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On the equation 1^k+2^k+...+x^k=y^n for fixed x

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On the equation 1^k+2^k+...+x^k=y^n for fixed x

Author: Bérczes, Attila; Hajdu, Lajos; Miyazaki, Takafumi; Pink, István
Year: 2016
Source: https://dea.lib.unideb.hu/bitstreams/e9c40910-a654-427b-bb38-6980615db216/download
ON THE EQUATION 1k+ 2k+···+xk=ynFOR FIXED x
A. B´
ERCZES, L. HAJDU, T. MIYAZAKI, AND I. PINK
Dedica ed o K´alm´an Gy˝o y on he occassion o his 75 h bi hday.
Abs ac . We p o ide all solu ions o he i le equa ion in pos-
i i e in ege s x, k, y, n wi h 1 ≤x < 25 and n≥3. Fo hese
alues o he pa ame e s, ou esul gi es an a i ma i e answe o
a ela ed, classical conjec u e o Sch¨a e . In ou p oo s we com-
bine se e al ools: Bake ’s me hod (in pa icula , sha p bounds o
he linea combina ions o loga i hms o wo algeb aic numbe s),
polynomial-exponen ial cong uences and compu a ional me hods.
1. In oduc ion
Le xand kbe posi i e in ege s. W i e
Sk(x)=1k+ 2k+···+xk.
The equa ion
(1) Sk(x) = yn
in unknown posi i e in ege s k, n, x, y wi h n≥2 has a long his o y.
The case (k, n) = (2,2) has al eady been conside ed by Lucas [9], [10],
Wa son [17] and o he s. He e we do no gi e de ails; he in e es ed
eade may consul o he book [15], he pape s [4], [1], [3] and he
e e ences gi en he ein.
I is long known ha when (k, n) is one o he pai s
(2) (1,2),(3,2),(3,4),(5,2),
hen (1) has in ini ely many solu ions. These solu ions can be desc ibed
easily. As he i s deep gene al esul , in 1956 Sch¨a e [14] p o ed ha
2010 Ma hema ics Subjec Classi ica ion. 11D61, 11D41.
Key wo ds and ph ases. Powe sums, powe s, Sch¨a e ’s conjec u e, polynomial-
exponen ial equa ions and cong uences.
Resea ch suppo ed in pa by he Uni e si y o Deb ecen, by he OTKA
g an s K100339 and NK101680, and by he T´
AMOP-4.2.2.C-11/1/KONV-2012-
0001 p ojec . The p ojec has been suppo ed by he Eu opean Union, co- inanced
by he Eu opean Social Fund. This pape was suppo ed by he J´anos Bolyai Schol-
a ship o he Hunga ian Academy o Sciences. The esea ch was also g an ed by he
Aus ian science ound (FWF) unde he p ojec P 24801-N26. The hi d au ho
was suppo ed by G an in Aid o JSPS Fellows (No.25484).
1
2 A. B´
ERCZES, L. HAJDU, T. MIYAZAKI, AND I. PINK
i (k, n) is ixed and is no in he lis (2), hen equa ion (1) has only
ini ely many solu ions. Sch¨a e ’s p oo was ine ec i e. S ill, o some
(small) pai s (k, n) he was able o show ha equa ion (1) has only
he i ial solu ion (x, y) = (1,1). Beside his, he conjec u ed ha o
(k, n) no in he lis (2), equa ion (1) has he only non i ial solu ion
(x, k, y, n) = (24,2,70,2).
Conside ably la e , Gy˝o y, Tijdeman and Voo ho e [5] ga e an e ec-
i e p oo o Sch¨a e ’s esul , in he much mo e gene al case whe e he
exponen nis also unknown. Mo eo e , Pin ´e [12] (unde some mild
assump ions) p o ed ha o he non i ial solu ions n < ck log(2k)
holds, whe e cis an e ec i ely compu able absolu e cons an . Fo u -
he esul s abou equa ion (1) and i s gene aliza ions we e e o he
book [15] and he pape s [4], [1], [3], and he e e ences he e.
