International Journal of Pure and Applied Mathematics
Volume 107 No. 4 2016, 1063-1072
ISSN: 1311-8080 (printed version); ISSN: 1314-3395 (on-line version)
url: http://www.ijpam.eu
doi: 10.12732/ijpam.v107i4.23
AP
ijpam.eu
ON THE DIOPHANTINE EQUATION x4 + y 4 = pk z 7
Keng Yarn Wong1 § , Hailiza Kamarulhaili2
1,2 School
of Mathematical Sciences
Universiti Sains Malaysia
11800, Penang, MALAYSIA
Abstract: This paper solves the Diophantine equation x4 +y 4 = pk z 7 nontrivially in the case
of x = y where p is prime and k ∈ Z+ . The parametric solutions are formulated using number
theory theorems, especially those concerning divisibility of integers, linear Diophantine equations, properties of prime numbers, and properties of congruence. There exist infinitely many
nontrivial integral solutions to this Diophantine equation, where the parametric solutions
found solve completely for different values of p and k in the case of x = y.
AMS Subject Classification: 11D41
Key Words: Diophantine equation, congruence, septic degree
1. Introduction
In 2011, Ismail (see [2]) sought the integral solutions for x4 + y 4 = pk z 3 where
p is prime, 2 ≤ p ≤ 13, and k ∈ Z+ . In her studies, she claimed to have found
all integral solutions to this equation. However, there are solutions that cannot
be obtained using her parametric solutions; thus not a complete solution to her
choice of Diophantine equation. For example, in the case of x4 + y 4 = 34 z 3
where x = y, her parametric solution could not find (12, 12, 8), (96, 96, 128),
(324, 324, 648), (768, 768, 2048), and (1500, 1500, 5000) which are also solutions
to the equation.
Received: February 23, 2016
Published: May 9, 2016
§ Correspondence
author
c 2016 Academic Publications, Ltd.
url: www.acadpubl.eu
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K.Y. Wong, H. Kamarulhaili
This is due to her assumption in her proofs that z must always contain
the prime p when represented as product of primes. Although her parametric
solutions can fulfill her Diophantine equation, they would not yield the complete
solutions to the equation whenever k ≡ 1 (mod 4) (for p = 2) and k ≡ 0
(mod 4) (for p > 2) are concerned. Having identified this shortcoming, this
motivates the study of
x4 + y 4 = p k z 7
(1)
where p is any prime and k ∈ Z+ , where the prime concern is to seek nontrivial
parametric solutions to this Diophantine equation by taking congruence consideration into account. This paper focuses on the case where x = y. The idea
of solving this Diophantine equation can be used on other similar Diophantine equations for x = y. This includes Ismail’s; thus closing the gap in her
parametric solutions for such case.
This paper is organized as follows: Section 2 gives the parametric solutions
for (1) where x = y and p = 2. It considers three cases of k; k = 1, k > 1 and
k ≡ 1 (mod 4), and k > 1 and k 6≡ 1 (mod 4). Section 3 gives the parametric
solutions for (1) where x = y and p > 2. It considers two cases of k; k ≡ 0
(mod 4) and k 6≡ 0 (mod 4). Finally, Section 4 concludes the paper with a
summary of all parametric solutions for (1) where x = y.
2. On the Diophantine Equation x4 + y 4 = pk z 7 ,
where x = y and p = 2
The following theorem gives the nontrivial parametric solutions to (1) where
x = y and p = 2.
Theorem 1. Suppose that (x0 , y0 , z0 ) is a nontrivial integral solution to
x4 + y 4 = 2k z 7 where x0 = y0 and k ∈ Z+ . If k = 1, then (x0 , y0 , z0 ) =
(±n7 , ±n7 , n4 ) where n ∈ Z+ . If k > 1 and k ≡ 1 (mod 4), then (x0 , y0 , z0 ) =
(±2v n7 , ±2v n7 , n4 ) where n ∈ Z+ and 4v + 1 = k. If k > 1 and k 6≡ 1 (mod 4),
then (x0 , y0 , z0 ) = (±22k−2 n7 , ±22k−2 n7 , 2k−1 n4 ) where n ∈ Z+ .
Proof. Let (x0 , y0 , z0 ) be a solution to x4 + y 4 = 2k z 7 where x0 = y0 . Then
x40 = 2k−1 z07 .
