Math. J. Okayama Univ. 63 (2021), 53–60

REMARK ON A PAPER
BY IZADI AND BAGHALAGHDAM
ABOUT CUBES AND FIFTH POWERS SUMS
Gaku IOKIBE
Abstract. In this paper, we refine the method introduced by Izadi andBaghalaghdam to search integer solutions to the Diophantine equation
X15 + X25 + X35 = Y13 + Y23 + Y33 . We show that the Diophantine equation
has infinitely many positive solutions.

1. Introduction
In [2], Izadi and Baghalaghdam consider the Diophantine equation:(1)a(X1′5 + X2′5 ) +nXai Xi5 = b(Y1′3 + Y2′3 ) +

i=0

m
X

bi Yi3i=0where n, m ∈ N ∪ {0}, a, b 6= 0, ai , bi are fixed arbitrary rational numbers.
They use theory of elliptic curves to find nontrivial integer solutions to (1).
In particular, they discuss the equation:
X15 + X25 + X35 = Y13 + Y23 + Y33(2)

and obtain integer solutions, for example:
85 + 65 + 145 = (−110)3 + 1243 + 143 ,
1281225 + (−79524)5 + 485985 = 3592275803 + (−251874598)3 + 1073529823 .
However, no positive solutions are presented in their paper [2]. In this paper,
we refine their method to find positive solutions to (2).
Consider the Diophantine equation (2). Let:(3)

X1 = t + x1 ,Y1 = t + v,

Then we get a quartic curve:

X2 = t − x1 ,Y2 = t − v,

X3 = αt,Y3 = βt.

2 + α5 4 20x21 − 2 − β 3 2 5x41
t +
t +
6
6
3
with parameters x1 , α, β ∈ Q. If we get a rational point (t, v) on C, we
can compute a rational solution to (2) (see [2]).

(4)

C : v2 =Mathematics Subject Classification. Primary 11D41; Secondary 11D45, 14H52.
Key words and phrases. Diophantine equations, Elliptic Curves.
53

54

G. IOKIBE

Once we obtain rational solutions to (2), we can obtain integer solutions
by multiplying an appropriate value to Xi , Yi . In the same way, in orderto obtain solutions in positive integers, it suffices to search positive rational
solutions to equation (2).
2. Additional Requirements for Positive Solutions
Suppose that a positive rational solution (Xi , Yi )1≤i≤3 to (2) is obtained
from a given point (t, v) on the quartic C.
Proposition 2.1. Let α, β, x1 ∈ Q and2 + α5 4 20x21 − 2 − β 3 2 5x41
F (t) =
t +
t +
.
6
6
3
A rational point (t, v) on the curve C : v 2 = F (t) in (4) produces a positive
rational solution to (2) by (3) if and only if
(5)
hold.

α, β > 0,

0 ≤ F (t) < t2 ,t > |x1 |Proof. If Xi and Yi are positive in the solution in the form (3), we havet = (X1 + X2 )/2 > 0, α = 2X3 /(X1 + X2 ) > 0 and β = 2Y3 /(Y1 + Y2 ) > 0.
For (t, v) ∈ C, one has that 0 ≤ v 2 = F (t) < v 2 + Y1 Y2 = t2 . It follows
from x21 < x21 + X1 X2 = t2 that t > |x1 | for t > 0. Conversely, suppose
the inequalities in (5) hold. Then the given point (t, v) on C satisfies v 2 =F (t) < t2 . This and (5) immediately imply Xi , Yi > 0 in (3).