The conjec u e o Sch¨a e has been e i ied unde ce ain assump-
ions o he pa ame e s in ol ed. Beside he ”small” ixed pai s (k, n)
conside ed by Sch¨a e [14], Jacobson, Pin ´e and Walsh [7] e i ied he
conjec u e o n= 2 and e en alues o kwi h 2 ≤k≤58. La e ,
Benne , Gy˝o y and Pin ´e [1] p o ed ha he conjec u e holds o any
n≥2 wi h 1 ≤k≤11. Fu he , Pin ´e [13] e i ed Sch¨a e ’s conjec-
u e o he e en alues o nwi h n > 4, p o ided ha kis odd wi h
1≤k < 170.
Recen ly, Hajdu [6] p o ed ha Sch¨a e ’s conjec u e holds unde
ce ain assump ions made on x, le ing all he o he pa ame e s ee.
Among o he esul s, he has p o ed ha he conjec u e is ue i x≡
0,3 (mod 4) and x < 25. The main ools in he p oo o his esul
we e he 2-adic alua ion o Sk(x) and local me hods o polynomial-
exponen ial cong uences.
The pu pose o he p esen pape is o ex end he esul s in [6] o
all alues o xwi h x < 25. I is impo an o men ion ha o his
pu pose we need di e en ools han hose used in [6]. The eason
is ha o he emaining alues o xwi h x < 25 (i.e. hose wi h
x≡1,2 (mod 4)) he me hods used in [6] a e no applicable. To p o e
ou main heo em, we need o combine sha p uppe bounds o linea
o ms in wo loga i hms and polynomial-exponen ial cong uences, and
we also make use o in ol ed compu a ional acili ies. The eason why
we s op a x < 25 ( hough ou me hod in p inciple is capable o co e
la ge in e als o x) is he ollowing. The o al unning ime o ou
compu e calcula ions o x= 21 ( he alue o x equi ing hea y com-
pu a ions) was al eady a ound six days. Fo la ge alues o x, he
bounds appea ing in Table 1 would be signi ican ly wo se, esul ing in
much longe unning imes in he compu a ional pa . Since sol ing
he equa ion o such alues o xwould ise ques ions mo e o echnical
ON THE EQUATION 1k+ 2k+· · · +xk=ynFOR FIXED x3
and compu a ional ype, and also because o a nice p ope y o x= 24
(being he only alue wi h a non- i ial solu ion), we decided o s op
a his poin .
The s uc u e o he pape is he ollowing. In he nex sec ion we
gi e ou main esul . In he hi d sec ion we gi e an o e iew o ou
s a egy o p o e ou main heo em, and we p o ide se e al lemmas.
Finally, in he las sec ion we gi e he p oo o ou main esul .
2. The main esul
Ou main esul is he ollowing.
Theo em 2.1. All solu ions o equa ion (1) in posi i e in ege s x, k, y, n
wi h x < 25 and n≥3a e gi en by
(x, k, y, n) = (1, k, 1, n),(8,3,6,4).
As a simple consequence we ob ain he ollowing immedia e
Co olla y 2.1. Fo x < 25 and n≥3, Sch¨a e ’s conjec u e is ue.
Rema k. We men ion ha in case o n= 2, in iew o he iden i y
S3(x) = x(x+1)
22, equa ion (1) has many mo e solu ions wi h x < 25.
3. Lemmas
In his sec ion we gi e some lemmas which a e needed in he p oo o
Theo em 2.1. Fi s we ge id o hose alues o x o which equa ion
(1) is al eady sol ed.
Lemma 3.1. Suppose ha
x∈ {1,2,3,4,7,8,11,12,15,16,19,20,23,24}.
Then equa ion (1) wi h n≥3has only he i ial solu ion wi h (x, y) =
(1,1).
P oo . The case x= 1 is i ial. When x= 2, he only solu ion o (1)
is gi en by (x, k, y, n) = (2,3,3,2) (whe e we ha e n= 2). This ac
is well-known; i ollows e.g. om he nice esul o Mih˘ailescu [11]
conce ning he Ca alan equa ion. All he o he cases a e handled by
Hajdu [6]. 
In iew o he abo e lemma, we may assume ha we ha e
x∈ {5,6,9,10,13,14,17,18,21,22}.
