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Noname manuscript No.
(will be inserted by the editor)
Fourth Power Diophantine Equations in Gaussian
Integers
Farzali Izadi · Rasool Naghdali
Forooshani · Amaneh Amiryousefi
Varnousfaderani .
Received: date / Accepted: date
Abstract In this paper we examine a class of fourth power Diophantine equations of the form x4 + kx2 y 2 + y 4 = z 2 and ax4 + by 4 = cz 2 , in the Gaussian
Integers, where a and b are prime integers.
Keywords Quartic Diophantine equation · Gaussian integers · Elliptic
curve · rank · torsion group
Mathematics Subject Classification (2010) MSC 11D25 · MSC 11G05
1 Introduction
Lebesque noted that x4 ± 2m y 4 = z 2 has integral solutions only when m =
4n ± 3. Likewise, the Diophantine equation 2m x4 − y 4 = z 2 has solution only
when m = 4n + 1, but x4 ± y 4 = 2m z 2 is impossible in the integers [2]. L.
Euler proved that 2x4 ± 2y 4 = z 2 has no integer solution for x 6= y by means
of the fact that x4 ∓ y 4 are not squares[2]. W. Mantel proved by descent that
x4 + 2m y 4 6= z 2 unless n ≡ 3 (mod 4).
The method of Yasutaka Suzuki [9] determined all solutions of the equation
2a X r +2b Y s = 2c Z t in nonzero integers X, Y, Z, where a, b, c are non-negative
integers, and r, s, t are 2 or 4, and X, Y, Z are pairwise relatively primes. In [8]
Suzuki show that if the equation 2a X 4 + 2b Y 4 = 2c Z 4 has an integer solution
Farzali Izadi
Department of Pure Mathematics, Faculty of Science, Urmia University, Urmia 165-57153,
Iran. E-mail: [email protected]
Rasool Naghdali Forooshani
Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz 53751-71379,
E-mail: [email protected]
Amaneh Amiryousefi Varnousfaderani
Department of Mathematics, Isfahan University of technology, Isfahan, Iran. E-mail:
aa1364 [email protected]
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F. Izadi et al.
then a + 1 = b + 1 = c. We prove all of these cases by means of elliptic curve
method.
F. Najman [4] showed that the equation x4 − y 4 = iz 2 has only trivial
solutions in the Gaussian integers. He also showed that the only nontrivial
Gaussian solutions of the equation x4 + y 4 = iz 2 , are x, y ∈ {±i, ±1} and
z = ±i(1 + i). Using these results, we solve some new Diophantine equations of
the form x4 +kx2 y 2 +y 4 = z 2 where, k = ±6. Also we examine the Diophantine
equations x4 ± p2 y 4 = iz 2 and x4 ± p2 y 4 = iz 2 for some prime integers p.
2 Elliptic curves method
In this section we describe the method which we use for proving our results.
Let E(Q) denote the group of rational points on elliptic curve with Weierstrass
equation E : y 2 = x3 +ax2 +bx. Let Q∗ be the multiplicative group of non-zero
rational numbers and Q∗ 2 denote the subgroup of squares of elements of Q∗ .
Define the group 2-descent homomorphism α from E(Q) to Q∗ /Q∗ 2 as follows:

2
 1 (mod Q∗ ) if P = ∞,
2
α(P ) = b (mod Q∗ ) if P = (0, 0),
(1)

x (mod Q∗ 2 ) if P = (x, y) with x 6= 0.
b : y 2 = x3 − 2ax2 + (a2 − 4b)x with group
Similarly, take the isogenous curve E
b
b
of rational points E(Q).
The group 2-descent homomorphism α
b from E(Q)
to
∗
∗2
Q /Q given by


(mod Q∗ 2 ) if Pb = ∞,
1
2
(2)
α
b(Pb ) = a − 4b (mod Q∗ 2 ) if Pb = (0, 0),

x
∗2
b
(mod Q ) if P = (x, y) with x 6= 0.
Proposition 1 By the above notations, we have the following equality for the
rank r of E(Q)
|Im(α)||Im(b
α)|
.