(2)
x0 is of quartic degree. Thus, without loss of generality, let x0 be a positive
integer. By Fundamental Theorem of Arithmetic, let x0 and z0 be represented
ON THE DIOPHANTINE EQUATION x4 + y 4 = pk z 7
1065
as a product of primes in their canonical forms, respectively;
r
Y
x0 = pα1 1 pα2 2 . . . pαr r =
pαi i
(3)
β
(4)
i=1
z0 = q1β1 q2β2 . . . qsβs =
s
Y
qj j
j=1
where pi and qj are primes, p1 < p2 < · · · < pr , q1 < q2 < · · · < qs , r, s are
nonnegative integers, and αi , βj are positive integers.
Plugging (3) and (4) into (2),
r
Y
i
p4α
= 2k−1
i
i=1
s
Y
7β
qj j .
(5)
j=1
There are two cases to be considered: k = 1 and k > 1.
Case 1 : Suppose k = 1. Then (5) becomes
r
Y
i
p4α
=
i
s
Y
7β
qj j .
(6)
j=1
i=1
In this case, due to the uniqueness of canonical representation of integers,
r = s, pi = qj , and 4αi = 7βj for 1 ≤ i = j ≤ r = s. Note that 7|4αi , 4|7βj , and
gcd(4,7) = 1. We must have positive integers vi and wj such that αi = 7vi and
βj = 4wj . Thus 4(7vi ) = 7(4wj ) and hence vi = wj . So, (3) and (4) become
x0 = y 0 =
r
Y
i
p7v
=
i
i=1
i=1
z0 =
s
Y
j=1
4wj
qj
r
Y

=
s
Y
j=1
w
pvi i
!7
4
qj j  .
(7)
(8)
Q
Q
w
Let n = ri=1 pvi i = sj=1 qj j , then from (7) and (8), x0 = y0 = n7 and
z0 = n4 . Note that n is any positive integer. Also, x0 being positive or negative
has no effect on the balance of the equation since it is of quartic degree in (2).
Thus we have (x0 , y0 , z0 ) = (±n7 , ±n7 , n4 ) where n ∈ Z+ .
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K.Y. Wong, H. Kamarulhaili
Case 2 : Suppose k > 1. Then from (2), 2k−1 |x40 . Thus, x0 =
contain a prime p1 = 2. (3) becomes
x0 = 2α1
r
Y
pαi i
Qr
αi
i=1 pi
must
(9)
i=2
and (5) becomes
24α1
r
Y
i
p4α
= 2k−1
i
s
Y
7β
qj j .
(10)
j=1
i=2
Comparing the indices of 2 on both sides of (10), 4α1 = k − 1 suggests that only
positive integers k where k ≡ 1 (mod 4) allow (10) to be consistent. However,
in the case of k 6≡ 1 (mod 4), (10) still holds if z0 contains a prime factor 2.
Hence, there are two cases that need to be considered: k 6≡ 1 (mod 4) and
k ≡ 1 (mod 4).
Case a : Suppose k 6≡ 1 (mod 4). (10) will not be consistent unless there exists
Q
β
a prime factor q1 = 2 in z0 . Thus z0 = 2β1 sj=2 qj j and (10) becomes
24α1
r
Y
i
p4α
= 2k−1+7β1
i
s
Y
7β
qj j .
(11)
j=2
i=2
We now have a linear Diophantine problem to be solved in order for (11) to be
consistent: 4α1 −7β1 = k −1. It is easy to check that (α1 , β1 ) = (2k −2, k −1) is
a particular solution to this equation, and its general solution is thus α1 = 2k −
2 − 7t and β1 = k − 1 − 4t where t ≤ 0. Any integer represented in its canonical
form is unique. Thus r = s, pi = qj , and 4αi = 7βj for 2 ≤ i = j ≤ r = s.
Note that 7|4αi , 4|7βj , and gcd(4,7)=1. We must have positive integers vi and
wj such that αi = 7vi and βj = 4wj . Thus 4(7vi ) = 7(4wj ) and hence vi = wj .
Using these new forms, we have
!7
r
Y
vi
−t
2k−2
pi
(12)
2
x0 = y 0 = 2
i=2

z0 = 2k−1 2−t
s
Y
j=2
w
4
qj j  .
