − k = 2 · Tn ( b
(2.8)
2
x −k =
q
2 +2
2
), we obtain the equation
2 + 4,
5 · F6n
n∈N,
where (Fn )n≥0 is the sequence of the Fibonacci numbers:
Fn+1 = Fn + Fn−1 ,
F0 = 0, F1 = 1 .
In (2.8) we utilized the identities
F6n = 8 · Un−1 (9) ,
∀n ∈ N ,
and
(2.10)
2
Tn2 (x) − (x2 − 1) · Un−1
(x) = 1 ,
∀ x ∈ C, ∀ n ∈ N ∗ ,
where (Un (x))n≥0 is the sequence of Chebyshev polynomials of the second kind
(2.11)
Un+1 (x) = 2x · Un (x) − Un−1 (x) ,
with U0 (x) = 1 and U1 (x) = 2x.
x ∈ C, n ∈ N ∗ ,
THE DIOPHANTINE EQUATIONS x2 − k = 2 · Tn ( b
2
±2
)
2
259
δ) If we put k = 0 in (2.8) one obtains (see (2.5)) that the equation
(2.12)
2
x4 − 5 · F6n
=4
has not solutions (x, n) ∈ Z × N.
REFERENCES
[1] Cohn, J.H.E. – The diophantine equation (x2 − c)2 = dy 2 + 4, J. London Math.
Soc., 8(2) (1974), 253.
[2] Cohn, J.H.E. – The diophantine equations (x2 − c)2 = (t2 ± 2) · y 2 + 1, Acta
Arithmetica, 30 (1976), 253–255.
Gheorghe Udrea,
Str. Unirii-Siret, Bl. 7A, Sc. 1, Ap. 17, Tg-Jiu,
Cod 1400, Judet Gorj – ROMANIA