Positive Integer Solutions of Certain Diophantine Equations
BIJAN KUMAR PATEL
International Institute of Information Technology
Bhubaneswar-751003, India
[email protected]
PRASANTA KUMAR RAY
Sambalpur University, Jyoti-Vihar
Burla-768019, India
[email protected]
MANASI K SAHUKAR
National Institute of Technology
Rourkela-769008, India
[email protected]
Abstract
In this study, the Diophantine equations x2 −32Bn xy−32y 2 = ±32r , x4 −32Bn x2 y−32y 2 = ±32r
and x2 − 32Bn xy 2 − 32y 4 = ±32r are considered and determine when these equations have
positive integer solutions. Moreover, all positive integer solutions of these Diophantine equations
in terms of balancing and Lucas-balancing numbers are also found out.
AMS 2010 Subject Classification : 11B37, 11B39
Keywords : Diophantine equations; Balancing numbers; Balancers; Lucas-balancing numbers
1
Introduction
An important area of number theory is devoted to finding solutions of equations where the solutions
are restricted to the set of integers. Diophantine equations get their name from Diophantus of
Alexandria and they are algebraic equations for which rational or integer solutions are sought.
Many researchers considered the Diophantine equations ax2 + bxy + cy 2 = d for different fixed
integer values of a, b, c and d [2–4]. In [6], Keskin et al. consider the values of a = 1, b = −5Fn ,
c = −5(−1)n , d = ±5r which gives the Diophantine equation
x2 − Fn xy − 5(−1)n y 2 = ±5r ,
and they determined the values of n for which the equation has positive integer solutions x and y.
In a subsequent paper Keskin et al. considered a = 1, b = −Ln , c = (−1)n , d = ±5r and determine
when the equations x2 − Ln xy + (−1)n y 2 = ±5r , have positive integer solutions and then find all
positive integer solutions of the given equations in terms of Fibonacci and Lucas numbers [7].
Another interesting number sequence which is closely related to the sequence of Fibonacci
numbers is the sequence of balancing numbers. In 1999, Behera et al. introduced balancing
numbers, which satisfy the Diophantine equation 1+2+· · ·+(x−1) = (x+1)+(x+2)+· · ·+(x+y).
Here y is the balancer corresponding to balancing number x [1]. Balancing numbers initiated from
the recurrence relation Bn+1 = 6Bn − Bn−1 with B0 = 0 and B1 = 1, where Bn is the nth balancing
number. For each balancing number B, the square root of 8B 2 + 1 is also generate a sequence
of numbers Cn called as sequence of Lucas-balancing numbers, where Cn denotes the nth Lucasbalancing number. Lucas-balancing numbers recursively defined as Cn+1 = 6Cn −Cn−1 with C0 = 1
1
and C1 = 3. Pell and associated Pell numbers are well known number sequences whose recurrence
relations are given by
Pn+1 = 2Pn + Pn−1 , P0 = 0, P1 = 1 and
Qn+1 = 2Qn + Qn−1 , Q0 = 1, Q1 = 2.
Here, Pn and Qn denote the nth Pell and nth Lucas Pell numbers respectively. The balancing
and Lucas-balancing numbers are closely associate with Pell and associated Pell numbers [11]. In
particular, P2n = 2Bn and Q2n = Cn . An interesting connection concerning balancing numbers,
Pell numbers and Lucas Pell numbers is that, the nth balancing number equals the product of nth
Pell number and nth Lucas Pell number [11].
Consider generalized Fibonacci sequence Un and generalized Lucas sequence Vn as follows:
Un+1 = P Un − QUn−1 for n ≥ 1 with U0 = 0, U1 = 1
(1.1)
Vn+1 = P Vn − QVn−1 for n ≥ 1 with V0 = 2, V1 = P,
(1.2)
where P and Q are positive integers. Moreover, generalized Fibonacci and Lucas numbers for
negative subscripts are given as
U−n =
Vn
−Un
and V−n = n ,
n
Q
Q
(1.3)
with n ≥ 1. Noting that, for P = 1 and Q = −1 in (1.1) and (1.2), Un = Fn and Vn = Ln . For
P = 2 and Q = −1, Un and Vn are Pell sequence Pn and Pell-Lucas sequence Qn respectively.