(13)
Q
Q
w
Let n = 2−t ri=2 pvi i = 2−t sj=2 qj j . Then (12) and (13) become x0 = y0 =
22k−2 n7 and z0 = 2k−1 n4 , respectively. x0 being positive or negative has no
ON THE DIOPHANTINE EQUATION x4 + y 4 = pk z 7
1067
effect on the balance of the equation since it is of quartic degree in (2). Thus, in
the case of k 6≡ 1 (mod 4), (x0 , y0 , z0 ) = (±22k−2 n7 , ±22k−2 n7 , 2k−1 n4 ) where
n ∈ Z+ .
Case b: Suppose k ≡ 1 (mod 4). Then there are another two subcases to be
considered: z0 contains a prime factor q1 = 2 and z0 does not contain a prime
factor 2.
Case b1 : Suppose z0 contains a prime factor q1 = 2. Then we have z0 =
Q
β
2β1 sj=2 qj j and (10) becomes (11). This subcase is solved using the same
steps in Case a, yielding the same forms of (x0 , y0 , z0 ) in the said case.
Case b2 : Suppose z0 does not contain a prime factor 2. Then z0 remains as
(4) and in order for (10) to be balanced, s = r − 1. Thus (10) becomes
24α1
r
Y
i
p4α
= 2k−1
i
r
Y
7β
qj j .
(14)
j=2
i=2
Looking at indices of 2 on both sides of (14), we have 4α1 = k − 1. 4|(k − 1), so
there exists a positive integer v such that k = 4v + 1. Thus, 4α1 = (4v + 1) − 1
which leads to α1 = v. Again, due to the uniqueness of canonical form of
integers, pi = qj for 2 ≤ i = j ≤ r. 7|4αi , 4|7βj , and gcd(4,7)=1. We must
have positive integers vi and wj such that αi = 7vi and βj = 4wj . This leads
to vi = wj . Using these new forms, we have
!7
r
Y
vi
v
pi
(15)
x0 = y 0 = 2
i=2

z0 = 
r
Y
j=2
w
4
qj j  .
(16)
Q
Q
w
Let n = ri=2 pvi i = rj=2 qj j , and thus (15) and (16) become x0 = y0 = 2v n7
and z0 = n4 , respectively. x0 being positive or negative has no effect on the
balance of the equation since x0 is of quartic degree in (2). Thus we have
(x0 , y0 , z0 ) = (±2v n7 , ±2v n7 , n4 ) where 4v + 1 = k and n ∈ Z+ . This form
of solution covers that of Case b1 where z0 contains prime factor 2 and k ≡
1 (mod 4). It is easy to show that for a choice of n = n1 in Case b1, the
same solution can be obtained at n = 2v n1 in Case b2. Thus, in the case of
k ≡ 1 (mod 4), the parametric form in Case b2 is sufficient to solve for all
(x0 , y0 , z0 ).
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K.Y. Wong, H. Kamarulhaili
3. On the Diophantine Equation x4 + y 4 = pk z 7 ,
where x = y and p > 2
The following theorem gives the nontrivial parametric solutions to (1) where
x = y and p > 2.
Theorem 2. Suppose that (x0 , y0 , z0 ) is a nontrivial integral solution to
x4 + y 4 = pk z 7 where x0 = y0 , k ∈ Z+ , and p is a prime where p > 2. If
k ≡ 0 (mod 4), then (x0 , y0 , z0 ) = (±25 pv n7 , ±25 pv n7 , 23 n4 ) where n ∈ Z+ and
4v = k. If k 6≡ 0 (mod 4), then (x0 , y0 , z0 ) = (±25 p2k n7 , ±25 p2k n7 , 23 pk n4 )
where n ∈ Z+ .
Proof. Let (x0 , y0 , z0 ) be a solution to (1) where x0 = y0 . Then,
2x40 = pk z07 .
(17)
x0 is of quartic degree. Thus, without loss of generality, let x0 be a positive
integer. By Fundamental Theorem of Arithmetic, let x0 and z0 be represented
Q
Q
β
as a product of primes ri=1 pαi i and sj=1 qj j , respectively, where pi and qj
are primes, r, s are nonnegative integers, and αi , βj are positive integers. Since
gcd(2,pk )=1, 2|z07 and pk |x40 . There must be a prime pi = p in x0 and qj = 2 in
z0 , where 1 ≤ i ≤ r and 1 ≤ j ≤ s. Let p2 = p and q1 = 2. Then
x0 = pα2
r
Y
pαi i
(18)
i=1,i6=2
z0 = 2β1
s
Y
β
qj j .
