ON A CONJECTURE OF SCH ¨
AFFER CONCERNING
THE EQUATION 1k+· · · +xk=yn
L. HAJDU
Dedica ed o N. Sa adha on he occasion o he 60 h bi hday.
Abs ac . We p o e Sch¨affe ’s conjec u e conce ning he solu-
ions o he equa ion in he i le unde ce ain assump ions on x,
le ing he o he a iables k, n, y be comple ely ee. We also p o-
ide uppe bounds o nunde mo e mode a e condi ions. Finally,
we gi e all solu ions o he equa ion in he i le o some conc e e
alues o x. Ou esul s ely on asse ions desc ibing he p ecise
exponen s o 2 and 3 appea ing in he p ime ac o iza ion o Sk(x)
and on he explici solu ion o polynomial-exponen ial cong uences.
1. In oduc ion
Fo posi i e in ege s kand x, w i e
Sk(x) = 1k+· · · +xk
o he sum o he k- h powe s o he fi s xposi i e in ege s. The
Diophan ine equa ion
(1) Sk(x) = yn
in posi i e in ege s k, n, x, y wi h n≥2 has a long his o y, going back
o Lucas [12, 13], Wa son [22] and o he s, who conside ed he case
(k, n) = (2,2). Fo de ails and mo e his o y we e e o he book [20]
and he pape s [2, 6, 7] and he e e ences gi en he e.
As i is well-known, in he case when (k, n) belongs o he se
(2) {(1,2),(3,2),(3,4),(5,2)}
(1) has infini ely many solu ions, which can be desc ibed easily. In
1956, Sch¨affe [18] p o ed ha o any fixed (k, n) no in he se (2),
equa ion (1) has only fini ely many solu ions. Sch¨affe ’s p oo was no
2010 Ma hema ics Subjec Classi ica ion. 11D61, 11D41.
Key wo ds and ph ases. Sch¨affe ’s conjec u e, powe sums, powe s, polynomial-
exponen ial cong uences.
Resea ch suppo ed in pa 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 sup-
po ed by he Eu opean Union, co-financed by he Eu opean Social Fund.
1
2 L. HAJDU
effec i e, hough o ce ain (small) pai s (k, n) he could show ha (1)
has only he i ial solu ion (x, y) = (1,1). Fu he , he conjec u ed
ha o (k, n) no in he se (2), equa ion (1) has he only non i ial
solu ion (k, n, x, y) = (2,2,24,70). F om his poin on, he solu ions
men ioned so a will be e e ed o as known solu ions.
La e , Gy˝o y, Tijdeman and Voo hoe e [8] p o ided an effec i e
p oo o a mo e gene al e sion o Sch¨affe ’s heo em, whe e he expo-
nen nis also unknown. Fu he , unde ce ain assump ions Pin ´e [16]
p o ed ha o he non i ial solu ions we ha e n < ck log(2k), whe e
cis an effec i ely compu able absolu e cons an . Fo mo e esul s con-
ce ning (1) and i s a ious gene aliza ions, we e e once again o he
book [20] and he pape s [2, 6, 7] and he e e ences he ein.
Beside he abo e men ioned spa se pai s (k, n) conside ed by Sch¨affe
himsel , Sch¨affe ’s conjec u e has been e ified o la ge se s o he
pa ame e s in ol ed. Jacobson, Pin ´e and Walsh [9] p o ed ha he
conjec u e is ue o n= 2 and e en alues o kwi h 2 ≤k≤58.
La e , Benne , Gy˝o y and Pin ´e [2] showed ha he conjec u e is
alid o any n≥2 o 1 ≤k≤11. Recen ly, Pin ´e [17] p o ed ha
Sch¨affe ’s conjec u e is also ue whene e nis e en wi h n > 4 and k
is odd wi h 1 ≤k < 170.
A common ea u e o all he abo e esul s is ha a leas one pa-
ame e in (1) is conside ed o be fixed, o belongs o a ela i ely small
fini e se . In his pape we p o e Sch¨affe ’s conjec u e unde ce ain
assump ions on x, le ing he o he a iables k, n, y o be comple ely
ee. As a as we know, his is he fi s esul o his ype in he li e a-
u e. We also men ion ha he assump ions imposed on xa e sa isfied
by a posi i e p opo ion o he posi i e in ege s. In pa icula , o he
e en alues o ki is sufficien o assume ha x≡3,4 (mod 8).