(3)
2r =
4
Theorem 1 The group α(E(Q)) is equal to the classes of 1, b and the positive
and negative divisors b1 of b modulo squares such that the quartic equation
N 2 = b1 M 4 + aM 2 e2 +
b 4
e
b1
has solution in integers with M ,N and e pairwise coprime such that M e 6= 0.
2
, b1 MN
If (M, N, e) is such a solution then the point P = ( b1eM
2
e3 ) is in E(Q)
and α(P ) = b1 .
Remark 1 A similar theorem is true for α
b.
For more details and the the proof of the above proposition and theorems see
[1, Section 8.2.4].
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Fourth Power Diophantine Equations in Gaussian Integers
3
Remark 2 It is well-known (see e.g. [7]) that if an elliptic curve E is defined
over Q, then the rank of E over Q(i) is given by
rank(E(Q(i))) = rank(E(Q)) + rank(E−1 (Q))
where E−1 is the (−1)-twist of E over Q. We use it during the proofs of this
article.
In order to determine the torsion subgroup of E(Q(i)), we use the extended
Lutz-Nagell theorem [10, Chapter 3], which is a generalization of the LutzNagell theorem from E(Q) to E(Q(i)). Therefore throughout this article, the
following extension of the Lutz-Nagell theorem is used to compute the torsion
groups of elliptic curves.
Theorem 2 (Extended Lutz-Nagell Theorem) Let E : y 2 = x3 + Ax + B
with A, B ∈ Z[i]. If a point (x, y) ∈ E(Q(i)) has finite order, then
1. Both x, y ∈ Z[i], and
2. Either y = 0 or y 2 |(4A3 + 27B 2 ).
Now we are ready to state our result about the elliptic curves.
Theorem 3 1. Let Fp2 : Y 2 = X 3 + p2 X, where p ≡ 5 (mod 8) is a prime
integer. Then Fp2 (Q(i)) = {∞, (0, 0), (ip, 0), (−ip, 0)}.
2. For prime integers p ≡ 3 (mod 8) let Ep2 : Y 2 = X 3 − p2 X. Then
Ep2 (Q(i)) = {∞, (0, 0), (−ip, 0), (ip, 0)}.
Proof 1. The biquadratic equation of the homogeneous space of the elliptic
2
curve Fp2 is N 2 = b1 M 4 + pb1 e4 where b1 ∈ {±1, ±p ± p2 } and 1 ∈ Im(α).
Clearly, the equation has no solutions for negative b1 . Considering b1 mod
squares, we have to examine b1 = p and hence we have pM 4 + pe4 = N 2 .
Considering this equation mod 8, we see that the left hand side is equivalent
to 5, while the right hand side is equivalent to 0, 1, 4. Therefore, we get
2 b
d
b2
b3
Im(α) = {1}. Next, we consider the isogenous curve F
p2 : Y = X −4p X.
The biquadratic equation of the homogeneous space of this curve is
where
2
c4 − 4p eb4
b 2 = b1 M
N
b1
b1 ∈ {±1, ±2, ±4, ±p, ±p2, ±2p, ±4p, ±2p2, ±4p2 }.
We have 1, −1 ∈ Imb
α. Considering b1 mod square, we have to examine the
equation for b1 = ±2, ±p, ±2p. For b1 = 2 we have
c4 − 2p2 eb4 = N
b 2 ⇒ 2M
c4 = N
b2
2M
mod p
but then 2 is a square mod p so p ≡ ±1 mod 8 which is false.