(19)
j=2
Plugging (18) and (19) into (17),
2p4α2
r
Y
i
p4α
= 27β1 pk
i
i=1,i6=2
s
Y
7β
qj j .
(20)
j=2
Observe the indices of 2 in (20) that in order for (20) to be consistent, 7β1 = 1
but β1 6∈ Z+ (contradiction). Thus, x0 must contain a Q
prime pi = 2 where
1 ≤ i ≤ r, i 6= 2. Let p1 = 2 and we have x0 = 2α1 pα2 ri=3 pαi i . Then (20)
becomes
s
r
Y
Y
7β
4αi
7β1 k
4α1 +1 4α2
qj j .
pi = 2 p
(21)
2
p
i=3
j=2
ON THE DIOPHANTINE EQUATION x4 + y 4 = pk z 7
1069
As for the indices of p in (21), 4α2 = k suggests that only positive integers k
where k ≡ 0 (mod 4) allow (21) to be consistent. However, in the case of k 6≡ 0
(mod 4), (21) still holds if z0 contains a prime factor p. Hence, there are two
cases that need to be considered: k 6≡ 0 (mod 4) and k ≡ 0 (mod 4).
Case a : Suppose that k 6≡ 0 (mod 4). Then (21) will not be consistent unless
there exists a prime qj = p where 2 ≤ j ≤ s. Let q2 = p and we have
Q
β
z0 = 2β1 pβ2 sj=3 qj j . Then (21) becomes
24α1 +1 p4α2
r
Y
i
p4α
= 27β1 p7β2 +k
i
s
Y
7β
qj j .
(22)
j=3
i=3
We now have two linear Diophantine problems to be solved in order for (22) to
be consistent:
−4α1 + 7β1 = 1
(23)
4α2 − 7β2 = k.
(24)
For (23), its solution is α1 = 5 − 7t1 and β1 = 3 − 4t1 where t1 ≤ 0. For (24),
its solution is α2 = 2k − 7t2 and β2 = k − 4t2 where t2 ≤ 0.
Let σ and γ be permutation functions defined respectively as
σ : {3, 4, . . . , r} 7−→ {3, 4, . . . , r}
(25)
where σ(i) = i′ , pi′ < pi′ +1 , i′ < i′ + 1, and
γ : {3, 4, . . . , s} 7−→ {3, 4, . . . , s}
(26)
where γ(j) = j ′ , qj ′ < qj ′ +1 , j ′ < j ′ + 1.
Applying them on the products of primes in (22), they arrange the primes in
Qr
Q
7β
4αi
and sj=3 qj j into their canonical forms:
i=3 pi
24α1 +1 p4α2
r
Y
i′ =3
4αi′
p i′
= 27β1 p7β2 +k
s
Y
7β
′
qj ′ j .
(27)
j ′ =3
Any integer represented in its canonical form is unique; r = s and pi′ = qj ′ for
3 ≤ i′ = j ′ ≤ r = s for (27) to be consistent. Consequently, 4αi′ = 7βj ′ . 7|4αi′ ,
4|7βj ′ , and gcd(4,7)=1. We must have positive integers vi′ and wj ′ such that
1070
K.Y. Wong, H. Kamarulhaili
αi′ = 7vi′ and βj ′ = 4wj ′ . Thus, 4(7vi′ ) = 7(4wj ′ ) and hence vi′ = wj ′ . Using
these new forms, we have
5 2k
x0 = y 0 = 2 p
−t1 −t2
2
p
r
Y
v
pi′i′
i′ =3

z0 = 23 pk 2−t1 p−t2
s
Y
j ′ =3
4
!7
w ′
qj ′ j  .
(28)
(29)
Q
Q
w ′
v′
Let n = 2−t1 p−t2 ri′ =3 pi′i = 2−t1 p−t2 sj′ =3 qj ′ j , and thus (28) and (29) become x0 = y0 = 25 p2k n7 and z0 = 23 pk n4 , respectively. Thus, in the case
of k 6≡ 0 (mod 4), we have (x0 , y0 , z0 ) = (±25 p2k n7 , ±25 p2k n7 , 23 pk n4 ) where
n ∈ Z+ .
Case b: Suppose that k ≡ 0 (mod 4). Then there are another two subcases to
be considered: z0 contains a prime factor qj = p where 2 ≤ j ≤ s and z0 does
not contain a prime factor p (see (21)).