Further, on substitution of P = 6 and Q = 1 into (1.1) and (1.2) gives un = Bn and the sequence
vn = 2Cn . This means that both sequence of balancing numbers and the sequence vn are special
cases of the generalized Fibonacci and Lucas sequences.
In the diophantine equation ax2 + bxy + cy 2 = d, setting a = 1, b = −8Bn , c = −2 and
d = ±2r , Karaatli et al. solved the diophantine equation x2 − 8Bn xy − 2y 2 = ±2r got solution
in terms of Pell and Pell-Lucas numbers [5]. In this article, we consider the diophantine equation
x2 − 32Bn xy − 32y 2 = ±32r by setting a = 1, b = −32Bn , c = −32 and d = ±32r . The aim of this
article is to find some new positive integer solutions of the equation given in the title in terms of
balancing and Lucas-balancing numbers.
2
Preliminaries
The following are some known results concerning the sequences Pn , Qn , Bn and vn that will be
needed later.
The following result is available in [5], that provides the information about the sum of the
squares of balancing numbers.
Lemma 2.1. Let Bk denotes the k th balancing number, then
n
X
k=1
Bk2 =
1
[B2n+1 − (2n + 1)] .
32
An immediate consequence of Lemma 2.1 arises the following results.
Corollary 2.2. If n be an odd positive integer, Bn ≡ n (mod 32).
2
Corollary 2.3. Let n be an even positive integer. Then Bn ≡ 3n (mod 32).
Corollary 2.4. Let Cn is the nth Lucas-balancing number, then
1 mod 16, if 2|n,
Cn =
3 mod 16, if 2 - n.
The following results are available in [5].
Lemma 2.5. Let n be a positive integer. There is no balancing number except 1 satisfying the
equation Bn = x2 .
Lemma 2.6. There is no positive integer x such that vn = x2 .
Lemma 2.7. If n ≥ 0 and x > 0 are integers such that vn = 2x2 , then (n, x) = (0, 1).
Lemma 2.8. There is no positive integer x such that Bn = vm x2 .
Lemma 2.9. There is no positive integer x such that Bn = 2vm x2 .
Lemma 2.10. If n ≥ 1 , then the equation Pn = x2 has positive solutions (n, x) = (1, 1) or (7, 13)
, where Pn is the nth pell number.
The following results are found in [12].
For m, n ∈ Z and n ≥ 1,
Lemma 2.11. Bn divides Bm if and only if n | m.
Lemma 2.12. Cn divides Bm if and only if n | m and
m
n
is an even integer.
Lemma 2.13. Cn divides Cm if and only if n | m and
m
n
is an odd integer.
Lemma 2.14. For any n, 4 - vn .
We state the following results from [10] and [8], respectively.
Lemma 2.15. Let P > 0 and Q = −1. If Un = wx2 with w ∈ {1, 2, 3, 6} and n ≤ 2 except when
(P, n, w) = (2, 4, 3), (2, 7, 1), (4, 4, 2), (1, 12, 1), (1, 3, 2), (1, 4, 3), (1, 6, 2), or (24, 4, 3).
Lemma 2.16. The generalized Lucas sequence has at most one non-trivial square-class. Furthermore, if P ≡ 2 (mod 4), then we have no non-trivial square-class except (v1 , v2 ) = (338, 114242)
when P = 338, Q = 1. If P ≡ 0 (mod 4), then we have no non-trivial square-classes when 2 - mn
or 2|(m, n).
3
Some new results of balancing and Lucas balancing numbers
In this section, we establish some new results of balancing and Lucas balancing numbers which will
be used subsequently. Following results are valid for balancing and Lucas balancing numbers.
Lemma 3.1. (Bn , vn /vm ) = 1 or 2 and (Bn , vn /Bm ) = 1 or 2.