Ou esul s mainly ely on asse ions desc ibing he exac alues o
ν2(Sk(x)) and ν3(Sk(x)), whe e νp(N) s ands o he exponen o he
p ime pappea ing in he p ime ac o iza ion o he posi i e in ege N.
The esul desc ibing ν2(Sk(x)) is due o MacMillan and Sondow [14],
while he asse ion conce ning ν3(Sk(x)) is new in i s ull gene ali y.
(The case when kis e en is gi en by Sondow and Tsuke man [21].)
No e ha many asse ions o a somewha simila ype a e also known;
see e.g. he pape s [10, 15] and he e e ences he e, dealing wi h
he E d˝os-Mose conjec u e conce ning he solu ions o he equa ion
Sk(x)=(x+ 1)k, o [19] abou a p oblem o Bedna ek asking o
desc ibing hose pai s (k, m) o which Sk(x) di ides Skm(x) o all
posi i e in ege s x; c . also [4, 5] and he e e ences he e o ce ain
o he ela ed p oblems.
ON A CONJECTURE OF SCH ¨
AFFER CONCERNING 1k+· · · +xk=yn3
Apa om his, a e bounding n, we shall also use local a gumen s
in o de o sol e equa ion (1) o fixed n. In pa icula , when we
conside x o be also fixed, (1) is a kind o exponen ial-polynomial
equa ion. Such equa ions a e o classical and ecen in e es . He e we
only e e o he pape s [1, 3] dealing wi h powe s ha ing ew digi s,
and he e e ences he e. We also men ion ha a esul o Lei ne [11]
implies he solu ion o (1) o n= 2 and x= 3.
In he nex sec ion we gi e ou esul s. Beside he al eady men ioned
heo em yielding a posi i e answe o Sch¨affe ’s conjec u e unde ce -
ain assump ions on x, we p o ide uppe bounds o nunde mo e
mode a e condi ions. Finally, we also gi e all solu ions o equa ion
(1) o some alues o x. In he hi d sec ion we gi e he o mula o
ν2(Sk(x)) om [14] and es ablish he o mula o ν3(Sk(x)). Finally, in
he las sec ion we gi e he p oo s o ou heo ems.
2. New esul s
Ou fi s esul p o ides a posi i e answe o he conjec u e o
Sch¨affe , unde ce ain cong uence condi ions on x, le ing he o he
h ee pa ame e s k, n, y o be comple ely ee.
Theo em 2.1. Assume ha x≡3,4 (mod 8). Then equa ion (1) has
no solu ions wi h k= 1 o ke en.
Fu he , i one o he cong uences x≡hi(mod mi)wi h hi∈Hi
(i= 1,2,3,4) is also alid, whe e
H1={2}, H2={5,7}, H3={2,7,9,14}, H4={18,22},
and
m1= 5, m2= 13, m3= 17, m4= 41,
hen equa ion (1) has only he known solu ions.
Rema k 1. No e ha i would be easy o p o ide u he cong uence
condi ions when he asse ion o Theo em 2.1 emains alid, e.g. based
upon he p oo o Theo em 2.3. Howe e , since i is clea ha ou
p esen me hod is no capable o sol e he conjec u e comple ely, we
do no wan o s ess his poin u he .
Ou nex esul p o ides uppe bounds o he exponen nin (1) in
e ms o he 2 and 3 alua ions ν2and ν3o some unc ions o xand
x, k. Recall ha νp(N) s ands o he exponen o he p ime pin he
p ime ac o iza ion o he posi i e in ege N.
4 L. HAJDU
Theo em 2.2. i) Suppose i s ha x≡0,3 (mod 4). Then o any
solu ion (k, n, x, y)o equa ion (1) we ha e
n≤{ν2(x(x+ 1)) −1,i k= 1 o kis e en,
2ν2(x(x+ 1)) −2,i k≥3is odd.
ii) Assume now ha x≡0,8 (mod 9). Then o any solu ion (k, n, x, y)
o equa ion (1) we ha e
n≤




ν3(x(x+ 1)),i k= 1,
ν3(x(x+ 1)(2x+ 1)) −1,i kis e en,
ν3(kx2(x+ 1)2)−1,i k≥3is odd.