Proposition 2.2. Under the same assumption as Proposition 2.1, let
2 + α520x21 − 8 − β 3
5
(6)
a=
, b=
, c = x41 .
66
3
2Then a, b, c satisfy b − 4ac > 0 and b < 0 if and only if there exists a real
number t such that F (t) < t2 .
Proof. Let F̃ (t) = F (t)− t2 . Since F̃ (0) = 5x41 /3 ≥ 0, and in this case a > 0,
it is easy to see that the following conditions are equivalent to each others:(i) There exists a real number t such that F (t) < t2 .
(ii) The equation F̃ (t) = 0 has four distinct solutions.
(iii) The quadratic equation ax2 +bx+c = 0 has two distinct non-negative
solutions.
(iv) The discriminant D = b2 − 4ac of the quadratic function f (x) =ax2 + bx + c is positive, and the axis of the quadratic function −b/2a ispositive, and f (0) ≥ 0.
The condition (iv) holds if and only if “b2 − 4ac > 0 and b < 0”, sincea > 0 and f (0) = c = 5x41 /3 ≥ 0.
CUBES AND FIFTH POWERS SUMS

55

3. Example for X15 + X25 + X35 = Y13 + Y23 + Y33
Let us first search parameters (x1 , α, β) such that0 < α, β,b < 0 < b2 − 4acwith a, b, c given by (6) and such that the quartic curve C of (4) has atleast one rational point. Note that these are necessary to satisfy conditionsof Proposition 2.1, 2.2. Then, the curve C is birationally equivalent to an
elliptic curve E over Q. If E has positive rank, then C has infinitely many
rational points.
Let (x1 , α, β) = (2, 1, 16). Then the quartic:
1 4 2009 2 80
t −
t + ,
2
3
3
has a rational point (t, v) = (44, 760). By T = t − 44, we transform C into
C : v2 =15415 2 334312
1
T +T + 7602
C ′ : v 2 = T 4 + 88T 3 +233
which is birationally equivalent over Q to the cubic elliptic curve (see [5,Theorem 2.17], [2]):E : y2 +41789
76876021 2
2460032672
xy + 133760y = x3 −x − 1155200x +,285
3249009where:
2

3343122 · 760(x + 15415
3 ) − 2·32 ·760,
T =y

T (T x − 334312
3 )v = −760 +.
2 · 760

Using the Sage software [3], we find that the cubic curve E is an elliptic
curve which has rank 2 and the generators of E are:
351379 47356344241
1802189 5513659679, P2 = −.
,,P1 = −
1521
417430363
2276010
We now consider the subset

C0 = (t, v) ∈ C | 0 ≤ F (t) < t2 ⊂ Cwhose points satisfy another condition (5) of Proposition 2.1. The twoquartic equations:2009 2 80
12009 2 80
1F (t) = t4 −t += 0, F̃ (t) = t4 −t +− t2 = 0
2
3
3
2
3
3
have respectively solutions:
q
q
√
√
1
2
t=±
6027 ± 3 4035601, t = ±1509 ± 3 252979.
3
3

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G. IOKIBE

Let us take larger solutions as:q√16027 + 3 4035601 ≃ 36.59635926...
a1 =
3q
√
2
1509 + 3 252979 ≃ 36.62367500...
a2 =3
If a point (t0 , v0 ) on C satisfies a1 ≤ t0 ≤ a2 , then (t0 , v0 ) lies on C0 . We now
make use of the composition law of points on the elliptic curve E. Since E
has positive rank, we can test infinitely many rational points of E till finding
a point (t0 , v0 ) on C0 . We find that the rational point
4767546475726965161322288395890039
Q = 2P1 − P2 = 304845381192111829037,−
58470412871306667
4652843756178203561643745770
on E corresponds to(t0 , v0 ) =690740884062625663919872925291699877683029096
170815619844155909156204
4664941095250009917983 , − 21761675422152362106175457381859866386788289

on C0 , and creates a positive rational solution:

X1 = 180145502034655928992170
4664941095250009917983 ≃ 38.61688676...
X2 = 161485737653655889320238
4664941095250009917983 ≃ 34.61688676...
X3 = 170815619844155909156204
4664941095250009917983 ≃ 36.61688676...

Y1 = 106103920658980331397442614601687483092587436
21761675422152362106175457381859866386788289 ≃ 4.875723886...

Y2 = 1487585688784231659237188465185087238458645628
21761675422152362106175457381859866386788289 ≃ 68.35804964...
Y3 = 2733049917506494546499264
4664941095250009917983 ≃ 585.8701882...