Proof. Suppose (Bn , vn /vm ) = k that implies k|Bn and k|vn /vm , then k|Bn and k|vn . but (Bn , vn ) =
1 or 2. Hence (Bn , vn /vm ) = 1 or 2. Similarly, (Bn , vn /Bm ) = 1 or 2.
3
Theorem 3.2. Bn = Bm x2 has no solution except (n, x) = (1, 1).
Proof. Let Bn = Bm x2 . Then Bn /Bm = x2 which follows that m|n. Therefore n = mk for some
positive integer k. Then two cases will arise, i.e. k is even or odd. Suppose k is even, then k = 2t for
some integer t. It follows that B2mt = Bm x2 . Since B2n = 2Bn Cn = Bn vn , have B2mt = Bmt vmt .
By virtue of Lemma 3.1,
vmt
) = 1 or 2.
d = (Bmt ,
Bm
For d = 1, then Bmt = a2 and vmt = b2 Bm for some integer a and b. Then by Lemma 2.5, x = 1
is the only solution. Now for d = 2, Bmt = 2a2 and vmt = 2b2 Bm . As P2n = 2Bn , P2mt = (2a)2 .
Therefore by Lemma 2.10, P2mt = (2a)2 has positive solutions (2mt, 2a) = (1, 1) or (7, 13), which
is not possible as 2mt is even. Suppose k is odd, If m is odd, then Bmk = Bm x2 which implies that
Bmk (mod 32) = Bm x2 (mod 32). By virtue of Corollary 2.2, mk (mod 32) = mx2 (mod 32) i.e.
x2 = k where k ≡ 1, 3, 5, 7 (mod 8), but x2 ≡ 0, 1, 4 (mod 8), which is not possible. In a similar
manner, it can be shown for m is even. Thus Bn = Bm x2 has no solution except x = 1.
Lemma 3.3. 2vm = vn x2 has no solution for any n.
Proof. Since (vn , vm ) = 2, so x is even. That implies 8|vn x2 . So 8|2vm which implies that 4|vm ,
which is not possible as 2 - Cm .
4
Positive integer solutions of some Diophantine equations
In this section, we consider the diophantine equations x2 − 32Bn xy − 32y 2 = ±32r , x4 − 32Bn x2 y −
32y 2 = ±32r and x2 − 32Bn xy 2 − 32y 4 = ±32r and find the positive integer solutions of these
equations. Indeed, the last two equations can be obtained from the first equation by replacing x
by x2 and y by y 2 .
4.1
Positive integer solutions of the Diophantine equations x2 − 32Bn xy − 32y 2 =
±32r
Consider the Diophantine equation x2 − 32Bn xy − 32y 2 = ±32r where n, r ≥ 0. Setting u =
x − 16Bn y and v = Cn y, this diophantine equation reduces to u2 − 32v 2 = ±32r . Then, the
following result can be easily obtained by induction using the identities Q2n − 8Pn2 = 4(−1)n and
Cn2 − 8Bn2 = 1.
Theorem 4.1. Let k ≥ 0 be an integer, then all positive integer solutions of the equation u2 −32v 2 =
32k are given by
( 5k
5k−2
(2 2 Cm , 2 2 Bm ),
if k is even;
(u, v) =
5k+1
5k−7
(2 2 P2m+1 , 2 2 Q2m+1 ), if k is odd.
and all positive integer solutions of the equation u2 − 32v 2 = −32k are given by
( 5k−2
5k−4
(2 2 Q2m+1 , 2 2 P2m+1 ), if k is even;
(u, v) =
5k+3
5k−5
(2 2 Bm , 2 2 Cm ),
if k is odd.
with m ≥ 0.
Using the above interesting facts, we have the following results.
4
Theorem 4.2. All positive integer solutions of x2 − 32Bn xy − 32y 2 = 32k where k is even, are
5k−2
5k
Bm
m
2
given by (x, y) = (2 2 Cm+n
Cn , 2
Cn ) with m ≥ 1 and n|m and n is even. For k is odd, all
positive integer solutions of the equation x2 − 32Bn xy − 32y 2 = 32k exist only if n = 0, are given
5k−7
5k+1
by (x, y) = (2 2 P2m+1 , 2 2 Q2m+1 ).