Rema k 2. No e ha om he p oo one can easily see ha in ac n
di ides he exp ession occu ing in he igh hand side in he inequal-
i ies in pa s i) and ii) o he heo em. We also men ion ha assum-
ing ha xsa isfies bo h cong uences x≡0,3 (mod 4) and x≡0,8
(mod 9), one can ce ainly combine he asse ions o pa s i) and ii) o
Theo em 2.2.
Finally, we gi e he comple e solu ion o equa ion (1) o alues o
xco esponding o pa i) o Theo em 2.2. The eason why we go up
o x= 24 is he exis ence o he “in e es ing” solu ion (k, n, x, y) =
(2,2,24,70).
Theo em 2.3. Suppose ha x≡0,3 (mod 4) and x < 25. Then
equa ion (1) has only he known solu ions.
3. Fo mulas o ν2(Sk(x)) and ν3(Sk(x))
One o ou main ools in he p oo s will be p ecise knowledge o
he alues o ν2(Sk(x)) and ν3(Sk(x)). The in o ma ion conce ning
ν2(Sk(x)) is due o MacMillan and Sondow [14], and is he ollowing.
Lemma 3.1. Le xbe a posi i e in ege . Then we ha e
ν2(Sk(x)) = {ν2(x(x+ 1)) −1,i k= 1 o kis e en,
2ν2(x(x+ 1)) −2,i k≥3is odd.
The desc ip ion o he alue o ν3(Sk(x)) is gi en by he ollowing
lemma. No e ha he case when kis e en has been p o ed by Sondow
and Tsuke man, see Co olla y 9 in [21]. Howe e , o he con enience
o he eade , ou p oo co e s his pa o he s a emen , as well.
ON A CONJECTURE OF SCH ¨
AFFER CONCERNING 1k+· · · +xk=yn5
Lemma 3.2. Le xbe a posi i e in ege . Then we ha e
ν3(Sk(x)) = 








ν3(x(x+ 1)),i k= 1,
ν3(x(x+ 1)(2x+ 1)) −1,i kis e en,
0,i x≡1 (mod 3) and k≥3is odd,
ν3(kx2(x+ 1)2)−1,i x≡0,2 (mod 3) and k≥3is odd.
P oo . Since S1(x) = x(x+1)/2 and S2(x) = x(x+1)(2x+1)/6 o any
posi i e in ege x, o k= 1 and 2 he s a emen is au oma ic. Hence
om his poin on we shall always assume ha k≥3.
Now we shall p oceed by induc ion on x. The s a emen is ob ious
o x= 1, and also o x= 2 i k= 1 o kis e en. When x= 2 and
k≥3 is odd, we can w i e
Sk(2) = 1 + (3 −1)k= 3k+
k
∑
i=2
(−1)k−i(k
i)3i.
By obse ing ha
ν3((k
i)3i)=ν3((k−1
i−1))+ν3(k)−ν3(i)+i > ν3(k)+1 (2 ≤i≤k),
he s a emen ollows in his case, as well.
Conside now he s a emen o some alue xwi h x≥3, and assume
ha he asse ion is alid o all x′wi h 1 ≤x′< x ( o all posi i e
in ege s k).