In hese cases, he s a egy o ou p oo is he ollowing. Fi s , using
Bake ’s me hod ( o linea o ms in wo loga i hms) we p o e ha one
4 A. B´
ERCZES, L. HAJDU, T. MIYAZAKI, AND I. PINK
o he exponen ial a iables kand nhas o be ”small”. Fo his we use
esul s o Lau en [8]. Then he emaining cases will be handled sepa-
a ely. I is impo an o men ion ha we need o p o ide a he sha p
uppe bounds o kand n(which makes he p oo s o ou co espond-
ing lemmas a he echnical). The eason is ha he ”small” alues
o kand nneed o be handled sepa a ely, one by one, by a nume ical
me hod, and he unning ime o ou algo i hm is e y sensi i e o he
ini ial uppe bounds o hese pa ame e s.
When kis small, since xis ixed, he le hand side o equa ion (1)
is ixed, and we only need o pe o m a simple check (which o ”la ge”
alues o kcan s ill be a he ime consuming). When nis ”small”
hen o each possible alues o n, we sol e (1) locally, as a polynomial-
exponen ial cong uence. A his s age we also make use o he p og am
package Magma [2]. We no e ha his is he poin whe e we need o
equi e he assump ion n > 2, since in case o n= 2 some o he
occu ing equa ions canno be handled locally.
So we s a wi h de i ing uppe bounds o he exponen ial a iables
k, n in equa ion (1). As we ha e men ioned, o his pu pose we use
Bake ’s me hod o linea o ms in loga i hms o wo algeb aic numbe s.
We need o in oduce some no a ion.
Fo an algeb aic numbe αo deg ee do e Q, we de ine he absolu e
loga i hmic heigh o αby he ollowing o mula:
h(α) = 1
d log |a0|+
d
X
i=1
log max1,|α(i)|!,
whe e a0is he leading coe icien o he minimal polynomial o αo e
Z, and α(1), α(2), ... , α(d)a e he conjuga es o αin he ield o complex
numbe s.
Le α1and α2be mul iplica i ely independen algeb aic numbe s
wi h |α1| ≥ 1 and |α2| ≥ 1. Conside he linea o m in wo loga i hms:
Λ=b2log α2−b1log α1,
whe e log α1,log α2a e any de e mina ions o he loga i hms o α1, α2
espec i ely, and b1, b2a e posi i e in ege s.
We shall use he ollowing esul due o Lau en [8].
Lemma 3.2 ([8], Theo em 2).Le ρand µbe eal numbe s wi h ρ > 1
and 1/3≤µ≤1. Se
σ=1+2µ−µ2
2, λ =σlog ρ.
ON THE EQUATION 1k+ 2k+· · · +xk=ynFOR FIXED x5
Le a1, a2be eal numbe s such ha
ai≥max {1, ρ|log αi|−log |αi|+ 2Dh(αi)}(i= 1,2),
whe e
D= [Q(α1, α2) : Q]/[R(α1, α2) : R].
Le hbe a eal numbe such ha
h≥max Dlog b1
a2
+b2
a1+ log λ+ 1.75+ 0.06, λ, Dlog 2
2.
We assume ha
a1a2≥λ2.
Pu
H=h
λ+1
σ, ω = 2 + 2 1 + 1
4H2, θ = 1 + 1
4H2+1
2H.
Then we ha e
log |Λ| ≥ −Ch02a1a2−√ωθh0−log C0h02a1a2
wi h
h0=h+λ
σ, C =C0
µ
λ3σ, C0=sCσωθ
λ3µ,
whe e
C0= ω
6+1
2sω2
9+8λω5/4θ1/4
3√a1a2H1/2+4
31
a1
+1
a2λω
H!2
.
Using his lemma, we show he ollowing.
Lemma 3.3. Le A={5,6,9,10,13,14,17,18,21,22}and conside
equa ion (1) wi h x∈Ain in ege unknowns (k, y, n)wi h k≥83, y ≥
2and n≥3a p ime. Then o y > x2we ha e n≤n0, o y > 106e en
n≤n1holds, and o y≤x2we ha e k≤k1, whe e n0=n0(x), n1=
n1(x)and k1=k1(x)a e gi en in Table 1.