Proof. Since k is even, by virtue of Theorem 4.1,
u = x − 16Bn y = 25k/2 Cm ,
Putting y = 2
5k−2
2
Bm
Cn ,
v = Cn y = 2
5k−2
2
Bm .
we get
5k−2
2
x − 16Bn 2
Bm
= 25k/2 Cm .
Cn
Further simplification gives
8Bn Bm + Cm Cn
.
Cn
5k
x=22
Since Cm Cn + 8Bn Bm = Cm+n [11], x = 25k/2 Cm+n
Cn . But these solutions will be positive integer
solutions if and only if Cn |Bm and Cn |Cm+n , that implies n|m and m
n is an even integer, which
follows from Lemma 2.12. Further for k is odd,
x − 16Bn y = 2
5k+1
2
P2m+1 ,
Cn y = 2
Since Cn = Q2n , we have
Q2n y = 2
5k−7
2
Q2m+1 = 2
0
5k−5
2
5k−7
2
Q2m+1 .
Q2m+1
.
2
0
0
0
In [9], McDaniel proved that for m = 2a m , n = 2b n , where m , n are odd and a, b ≥ 0 with
d = (m, n). Then
vd ,
if a = b
(Vm , Vn ) =
1 or 2, if a 6= b.
Using this result one can see that (Q2n , Q2m+1
) = 1. Q2n divides 2
2
n = 0 and we get the desired result.
5k−5
2
and it is possible only when
It is notice that, since vn = 2Cn , all positive integer solutions of x2 −32Bn xy −32y 2 = 32k ,where
5k−4
5k
Bm
m
2
k is even, are given by (x, y) = (2 2 vm+n
vn , 2
vn ) with m ≥ 1 and n|m and n is even .
Theorem 4.3. For k is even, all positive integer solutions of the equation x2 − 32Bn xy − 32y 2 =
5k−2
5k−4
−32k exist only for n = 0 are given by (x, y) = (2 2 Q2m+1 , 2 2 P2m+1 ). For k is odd, all
positive integer solutions of the equation x2 − 32Bn xy − 32y 2 = −32k are given by (x, y) =
5k+3
5k−5
+n
Cm
m
2
(2 2 Bm
Cn , 2
Cn ) with m ≥ 1 and n|m and n is an odd integer.
Proof. For k is even, by virtue of Theorem 4.1
u = x − 16Bn y = 2
5k−2
2
Q2m+1 ,
Cn y = 2
5k−4
2
In [9], McDaniel proved that for d = (m, n), then
(Pm , Qn ) = Qd , if m/d is even
(Pm , Qn ) = 1,
otherwise.
5
P2m+1 .
5k−4
Using the result one can see that (Q2n , P2m+1 ) = 1. Q2n divides 2 2 and it is possible when
5k+3
5k−5
n = 0 and we get the desired result. Further for k is odd, x − 16Bn y = 2 2 Bm , Cn y = 2 2 Cm .
Putting y = 25k−5/2 CCm
, we get x − 16Bn 25k−5/2 CCm
= 25k+3 Bm . Further simplification gives
n
n
x=2
5k+3
2
Bm Cn + Cm Bn
.
Cn
5k+3
m+n
Since Bm Cn +Cm Bn = Bm+n , x = 2 2 BC
. But these solutions will be positive integer solutions
n
if and only if Cn divides Cm , that implies n|m and m
n is an odd integer and the result follows.
4.2
Non-negative integer solutions of the Diophantine equations x4 − 32Bn x2 y −
32y 2 = ±32r and x2 − 32Bn xy 2 − 32y 4 = ±32r
The following two results identify the non-negative integer solutions of the Diophantine equations
x4 − 32Bn x2 y − 32y 2 = ±32r .