Since Im(b
α) is a multiplicative group, −2 ∈
/ Im(b
α). For b1 = −p the equac4 + 4pb
b 2 . Since M
c is odd, the left hand is 3 or 7 mod 8
tion is −pM
e4 = N
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while the right hand side is 0, 1, 4. Also p ∈
/ Im(b
α) since Im(b
α) is multiplicative. Therefore, |Im(b
α)| ≤ 4. By proposition 1 the rankFp2 (Q) = 0;
Using the Extended Lutz-Nagell theorem, ∆Fp2 = 4p6 and so if (X, Y ) is
a torsion point,
Y 2 = 0 or apk
where a = ±1, ±2, ±2i, ±4 and k = 0, 2, 4, 6. For k = 0, by comparing the
power of p in Y 2 = X 3 + p2 X we have Y 2 6= apk for all a = ±1, ±2, ±2i, ±4
and k = 2, 4, 6. For Y 2 = 4, suppose that q is a prime divisor of x
in Z[i]. Then q | 4 hence, q = ω = 1 + i. Comparing the powers of
ω in both sides, we deduce that Y 2 6= 4. In a similar way, we have
Y 2 6= ±1, ±2i. Only for Y = 0 do we have X = 0, ip which means that
Fp2 (Q(i))T or = {∞, (0, 0), (ip, 0), (−ip, 0)}.
2. It is similar to the part one.
3 Main results
In this section we study some quartic Diophantine equations in the Gaussian
integers.
3.1 On the Diophantine equations y 4 ± p2 x4 = z 2 and y 4 ± x4 = pz 2
Theorem 4 1. Let p ≡ 3 (mod 8). The Diophantine equations y 4 − p2 x4 =
±z 2 and y 4 + p2 x4 = ±iz 2 has only trivial solutions in Z[i].
2. For p ≡ 5 (mod 8), Diophantine equations y 4 +p2 x4 = ±z 2 and y 4 −p2 x4 =
±iz 2 has no nontrivial solution in Z[i].
Proof In the equations y 4 ± x4 = ±pz 2 , we divide both sides by x4 and put
s = xy , t = xz2 . We have s4 ± 1 = pt2 . Let r = s2 and multiplying two last terms
we have r3 ± r = p(st)2 , which leads to the elliptic curve Y 2 = X 3 ± p2 X using
X = pr and Y = p2 st. By theorem 3 the rank of theses curves is zero and the
torsion point lead to trivial solution. The other cases are similar.
Corollary 1 Let for n ∈ N ∪ {0}:
1. Let p ≡ 3 (mod 8) The Diophantine equations y 4 − p2 x4 = ±2n z 2 , y 4 −
p2 x4 = 2n z 4 , y 4 + p2 x4 = ±2n iz 2 and y 4 + p3 x4 = 2n iz 4 have only trivial
solutions in Z[i].
2. For p ≡ 5 (mod 8), the Diophantine equations y 4 + p2 x4 = ±2n z 2 , y 4 +
p3 x4 = 2n z 4 , y 4 − p2 x4 = ±2n iz 2 and y 4 − p2 x4 = 2n iz 4 have only trivial
solutions in Z(i).
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Fourth Power Diophantine Equations in Gaussian Integers
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3.2 On the Diophantine equations 2m x4 ± 2n y 4 = z 2 and 2m x4 ± 2n y 4 = iz 2
First we state results of Najman [4]:
1. x4 ± y 4 = z 2 has only trivial solutions in the Gaussian integers. (Hilbert)
2. The equation x4 − y 4 = iz 2 has only trivial solutions in the Gaussian
integers.
3. The only nontrivial Gaussian integer solutions of the equation x4 +y 4 = iz 2
are (x, y, z), where x, y ∈ {±i, ±1}, z = ±iω and gcd(x, y, z) = 1.
It is easy to change the Diophantine equations x4 ± y 4 = ±2m z 2 , x4 ± y 4 =
±2 iz 2 , x4 ± 4m y 4 = z 2 , x4 ± 4m y 4 = iz 2 , x4 ± 4m y 4 = 2n iz 2 , to one of
the above equations and study their solvability in the Gaussian integers. Note
that we have ω 2 = 2i and ω 4 = −4. For instance, Diophantine equation
x4 + y 4 = 2z 2 transforms to X 4 + Y 4 = iZ 2 by X = x, Y = y, Z = iωz. So,
the nontrivial solutions satisfying gcd(x, y, z) = 1 in the Gaussian integers of
the equation x4 + y 4 = 2z 2 are (x, y, z), where x, y ∈ {±i, ±1}, z = ±2i. In [3]
some quartic Diophantine equations of the form x4 + kx2 y 2 + y 4 = z 2 , with
trivial integer solutions were studied. Some of these equations have solutions
in the Gaussian integers. For instance, x4 + x2 y 2 + y 4 = z 2 is impossible in
the integers, while (x, y, z) = (1, i, 1) is a nontrivial Gaussian integer solution.
m
Corollary 2 1. The only nontrivial Gaussian integer solutions of the Diophantine equation x4 + 6x2 y 2 + y 4 = z 2 are (x, y, z) where gcd(x, y) = 1,
x, y ∈ {±ω, ±ω} and z ∈ {±4}.