We dis inguish wo cases. Assume fi s ha xis o he o m ε3αwi h
ε= 1,2 and α≥1. Now i kis e en, hen we ha e
Sk(3α) =
3α−1
2
∑
i=0
(ik+ (3α−i)k)≡2Sk(3α−1
2)(mod 3α)
and
Sk(2 ·3α) = 3αk +
3α−1
∑
i=0
(ik+ (2 ·3α−i)k)≡2Sk(3α−1) (mod 3α)
o ε= 1 and 2, espec i ely. Since he induc ion hypo hesis now
implies
ν3(Sk(3α)) = ν3(Sk(3α−1
2))=α−1
and
ν3(Sk(2 ·3α)) = ν3(Sk(3α−1)) = α−1,

6 L. HAJDU
we a e done in his case. On he o he hand, i kis odd hen w i ing
k:= 3γk′wi h γ≥0 and 3 -k′, using
ν3((3γ
u)3αu)≥γ−ν3(u) + αu ≥2α+γ o 2 ≤u≤3γ
and
(3α+γi3γ−1−i3γ)k′≡k′3α+γik−1−ik(mod 32α+γ)
by he induc ion hypo hesis o ε= 1 we ob ain
Sk(3α) =
3α−1
2
∑
i=0
(ik+ ((3α−i)3γ)k′)≡
3α−1
2
∑
i=0
(ik+ (3α+γi3γ−1−i3γ)k′)≡
≡
3α−1
2
∑
i=0
k′3α+γik−1≡ ±32α+γ−1(mod 32α+γ)
which p o es ou claim. In case o ε= 2 by a simila a gumen and
wi h he same no a ion we ge
Sk(2 ·3α) = 3αk +
3α−1
∑
i=0
(ik−((2 ·3α−i)3γ)k′)≡
≡
3α−1
∑
i=0
(ik+ (2 ·3α+γi3γ−1−i3γ)k′)≡
≡
3α−1
∑
i=0
k′2·3α+γik−1≡ ±32α+γ−1(mod 32α+γ)
and he s a emen ollows also in his case.
Suppose nex ha xis no o he o m ε3αwi h ε= 1,2 and α≥1.
Then, as x≥3, by he e na y expansion o x, we can w i e x=
η3β+ε3α, wi h ηa posi i e in ege no di isible by 3, ε= 1,2, and
in ege s βand αwi h β > α ≥0. Then we ha e
Sk(x) = Sk(η3β) +
ε3α
∑
i=1
k
∑
j=0 (k
j)(η3β)k−jij=
=Sk(η3β) +
k
∑
j=0 (k
j)(η3β)k−jSj(ε3α)
whe e S0(x) = x. Now as
ν3((k
j))=ν3(( k
k−j))≥max(ν3(k)−ν3(j), ν3(k)−ν3(k−j))
ON A CONJECTURE OF SCH ¨
AFFER CONCERNING 1k+· · · +xk=yn7
o 1 ≤j≤k−1, using induc ion one can easily see ha
ν3(Sk(x)) = ν3(Sk(ε3α)).
Hence he lemma ollows. 
4. P oo s o he heo ems
Now we a e eady o gi e he p oo s o ou heo ems. We s a wi h
Theo em 2.2, since i will be used in he p oo s o he o he s a emen s.
P oo o Theo em 2.2. i) Since x≡0,3 (mod 4), by Lemma 3.1 we
ha e ha ν2(Sk(x)) >0, ha is, Sk(x) is e en. Thus i (1) holds, hen
ν2(y)>0 and we ha e
nν2(y) = ν2(yn) = ν2(Sk(x)) = {ν2(x(x+ 1) −1,i kis e en,
2ν2(x(x+ 1) −2,i kis odd,
implying he s a emen in his case.
ii) As now x≡0,8 (mod 9), Lemma 3.2 implies ha ν3(Sk(x)) >0.
Hence (1) gi es ν3(y)>0, and no ing ha x≡0,2 (mod 3), we ha e
nν3(y) = ν3(yn) = ν3(Sk(x)) = 




ν3(x(x+ 1)),i k= 1,
ν3(x(x+ 1)(2x+ 1)) −1,i kis e en,
ν3(kx2(x+ 1)2)−1,i k≥3 is odd,
and he heo em is p o ed. 
P oo o Theo em 2.1. Obse e ha since x≡3,4 (mod 8), we ha e
ν2(x(x+ 1)) −1 = 1. Hence i k= 1 o kis e en hen by pa i) o
Theo em 2.2 we ha e n≤1, which is impossible. Thus he fi s pa
o he s a emen ollows.
So we may assume ha kis odd wi h k≥3. Then pa i) o Theo em
2.2 implies ha n= 2. As he cases (k, n) = (3,2),(5,2) gi e only
known solu ions, we may assume ha k≥7. Then one can easily
check ha
Sk(x)≡y2(mod 32)
is sol able i and only i k≡1 (mod 8). Howe e , one can also eadily
check ha in case o x≡hi(mod mi) o any hi∈Hi(i= 1,2,3,4)
Sk(x)≡y2(mod mi)
is no sol able whene e k≡1 (mod 8). This implies he s a emen .