P oo . In he cou se o he p oo we will always assume ha x∈Aand
we dis inguish h ee cases acco ding o y > x2,y > 106o y≤x2.
Case I. y > x2
We may suppose, wi hou loss o gene ali y, ha nis la ge enough,
ha is
(3) n>n0.
Fu he , by k≥83 we easily deduce ha o e e y x∈Awe ha e
(4) 1k+ 2k+···+xk<2xk<(x+ 1)k,

6 A. B´
ERCZES, L. HAJDU, T. MIYAZAKI, AND I. PINK
x n0(y > x2)n1(y > 106)k1(y≤x2)
5 14,000 6,100 78,000
6 21,000 10,100 121,000
9 52,000 28,000 304,000
10 65,000 36,000 381,000
13 111,000 64,000 651,000
14 129,000 75,000 754,000
17 187,000 113,000 1,099,000
18 209,000 127,100 1,224,000
21 278,000 174,100 1,633,000
22 244,000 168,000 1,466,000
Table 1. Bounding nand kunde he indica ed condi ions
and
(5) 1k+ 2k+···+ (x−1)k<2(x−1)k.
Since y > x2by (1), (4) and x≥5 we ge ha
(6) k≥2n.
Using (6) and he ac ha nis odd we may w i e kin he o m
(7) k=Bn + wi h B≥1,0≤ | | ≤ n−1
2.
We show ha in (7) we ha e 6= 0. On he con a y, suppose = 0.
Then, using (1) and (5) we in e by (7) ha
2(x−1)k>1+2k+···+ (x−1)k=yn−xk=yn−xBn
= (y−xB)(yn−1+···+xB(n−1))≥xB(n−1).
Hence
n < log x
log x
x−1+log 2
Blog x
x−1.
This oge he wi h x≤22 and B≥1 implies n < 82, which con adic s
(3). Thus, 6= 0.
On di iding equa ion (1) by ynwe ob iously ge
(8) 1 −xk
yn=s
yn,
whe e s= 1k+ 2k+. . . + (x−1)k. Using (7) and (8) we in e ha
(9) 
x ·xB
yn
−1
=s
yn.
ON THE EQUATION 1k+ 2k+· · · +xk=ynFOR FIXED x7
Pu
(10) Λ =( log x−nlog y
xBi > 0,
| |log x−nlog xB
yi < 0.
In wha ollows we ind uppe and lowe bounds o log |Λ |. We dis-
inguish wo subcases acco ding o
1−xk
yn≥0.795 o 1 −xk
yn<0.795,
espec i ely. I 1 −xk
yn≥0.795 hen by (1) and (4) we immedia ely
ob ain a con adic ion, so we may assume ha he la e case holds.
I is well known (see Lemma B.2 o [16]) ha o e e y z∈Rwi h
|z−1|<0.795 one has
(11) |log z|<2|z−1|.
On applying inequali y (11) wi h z=xk/ynwe ge by (8), (9), (10)
and xk6=yn ha
(12) |Λ |<2s
yn.
Obse e ha (1) implies
(13) k < nlog y
log x.
Thus by (12), (5) and (13) we in e ha
log |Λ |<−logx
x−1
log x(log y)n+ log 4.(14)
Nex , o a lowe bound o log |Λ |, we shall use Lemma 3.2 wi h
(α1, α2, b1, b2) = (y
xB, x, n, i > 0,
xB
y, x, n, | |i < 0.
Using (1) and (4) one can easily check ha α1>1 and α2>1. We show
ha α1, α2a e mul iplica i ely independen . Assume he con a y.
Then he se o p ime ac o s o ycoincides wi h ha o x. Since yis odd
(as x6≡ 0,3 (mod 4)), hence xis also odd, ha is, x∈ {5,9,13,17,21}.
I xis a p ime, i.e. x∈ {5,13,17}, hen yhas o be a powe o x, and
equa ion (1) can be w i en as
1k+ 2k+···+xk=xm, k ≥2, m ≥2.