2. Triples (x, y, z) where x, y ∈ {±ω} and z ∈ {±4}, are the only nontrivial
solutions of Diophantine equation x4 − 6x2 y 2 + y 4 = z 2 in the Gaussian
integers.
3. The equations x4 ± 6x2 y 2 + y 4 = 2z 2 have no nontrivial solutions in the
Gaussian integers.
4. The Diophantine equations x4 ±6x2 y 2 +y 4 = iz 2 have only trivial solutions
in the Gaussian integers.
Proof 1. This equation implies that (x+y)4 +(x−y)4 = 2z 2 . By our discussion
before the corollary, we have (x + y), (x − y) ∈ {±i, ±1} and z ∈ {±1}.
An elementary calculation rise to eight nontrivial solutions in Q(i) and
therefore eight solutions in Z[i].
2. Similar to the first part (ix + y)4 + (ix − y)4 = 2z 2 . So, it is sufficient to
change the x coordinate of the above solutions to ix.
3. If (x, y, z) is a solution of x4 + 6x2 y 2 + y 4 = 2z 2 then (x + y)4 + (x − y)4 =
(2z)2 . This equation has no solution in Z[i].
4. From x4 + 6x2 y 2 + y 4 = iz 2 we have the Diophantine equation (x + y)4 +
(x − y)4 = (ωz)2 with only trivial solution.
Note that by the mapping x 7→ ωx and y 7→ ωy, we can obtain many
Diophantine equations from the above equations and discuss about their solutions.
Now consider the equation x4 ± 2m y 4 = z 2 . Without loss of generality we
suppose that 0 ≤ m ≤ 3. The cases m = 0, 2 have been considered above. The
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Diophantine equation x4 − 2y 4 = z 2 and x4 + 8y 4 = z 2 have Gaussian integer
solutions such as (3i, 2, 7) and (i, i, 3), respectively. The equation x4 −8y 4 = z 2
can be written in the form x4 + 2(ωy)4 = z 2 . So, it is sufficient to consider
x4 + 2y 4 = z 2 .
Theorem 5 The solutions of the Diophantine equation x4 + 2y 4 = z 2 are
trivial in Z[i].
Proof Let (x, y, z) be a nontrivial solution of x4 + 2y 4 = z 2 . Dividing the
equation by y 4 and considering the change of variables s = xy and t = yz2 , we
have s4 + 2 = t2 for s, t ∈ Q(i). Let
X = s2
X + 2 = t2 .
2
Multiplying both sides of these equations together and letting Y = st, we have
the elliptic curve Y 2 = X 3 + 2X. The (−1) − twist of this curve is isomorphic
to itself. Using sage and remark 2, we found that the rank of this curve is
zero over Q(i). The only torsion point (0, 0) on this curve leads to the trivial
solution for the original equation.
Corollary 3 The Diophantine equations x4 + 2y 4 = 2iz 2 and x4 − 2y 4 = iz 2
have only trivial solutions in Z[i].
Proof The first one is obvious. For the second one we have:
x4 − 2y 4 = iz 2 ⇐⇒ (2y)4 − 8x4 = (2iωz)2
⇐⇒ (2y)4 + 2(ωx)4 = (2iωz)2 .
The last equation has only trivial Gaussian integer solutions.
Acknowledgements We are indebted to an anonymous reviewer of an earlier paper for
providing insightful comments and providing directions for additional work which has resulted in this paper.
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Fourth Power Diophantine Equations in Gaussian Integers
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