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General Mathematics Vol. 12, No. 1 (2004), 23–
Parametric solutions for some Diophantine
equations
Dorin Andrica and Gheorghe M. Tudor
Abstract
Under some hypotheses we show that the Diophantine equation
(1) has infinitely many solutions described by a family depending on
k + 2 parameters. Some applications of the main result are given and
some special equations are studied.
2000 Mathematical Subject Classification: 11D
1 Introduction
Consider the Diophantine equation
a 0 x
p 0
0
p 1
1
+... + a k x
p k
k
where a 0 , a 1 ,... , ak are integers, a 0 > 0, and p 0 , p 1 ,... , pk are positive in-
tegers. Concerning the equation (1) in the book [3] the following general
result is presented:
24 Dorin Andrica and Gheorghe M. Tudor
Assume that p is relatively prime to the product P k = p 1 p 2
... p k . Then:
a) if a 1
= 0, the equation (1) has infinitely many solutions
in integers;
b) if a 1
- a 2 +... + a k < 0 , the equation (1) has infinitely many solutions
in positive integers.
In both cases mentioned above, the solutions are described by a family
depending on a parameter.
In the paper [4], the second author gave a much general result without
restrictive conditions a) and b). Moreover, the solutions are described by a
family depending on k + 2 parameters. The main result in [4] is contained
in the following:
Theorem. Consider the equation (1) with a 0
0 and assume that p 0 is
relatively prime to m = lcm(p 1 , p 2 ,... , p k ). Then:
a) the equation (1) has infinitely many solutions in integers;
b) if ai < 0 , for some i ∈ { 1 , 2 ,... , k}, then the equation (1) has in-
finitely many solutions in positive integers.
In order to construct a family of solutions, let us denote
T
k = a
p 0 − 1
0
(−a 1 n
p 1
1
− a 2 n
p 2
2
−... − a k n
p k
k
where n 1 , n 2 ,... , n k are arbitrary integers. Taking into account that p and
m are relatively prime, it follows that for infinitely many pairs (q, r) of
positive integers the relation
(3) p 0 q = mr + 1
26 Dorin Andrica and Gheorghe M. Tudor
In what follows we will apply the result in the above mentioned Theorem
for some of these equations as well as for some other generalized equations.
2 The equations x
p
+ y
p
= z
np± 1
and
x
p
+ y
p
= z
p
n ± 1
First of all we will change the notations in order to apply in a direct way
the result in our Theorem.
Consider the equation
x
np± 1
0
− x
p
1
− x
p
2
where p and n are positive integers. In that case we have a 0 = 1, a 1 = a 2
−1 and p 0 = np ± 1 is relatively prime to p. There exist infinitely many
positive integers q and r such that
(7) (np ± 1)q = pr + 1
It is easy to show that
r(t) = (np ± 1)t ± n
q(t) = pt ± 1
where t is any positive integers and the signs + and − correspond. Using
formula (5) we find the following family of solutions to equation (6):
x 0 = (n
p
1
p
2
pt± 1
x 1 = n 1 (n
p
1
p
2
(np±1)t±n
x 2 = n 2 (n
p
1
p
2
(np±1)t±n
Parametric solutions for some Diophantine equations 27
Let us note that if n = 1, then we obtain the equations (a). If n 1
n 2 = k, t = 1 and n 1 = k, n 2 = 1, t = 1, respectively, we find the solutions
x 0 = k
p
p
p
when we consider the sign +, and
x 0 = (k
p
p− 1 , x 1 = (k
p
p− 2 , x 2 = k(k
p
p− 2
in case of the sign −. These solutions are given in the book [1].
Let us consider the equations
x
p
n ± 1
0 − x
p
1 − x
p
2
where p and n are positive integers. In that case we have p 1 = p 2 = p,
p 0 = p
n ± 1 and p 0 is relatively prime to p. Hence
(p
n (11) ± 1)q = pr + 1
for some positive integers r and q. All such pairs (r, q) are given by
r(t) = (p
n ± 1)t ± p
n− 1
q(t) = pt ± 1
where t is any positive integer and signs + and − correspond. From formula
(5) we find the following family of solutions to equation (10):
x 0 = (n
p
1
p
2
pt± 1
x 1 = n 1 (n
p
1
p
2
(p n ±1)t±p n− 1
x 2 = n 2 (n
p
1
p
2
(p n ±1)t±p n− 1
Parametric solutions for some Diophantine equations 29
In that case we have a 0 = 1, a 1 = a 2 = −1, p 1 = 2p − 1, p 2 = 2p + 1,
p 0 = p. It is clear that p 0 is relatively prime to p 1 p 2 = 4p
2 − 1. In the case
p = 2 we obtain equation (b) also studied in the book [1].
Because p 0 is relatively prime to 4p
2 − 1, we have
pq = (4p
2 (19) − 1)r + 1
and all pairs (r, q) of such positive integers are given by
r(t) = pt + 1
q(t) = 4p
2 t + 4p − t
for any positive integer t. Applying formula (4) we obtain the following
family of solutions to equation (18):
x 0 = n
4 p
2 − 1
0
(n
2 p− 1
1
2 p+
2
(4p 2 −1)t+4p
x 1 = n
p(2p+1)
0
n 1 (n
2 p− 1
1
2 p+
2
(2p+1)(pt+1)
x 2 = n
p(2p−1)
0
n 2 (n
2 p− 1
1
2 p+
2
(2p−1)(pt+1)
The family (21) depends on four parameters n 0 , n 1 , n 2 , t.