Theorem 4.4. For k ≡ 0, 1, 2 (mod 4), then the Diophantine equation x4 − 32Bn x2 y − 32y 2 = 32k
has no positive integer solution x and y. If k ≡ 3 (mod 4), then all positive integer solutions
5k−5
5k+1
of the equation x4 − 32Bn x2 y − 32y 2 = 32k are given by (x, y) = (2 4 , 2 2 ) or (x, y) =
5k−3
5k+1
(13.2 4 , 239.2 2 ).
Proof. Assume that k is even, then virtue of Theorem 4.2, we have
5k
(x2 , y) = (2 2
Cm+n 5k−2 Bm
,2 2
)
Cn
Cn
5k
Cm+n
2
2
with m ≥ 1 and n|m and m
n is even. Hence, we get x = 2
Cn .
Case 1 : Let k ≡ 0 (mod 4). We readily obtain from the equation Cn u2 = Cn+m for some u > 0.
By Lemma 2.13, this is possible only when n + m = n, implying that m = 0, which contradicts the
fact that m ≥ 1.
Case 2 : Let k ≡ 2 (mod 4). So, we immediately have Cn u2 = 2Cm+n , for some u > 0. Since 2 - Cn ,
this is impossible.
5k+1
5k−7
If k is odd, then by virtue of Theorem 4.2, we have (x2 , y) = (2 2 P2m+1 , 2 2 Q2m+1 ) with m ≥ 0.
5k+1
This implies that x2 = 2 2 P2m+1 .
5k+1
Case 1 : Let k ≡ 1 (mod 4). Then from x2 = 2 2 P2m+1 , we obtain u2 = 2P2m+1 for u > 0, which
has no positive integer solution.
Case 2 : Let k ≡ 3 (mod 4). Then from the equation we obtain u2 = P2m+1 for u > 0. By
Lemma 2.10, we get 2m + 1 = 1 or 7, implying that m = 0 or 3. Substituting these values of m
5k−7
5k+1
5k−5
5k+1
into (x2 , y) = (2 2 P2m+1 , 2 2 Q2m+1 ) with m ≥ 0, we conclude that (x, y) = (2 4 , 2 2 ) or
5k+1
5k−7
(13.2 4 , 239.2 2 ).
Theorem 4.5. The Diophantine equation x4 − 32Bn x2 y − 32y 2 = −32k has only one positive
5k−4
5k
integer solution (x, y) = (2 4 , 2 2 ) for k ≡ 0 (mod 4) and for k ≡ 1, 2, 3 (mod 4), the Diophantine
equation x4 − 32Bn x2 y − 32y 2 = −32k has no solutions.
Proof. Assume that k is even, then by virtue of Theorem 4.3, it follows that
(x2 , y) = (2
5k−2
2
Q2m+1 , 2
6
5k−4
2
P2m+1 ),
5k−2
with m ≥ 0. Hence, we get x2 = 2 2 Q2m+1 .
5k−2
Case 1 : Let k ≡ 0 (mod 4). Hence, we immediately have from x2 = 2 2 Q2m+1 that 2u2 = Q2m+1
5k−4
5k
for u > 0. By Theorem 2.15, we get m = 0. This yields that (x, y) = (2 4 , 2 2 ).
Case 2 : Let k ≡ 2 (mod 4), then from the equation we obtain u2 = Q2m+1 for u > 0. Since Q2m+1
is even as 2|Qn . It is clear that u is even and therefore 4|Q2m+1 , which is impossible because
Q2n − 8P n2 = 4(−1)n .
Let k is odd. Then by the Theorem 4.3, it follows that
(x2 , y) = (2
5k+3
2
Bm+n 5k−5 Cm
,2 2
),
Cn
Cn
with m ≥ 1 and n|m and m
n is an odd integer.
Case 1 : Let k ≡ 1 (mod 4). Then from the equation we obtain Cn u2 = Bm+n for u > 0. From
Lemma 2.12, it has no solution.
Case 2 : Let k ≡ 3 (mod 4). Then from the equation we obtain Cn u2 = 2Bm+n for u > 0, which
has no solution.