8 A. B´
ERCZES, L. HAJDU, T. MIYAZAKI, AND I. PINK
One can e i y ha his equa ion has no solu ion (since xk< xm<
2xk< xk+1). I x= 9, hen yhas o be a powe o 3, and equa ion (1)
can be w i en as
1k+ 2k+···+ 9k= 3m, k ≥2, m ≥2.
Taking his equa ion modulo 4, we ha e
3+2·(−1)k≡(−1)m(mod 4).
This implies ha mis e en, and we ind
1k+ 2k+···+ 9k= 9m/2,
which, as we al eady know, has no solu ion. I x= 21, hen he se o
p ime ac o s o he in ege
1k+ 2k+···+ 21k
should be {3,7}. Howe e , we can obse e ha he abo e in ege is
no di isible by 3 i kis e en, and ha i is di isible by 11 p o ided
ha kis odd. This is a con adic ion. To sum up, we may assume ha
α1, α2a e mul iplica i ely independen .
Now, we apply Lemma 3.2 wi h he ollowing choice o pa ame e s
(ρ, µ): o e e y x∈Awe choose µ= 0.57 uni o mly, and se
(15) ρ=(7.7 i x∈A {22},
7 i x= 22.
In wha ollows we shall de i e uppe bounds o he quan i ies
ρ|log αi|−log |αi|+ 2Dh(αi),(i= 1,2)
occu ing in Lemma 3.2. Since D= 1 and α2>1, o i= 2 we ge
(16) ρ|log α2|−log |α2|+ 2Dh(α2)=(ρ+ 1) log x.
Fo i= 1 we ob ain
(17) ρ|log α1|−log |α1|+ 2Dh(α1)<ρ+ 1
2log x+ 2 log y−2 log g,
whe e g= gcd(x, y). To e i y ha (17) is alid we shall es ima e
log α1and h(α1) om abo e, by using equa ion (1), i.e. s+xBn+ =yn.
Obse e
h(α1) = h xB
y≤log max{xB, y}−log g=(log y−log gi > 0,
log xB−log gi < 0.
I > 0, hen
αn
1=y
xBn=x +s
xBn =x 1 + s
xk<2x (as s<xk),
ON THE EQUATION 1k+ 2k+· · · +xk=ynFOR FIXED x9
so
log α1<log 2
n+
nlog x≤log 2
n+n−1
2nlog x,
whence
ρ|log α1|−log |α1|+ 2Dh(α1)<
<log 2
nlog x+n−1
2n(ρ−1) log x+ 2 log y−2 log g
which by (15), (3) and x≥5 clea ly implies (17).
I < 0, hen
αn
1=xB
yn
=x− 1−s
yn< x− =x| |,
so
log α1<| |
nlog x≤n−1
2nlog x,
and
log xB= log α1+ log y < n−1
2nlog x+ log y,
and we ge
ρ|log α1|−log |α1|+ 2Dh(α1)<
<n−1
2n(ρ−1) + n−1
nlog x+ 2 log y−2 log g,
which by (3) again implies (17).
In iew o (16) we can ob iously ake o e e y x∈A
(18) a2= (ρ+ 1) log x.
Fo he alues a1we do he ollowing. I we can calcula e he exac
alue o g= gcd(x, y) hen we use o a1 he uppe bound occu ing in
(17), while i we do no know he exac alue o g= gcd(x, y) we use
(17) wi h g= 1. Namely, we can ake a1as
(19) a1=(ρ+1
2log x+ 2 log yi x∈A {22},
ρ+1
2log x+ 2 log y−2 log 11 i x= 22.
To see ha he choice o a1 o x= 22 is alid we obse e ha
(20)
(Sk(22) ≡0 (mod 3) and Sk(22) 6≡ 0 (mod 9) i kis e en,
Sk(22) ≡0 (mod 11) i kis odd,
Since by (3) nis la ge in equa ion (1), we may assume ha kis odd,
hence (20) oge he wi h (1), implies y≡0 (mod 11). Thus, since y
16 A. B´
ERCZES, L. HAJDU, T. MIYAZAKI, AND I. PINK
Lemma 3.5. Le A={5,6,9,10,13,14,17,18,21,22}. Assume ha
equa ion (1) has a solu ion (x, k, y, n)wi h x∈Asuch ha ei he :
(i) k≥83 and x2< y ≤106, o
(ii) k≥83 and y≤x2.