In the case p = 2 we obtain a family of solutions to equation (b):
x 0 = n
15
0
(n
3
1
5
2
15 t+
x 1 = n
10
0
n 1 (n
3
1
5
2
10 t+
x 2 = n
6
0
n 2 (n
3
1
5
2
6 t+
where n 0 , n 1 , n 2 , t are any positive integers.
30 Dorin Andrica and Gheorghe M. Tudor
4 The equation
b
m
x
2 n+m
m
+ b
m+
x
2 n+m+
m+
+... + b
m+p
x
2 n+m+p
m+p
In the above equation n and p are positive integers and m is an integer.
The coefficients b i , i = m, m + 1,... , m + p, are integers. In what follows
we will study three special cases of this equation. We use the notations in
our Theorem.
4.1. Let us consider the equation
a 0 x
2 n+
0
2 n− 1
1
2 n
2
2 n+
3
2 n+
4
where a 0
0, a
2
1
2
2
2
3
2
4
= 0 and n ≥ 2 is a positive integer.
We have p 0 = 2n + 1, p 1 = 2n − 1, p 2 = 2n, p 3 = 2n + 2, p 4 = 2n + 3
and p 0 is relatively prime to each of the integers p 1 , p 2 , p 3 , p 4
. Applying the
result in our main Theorem we obtain:
Proposition 1. a) The equation (23) has infinitely many solutions in
integers.
b) If a i < 0 for some i ∈ { 1 , 2 , 3 , 4 }, then the equation (23) has infinitely
many solutions in positive integers.
Let us indicate how we can construct an infinite family of solutions.
Because p 0 = 2n + 1 is relatively prime to each p 1 , p 2 , p 3 , p 4 it follows that
(24) (2n + 1)q = (2n − 1)2n(2n + 2)(2n + 3)r + 1
for some positive integers r and q. That is equivalent to
p 0 q = (p
2
0
− 4)(p
2
0
(25) − 1)r + 1
From (25) it follows that 4r + 1 = p 0 s, where s is a positive integer. We
32 Dorin Andrica and Gheorghe M. Tudor
for some positive integers r and q. That is equivalent to
p 0 q = (p
2
0
− 16)(p
2
0
− 4)(p
2
0
(30) − 1)r + 1
It follows 64r − 1 = p 0 s, where s is a positive integer. In order to
find convenient pairs (r, q) of positive integers satisfying (30) let us use the
following obvious property: For any positive integers n, k ≥ 1 , the integer
(2n + 1)
2 k
− 1 is divisible by 2
k+
. In that case we can consider
r(t) =
(64t − 1)p
16
0
q(t) =
p 0
[(p
2
0
− 16)(p
2
0
− 4)(p
2
0
− 1)r(t) + 1],
where t is any positive integer.
In the particular case n = 3, we have
T
6 = a
6
0
(−a 1 n
3
1
− a 2 n
5
2
− a 3 n
6
3
− a 4 n
8
4
− a 5 n
9
5
− a 6 n
11
6
and
r(t) =
[(64t − 1)
16
(33 · 45 · 48 r(t) + 1)
A family of integral solutions to equation (28) can be obtained by using
formula (4) or (5).
4.3. Let us consider the equation
a 0 x
2 n+
0
2 n− 1
1
2 n
2
2 n+
3
2 n+
4
+a 5 x
2 n+
5
2 n+
6
2 n+
7
2 n+
8
where n ≥ 2 is a positive integer, the coefficients ai are integers, a 0 > 0 and
a
2
1
2
2
+... + a
2
8
= 0. Assume that n is not divisible by 3. Then p 0 = 2n + 3
is relatively prime to all pi, i = 1, 2 ,... , 8. We have many integral solutions.
Parametric solutions for some Diophantine equations 33
b) If a i < 0 for some i ∈ { 1 , 2 ,... , 8 }, then equation (32) has infinitely
many solutions in positive integers.
We will indicate the effective construction of an infinite family of integral
solutions. Taking into account that p 0 = 2n + 3 is relatively prime to the
product p 1 p 2
... p 8 , we have
(33) (2n + 3)q =
= (2n − 1)(2n)(2n + 1)(2n + 2)(2n + 4)(2n + 5)(2n + 6)(2n + 7)r + 1,
for some positive integers r and q. The relation (33) is equivalent to
p 0 q = (p
2
0
− 16)(p
2
0
− 9)(p
2
0
− 4)(p
2
0
(34) − 1)r + 1
Therefore, the relation 9 · 64 r + 1 = p 0 s, for a positive integer s.
Taking into account that (2n + 3)
16 − 1 = (2(n + 1) + 1)
16 − 1 is divisible
by 64 (see the general property in 4.2) and [(2n + 3)
2 − 1]
2 is divisible by 9,
it follows that we can choose r(t) and q(t) as
r(t) =
(p 0 t + 1)(p
16
0
− 1)(p
2
0
2
q(t) =
p 0
[(p
2
0
− 16)(p
2
0
− 9)(p
2
0
− 4)(p
2
0
− 1)r(t) + 1]
where t is any positive integer. Using (3) we can derive an infinite family
of integral solutions to equation (32) directly from formula (4) or (5).
References
[1] Andreescu, T., Andrica, D., An Introduction to Diophantine Equations,
GIL Publishing House, 2002.