Case b1 : Suppose that z0 contains a prime factor qj = p where 2 ≤ j ≤ s. Let
Q
β
q2 = p and we have z0 = 2β1 pβ2 sj=3 qj j and (21) becomes (22). This subcase
is solved using the same steps in Case a, yielding the same forms of (x0 , y0 , z0 )
in the said case.
Case b2 : Suppose that z0 does not contain a prime factor p. Then z0 is in
the form of (19) and in order for (21) to be consistent, s = r − 1. Thus (21)
becomes
r
r
Y
Y
7β
4αi
7β1 k
4α1 +1 4α2
qj j .
pi = 2 p
(30)
2
p
i=3
j=3
Comparing the indices of 2 and p on both sides in (30), we have two linear
Diophantine problems to solve:
−4α1 + 7β1 = 1
(31)
4α2 = k.
(32)
(31) is the same as (23) in Case a with the solutions of α1 = 5 − 7t and
β1 = 3 − 4t where t ≤ 0. For (32), note that 4|k, so there exists a positive
integer v such that k = 4v. So 4α2 = 4v and hence α2 = v. Applying (25)
ON THE DIOPHANTINE EQUATION x4 + y 4 = pk z 7
1071
and (26) (where the domain and codomain of (26) are now {3, 4, . . . , r}) on the
products of primes in (30) to arrange them into canonical forms, we get
4α1 +1 4α2
2
p
r
Y
4α
pi′ i′
7β1 k
=2
p
i′ =3
r
Y
7β
′
qj ′ j .
(33)
j ′ =3
Any integer represented in its canonical form is unique, so pi′ = qj ′ and 4αi′ =
7βj ′ for 3 ≤ i′ = j ′ ≤ r. 7|4αi′ , 4|7βj ′ , and gcd(4,7)=1. We must have positive
integers vi′ and wj ′ such that αi′ = 7vi′ and βj ′ = 4wj ′ . So, 4(7vi′ ) = 7(4wj ′ )
and hence vi′ = wj ′ . Using these new forms, we have
!7
r
Y
v
′
x0 = y0 = 25 pv 2−t
pi′i
(34)
i′ =3

z0 = 23 2−t
r
Y
j ′ =3
4
w ′
qj ′ j  .
(35)
Q
Q
w ′
v
Let n = 2−t ri′ =3 pi′i′ = 2−t rj′ =3 qj ′ j , and thus (34) and (35) become x0 =
y0 = 25 pv n7 and z0 = 23 n4 , respectively. Thus, in the case of k ≡ 0 (mod 4),
we have (x0 , y0 , z0 ) = (±25 pv n7 , ±25 pv n7 , 23 n4 ) where 4v = k and n ∈ Z+ .
This form of solution covers that of Case b1 where z0 contains prime factor p
and k ≡ 0 (mod 4). It is easy to show that for a choice of n = n1 in Case b1,
the same solution can be obtained at n = pv n1 in Case b2. Thus, in the case
of k ≡ 0 (mod 4), the parametric form in Case b2 is sufficient to solve for all
(x0 , y0 , z0 ).
4. Conclusion
There exist infinitely many nontrivial integral solutions to (1) in the case of
x = y, where the parametric solutions are (x, y, z) = (±n7 , ±n7 , n4 ) for p = 2
and k = 1, (x, y, z) = (±2v n7 , ±2v n7 , n4 ) where 4v +1 = k for p = 2, k > 1, and
k ≡ 1 (mod 4), (x, y, z) = (±22k−2 n7 , ±22k−2 n7 , 2k−1 n4 ) for p = 2, k > 1, and
k 6≡ 1 (mod 4), (x, y, z) = (±25 pv n7 , ±25 pv n7 , 23 n4 ) where 4v = k for p > 2
and k ≡ 0 (mod 4), and (x, y, z) = (±25 p2k n7 , ±25 p2k n7 , 23 pk n4 ) for p > 2 and
k 6≡ 0 (mod 4); n ∈ Z+ . These parametric solutions solve (1) completely in
such case.
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K.Y. Wong, H. Kamarulhaili
References
[1] D. Burton, Elementary Number Theory, McGraw-Hill, USA (2007).
[2] S. Ismail, Solutions of Diophantine equation x4 + y 4 = pk z 3 for primes p, 2 ≤ p ≤ 13,
Universiti Putra Malaysia, Malaysia (2011).