Then we ha e n < 12.
P oo . In he case (ii), by Lemma 3.3 we immedia ely ge k≤k1. In
he case o (i), by Lemma 3.3 we ge n≤n0, which oge he wi h (1)
and he assump ion y≤106gi es he es ima e
k < nlog y
log x≤6n0log 10
log x.
Pu k0:= max n6n0log 10
log x, k1o, which o gi en xis a ixed numbe .
Fo gi en x∈Ale us ake any ix alue o kwi h 2 ≤k≤k0.
Fo e e y p ime 2 ≤p≤106we check whe he Sk(x) is di isible by p
(in ac we compu e Sk(x) (mod p) and we check i i is 0 o no ). I
pdi ides Sk(x), hen we check i Sk(x) is also di isible by p12 o no .
Du ing ou compu a ions o e e y possible pai (x, k) we ei he ound
ha he e is no p ime p≤106di iding Sk(x) a all, which by y≤106
p o es he e is no solu ion, o we could ind a p ime di iso p≤106
o Sk(x) wi h he p ope y ha p12 does no di ide Sk(x) leading o
he conclusion ha n < 12. The compu a ions we e pe o med again
in Magma [2]. 
Lemma 3.6. The only solu ion o equa ion (1) wi h 5≤x < 25 and
n≥3unde he assump ion k≤100 is (x, k, y, n) = (8,3,6,4).
P oo . A di ec compu a ion o Sk(x) o all possible pai s (k, x) co e-
sponding o he equi emen s o Lemma 3.6, and checking whe he i
is a pe ec powe can be done in Magma [2] in a ew seconds. 
4. P oo o Theo em 2.1
In p inciple he p oo is a simple combina ion o he abo e p o ed
lemmas, namely o Lemma 3.3, Lemma 3.4, Lemma 3.5 and Lemma
3.6.
P oo o Theo em 2.1. Clea ly, i is enough o p o e he heo em o
n= 4 and o odd p ime alues o n. Fu he , he cases
x∈ {1,2,3,4,7,8,11,12,15,16,19,20,23,24}
a e handled by Lemma 3.1. So now we only need o p o e Theo em
2.1 o x∈A={5,6,9,10,13,14,17,18,21,22}.

ON THE EQUATION 1k+ 2k+· · · +xk=ynFOR FIXED x17
We spli he p oo in o se e al subcases. The case k≤100 is com-
ple ely co e ed by Lemma 3.6, so o he es o he p oo we may
assume k > 100. I y > 106 hen by Lemma 3.3 we ha e n≤n1,
and by Lemma 3.4 we know ha he e is no solu ion o 3 ≤n≤
n1, n p ime o n= 4. This concludes he p oo o Theo em 2.1 when-
e e y > 106.
Fo y≤106by Lemma 3.5 we ha e n < 12. Thus by x≥5 om (1)
we ge he es ima e
k < nlog y
log x≤66 log 10
log 5 <100,
which has been ea ed al eady. 
Re e ences
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18 A. B´
ERCZES, L. HAJDU, T. MIYAZAKI, AND I. PINK
[17] G. N. Wa son, The p oblem o he squa e py amid, Messenge o Ma h. 48
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A. B´
e czes and L. Hajdu,
Uni e si y o Deb ecen, Ins i u e o Ma hema ics
H-4010 Deb ecen, P.O. Box 12.
Hunga y
E-mail add ess:[email p o ec ed]
E-mail add ess:[email p o ec ed]
T. Miyazaki
Facul y o Science and Enginee ing, Gunma Uni e si y,
Gunma, 376-8515,
Japan
E-mail add ess:[email p o ec ed]
I. Pink
Ins i u e o Ma hema ics, Uni e si y o Deb ecen
H-4010 Deb ecen, P.O. Box 12, Hunga y
and
Uni e si y o Salzbu g
Hellb unne s asse 34/I
A-5020 Salzbu g, Aus ia
E-mail add ess:[email p o ec ed]; [email p o ec ed]