Next we shall prove that the Diophantine equation (2) has infinitely many
positive solutions. The real locus of elliptic curve E(R) can be regarded as
a compact topological subspace of complex projective variety E.
Lemma 3.1. If the rank of elliptic curve E over Q is positive, every pointof E(Q) is an accumulation point in E(R).
Proof. Since E(R) is a compact topological group, and E(Q) is an infinite
subgroup of E(R), there is at least one accumulation point of E(Q) in E(R).
The group operations are homeomorphisms from E(R) to itself. Therefore
all points of E(Q) are accumulation points of E(R).

Theorem 3.2. The Diophantine equation (2) has infinitely many positive
solutions.
Proof. The part of C0 has one rational point (t0 , v0 ) which corresponds to theabove point Q. By Lemma 3.1, the point Q is an accumulation point of E(Q)in E(R), and (t0 , v0 ) is that of C(Q) in C(R). Thus the part of C0 includes
infinitely many rational points. Since 2 = |x1 | < a1 = 36.59635926..., theycorrespond to positive rational solutions to (2).
CUBES AND FIFTH POWERS SUMS

57

4. Example for X15 + X25 = Y13 + Y23 + Y33
Let α = 0. Then (2) gives another Diophantine equation:(7)

X15 + X25 = Y13 + Y23 + Y33 .

In the same way, we can obtain a rational or positive rational solutions of
it. For example, let x1 = 10, β = 18. Then the quartic curve:
50000
1
C : v 2 = t4 − 639t2 +3
3
has a rational point (t, v) = (−5, 30) and can be regarded as an elliptic curve
over Q that has rank 2. It is birationally equivalent to:1867
3676525 2367652500
E : y2 +xy − 400y = x3 −x − 1200x +
.
9
324
27
From this, we can compute positive rational solutions to (7). For example,
there is a point Q = (x0 , y0 ) on E withx0 = 9233921838917810856046138588468998730
71226852166762122405616706766475947
corresponding to (t0 , v0 ) on C with
t0 = 7869911761727476320751662986237524106650
180965667579279848488380712753242417827
which creates the following solution to (7):X1 = 9679568437520274805635470113769948284920
180965667579279848488380712753242417827 ,
X2 = 6060255085934677835867855858705099928380
180965667579279848488380712753242417827 ,Y1 = 2102579397586077496858869804126511993988094601307100986270258503567645177035000
32748572842414417658282657731373155447687070419319181813277645661864847401929 ,Y2 = 745788273916000738265027095213285105579870143595196644344754733646843241464100
32748572842414417658282657731373155447687070419319181813277645661864847401929 ,Y3 = 141658411711094573773529933752275433919700
180965667579279848488380712753242417827 .
The case of β = 0 will be discussed briefly in 5.2 below.
5. Parameters (x1 , α, β) from Trivial Solutions
5.1.

There are several trivial solutions; for example:
15 + 15 + 15 = 13 + 13 + 13 .

We call solutions to (2) which consist of 0, ±1 trivial. We are going to checksome of them to search integer (or positive) solutions.
A solution to (2) may decide parameter. For example, when Xi = Yi = 1(i = 1, 2, 3), we get (x1 , α, β) = (0, 1, 1). Then:
1
1
C : v 2 = t4 − t2
2
2

58

G. IOKIBE

has a singular point (t, v) = (0, 0)
and can be parametrized by one parameter. Let us divide both sides of C by t4 and substitute s, w for 1/t, v/t2
respectively. Then:
1 1
C ′ : w 2 = − s2
2 2
has a rational point (s, w) = (1, 0). Hence we can parametrize rational
points on C ′ and integer solutions to (2). That is to say we have:5  25  25
 2
2k + 1
2k + 1
2k + 1++2k2 − 12k2 − 12k2 − 1 4
3  43  23
4k − 4k 3 − 2k − 1
4k + 4k 3 + 2k − 1
2k + 1=++(2k2 − 1)2(2k 2 − 1)22k2 − 1
where k ∈ Q. We can see that large enough k give positive solutions to (2).
For example: 5  5  5  3  3  3
9
9
27
99
9
9
+
+
=
+
+
7
7
7
49
49
7

where k = 2. Since X1 = X2 = X3 = Y3 , this solution also gives positive
solution to another Diophantine equation 3X 5 = Y13 + Y23 + X 3 . Moreoverit satisfies X1 + X2 + X3 = Y1 + Y2 + Y3 because α = β.
5.2.