The following Theorems are the results concerning the non-negative integer solutions of the
Diophantine equations x2 − 32Bn xy 2 − 32y 4 = ±32r .
Theorem 4.6. The Diophantine equation x2 − 32Bn xy 2 − 32y 4 = 32k has no positive integer
solution for k ≡ 0, 2, 3 (mod 4). If k ≡ 1 (mod 4), then the equation has positive integer solutions
5k+1
5k−5
only when n = 0 and the solution is (x, y) = (2 2 , 2 4 ).
Proof. Assume that k is even. Then by Theorem 4.2, it follows that
5k
(x, y 2 ) = (2 2
Cm+n 5k−2 Bm
,2 2
)
Cn
Cn
5k−2
2
2 Bm .
with m ≥ 1 and n|m and m
n is even. Hence we obtain Cn y = 2
Case 1 : Let k ≡ 0 (mod 4), then from the equation we obtain Bm = 2Cn u2 = vn u2 , which is
impossible by lemma 2.8.
Case 2 : Let k ≡ 2 (mod 4). Then we have Bm = u2 Cn = u2 Q2n , which is impossible as (Bm , Q2n ) =
1.
5k+1
5k−7
Let k is odd. Then the solution will be (x, y 2 ) = (2 2 P2m+1 , 2 2 Q2m+1 ). Hence, we obtain
5k−7
y 2 = 2 2 Q2m+1 .
Case 1 : Let k ≡ 1 (mod 4), then from the equation we obtain 2u2 = Q2m+1 for u > 0. By Theorem
5k+1
5k−5
2.15, we get m = 0. Thus, x = 2 2 and y = 2 4 .
Case 2 : Let k ≡ 3 (mod 4), then from the equation we obtain u2 = Q2m+1 for some u > 0. Since,
Q2m+1 is even, it follows that 2|u and therefore 4|Q2m+1 , which is impossible as 4 - Qn .
Theorem 4.7. The Diophantine equation x2 − 32Bn xy 2 − 32y 4 = −32k has no positive integer
solutions for k ≡ 2, 3 (mod 4). For k ≡ 0, 1 (mod 4), the Diophantine equation x2 − 32Bn xy 2 −
5k−4
5k−4
5k+5
5k−5
5k
5k
32y 4 = −32k has (x, y) = (2 2 , 2 4 ) or (239.2 2 , 13.2 4 ) and (x, y) = (2 2 , 2 4 ).
5k−2
5k−4
Proof. Assume k is even, then the solution will be (x, y 2 ) = (2 2 Q2m+1 , 2 2 P2m+1 ) with m ≥ 0.
5k−4
Hence we obtain y 2 = 2 2 P2m+1 .
Case 1 : Let k ≡ 0 (mod 4). Then from the equation we obtain u2 = P2m+1 for some u > 0. By
5k
Lemma 2.10, we get 2m + 1 = 1 or 7 i.e. m = 0 or 3. If m = 0, then we immediately have x = 2 2
7
5k−4
5k−4
5k
and y = 2 4 . If m = 3, we obtain x = 239.2 2 and 13.2 4 .
Case 2 : Let k ≡ 2 (mod 4), then from the equation we obtain u2 = 2.P2m+1 for some u > 0, which
is not possible.
5k−5
5k+3
m+n
Now assume k is odd, then the solution will be (x, y 2 ) = (2 2 BC
, 2 2 CCm
) with m ≥ 1 and
n
n
n|m and m
is
an
odd
integer.
n
Case 1 : Let k ≡ 1 (mod 4), then from the equation we obtain vn u2 = vm for some u > 0. By
5k+5
5k−5
5k+3
= 2 2 Bn and y = 2 4 .
Lemma 2.13, this is possible when n = m. Hence, we get x = 2 2 BC2n
n
Case 2 : Let k ≡ 3 (mod 4), then from the equation we obtain Cn+m u2 = 2Cm for u > 0, which is
impossible by Lemma 3.3.
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