From another trivial solution:
15 + 05 + 05 = 13 + 03 + 03 ,we can derive parameters (x1 , α, β) = ( 21 , 0, 0). Then:1
5
1
C : v 2 = t4 + t2 +3
2
48
is an elliptic curve defined over Q with rational point (t, v) = ( 21 , 21 ). It isbirationally equivalent to:
4
2
5
1
5
E : y 2 + xy + y = x3 + x2 − x −
3
3
9
3
27
over Q and has rank 1. Hence we can apply the method of Section 3 tocompute positive solutions to(8)

X15 + X25 = Y13 + Y23

as a special case of (2) with X1 , X2 , Y1 , Y2 > 0, X3 = Y3 = 0 (where α =β = 0 in (3)). For example, a point29224609136538294659462738431
Q = 10017045137918654785
165672066306928896 , 67433225470590933809197056

on E corresponding to the point(t0 , v0 ) =2806052350871126431439 5797926783162005502807971914786692611082209
4379016004568066987998 , 9587890584131638439948667971418559938024002

CUBES AND FIFTH POWERS SUMS

59

on C creates the positive solution to (8):2497780176577579962719
308272174293546468720
X1 = 2189508002284033493999, X2 = 2189508002284033493999,

Y1 = 5970900430111130674379700360675596051258385
4793945292065819219974333985709279969012001 ,172973646949125171571728445888903440176176.
Y2 = 4793945292065819219974333985709279969012001

5.3. There exists one more parameter with β = 0, (x1 , α, β) = (0, 0, 0),
which is derived from the trivial solution:
15 + 15 + 05 = 13 + 13 + 03 .
Then the rational points on:
1
1
C : v 2 = t4 − t2
33
can be parametrized. Thus we have:

3k 2 + 1
3k 2 − 1

5+

3k 2 + 1
3k 2 − 1

5=



9k 4 − 6k 3 − 2k − 1(3k 2 − 1)2

3

+



9k 4 + 6k 3 + 2k − 1(3k 2 − 1)2

3,

where k ∈ Q. For example, substituting 2 for k, we have:
3 
3

 5  5
13
91
195
13
+
+ 05 =++ 03 .
11
11
121
121
The solutions which are obtained in these ways give solutions to another
Diophantine equation 2X 5 = Y13 + Y23 .
5.4. It is not simple to find parameters (x1 , α, β) that produce elliptic
curves for non-trivial solutions (Xi , Yi )1≤i≤3 . In particular, the author could
not find a good parameter for β = 0, α 6= 0:
Question 5.1. Find (a good method for) positive solutions to:
X15 + X25 + X35 = Y13 + Y23 .
Acknowledgement: The author would like to thank the referee for many
valuable suggestions to improve this article.
References[1] G.Iokibe: Search for positive solutions to Diophantine equations with cubes and fifth
powers sums, Master thesis, Department of Mathematics, Osaka University, February
2018.
[2] F.Izadi and M.Baghalaghdam: On the Diophantine equation in the form that asum of cubes equals a sum of quintics, Math. J. Okayama Univ. 61 (2019), 75–84.
(arXiv:1704.00600v1 [math.NT] 30 Mar 2017)[3] Sage software, available from http://www.sagemath.org/[4] J. Silverman: The Arithmetic of Elliptic Curves, Graduate Texts in Mathematics, vol.
106, Springer-Verlag, New York, second edition, 2008.

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G. IOKIBE[5] L. C. Washington: Elliptic Curves: Number Theory and Cryptography, Chapman &Hall/CRC, 2003.
Gaku Iokibe
Department of Mathematics, Graduate School of Science, Osaka University,
Toyonaka, Osaka 560-0043, Japane-mail address: u325137g@alumni.osaka-u.ac.jp(Received December 16, 2018 )(Accepted June 19, 2019 )