8 L. HAJDU
P oo o Theo em 2.3. Th oughou he p oo , we shall assume ha k≥
9. Since x < 25, he alues k < 9 can be easily checked.
To p o e he heo em o he sepa a e alues o x, fi s we gi e a
bound o he exponen nusing pa i) o Theo em 2.2, hen we handle
he emaining exponen s by cong uences using app op ia e moduli. We
summa ize he esul s o ou calcula ions in Table 1. In ac he cases
x= 12 and 20 a e co e ed by Theo em 2.1, howe e , o he sake o
comple eness we include hem also he e. Fu he , no e ha he moduli
occu ing in Table 1 could ce ainly be “me ged” in o one la ge modulus
in each case. Howe e , we p e e he “sepa a e” p esen a ion because
i makes he a gumen mo e anspa en . Since ou me hod is simila
o each case, we only illus a e i (and also explain ou no a ion in
Table 1) h ough wo pa icula ins ances.
Fi s conside he case x= 4. Then pa i) o Theo em 2.2 gi es
n= 2. Conside ing equa ion (1) modulo 16, 7 and 13 we ob ain ha
k≡1 (mod 4), k≡ 1,5 (mod 6) and k≡ 9 (mod 12), espec i ely.
Howe e , combining hese cons ain s on kyields a con adic ion.
Nex conside he ( echnically mo e complica ed) case x= 16. Now
pa i) o Theo em 2.2 gi es n≤6. Hence i is sufficien o p o e he
insol abili y o equa ion (1) o n= 2,3,5. Since he cong uence
Sk(16) ≡y5(mod 128)
has no solu ions (unde ou assump ion k≥9), we ge ha equa ion
(1) has no solu ion wi h n= 5 in his case. When n= 3, conside ing
ou equa ion modulo 9 and 13, we deduce ha kis odd and kis e en,
espec i ely. This o cou se immedia ely shows ha n= 3 is impos-
sible. Finally, i n= 2 hen checking equa ion (1) modulo 512, 7 and
73 we ge ha k≡1 (mod 8), k≡3 (mod 6) and k≡ 9 (mod 24),
espec i ely. Howe e , hese oge he yield a con adic ion, and ou
claim ollows also in his case. No e ha when he exponen nis no
indica ed in Table 1, we assume ha n= 2.
In all he o he cases a simila a gumen wo ks, he de ails a e sum-
ma ized in Table 1, o be unde s ood in a simila way as abo e. 
5. Acknowledgemen s
The au ho is g a e ul o he e e ee o he use ul and help ul com-
men s.
Re e ences
[1] M. A. Benne , Y. Bugeaud, M. Migno e, Pe ec powe s wi h ew bina y
digi s and ela ed Diophan ine p oblems, II, Ma h. P oc. Camb. Phil. Soc.
153 (2012), 525–540.
ON A CONJECTURE OF SCH ¨
AFFER CONCERNING 1k+· · · +xk=yn9
xbound o nmoduli and in o ma ion deduced
3 2 mod 16: k≡1 (mod 4); mod 9: k≡3 (mod 6);
mod 13: k≡ 9 (mod 12)
4 2 mod 16: k≡1 (mod 4); mod 7: k≡ 1,5 (mod 6);
mod 13: k≡ 9 (mod 12)
7 4 mod 32: n= 3; mod 3: kis odd; mod 5: k≡ 1
(mod 4); mod 128: k≡ 3 (mod 4)
8 4 mod 32: n= 3; mod 81: k≡3 (mod 6); mod 128:
k≡0,1 (mod 4); mod 13: k≡ 9 (mod 12)
11 2 mod 32: k≡1 (mod 8); mod 9: k≡3 (mod 6);
mod 97: k≡ 9 (mod 24)
12 2 mod 16: k≡1 (mod 4); mod 5: k≡ 1 (mod 4)
15 6
mod 128: n= 5; mod 9: k≡0 (mod 3) i n= 3;
mod 13: k≡ 3 (mod 6) i n= 3; mod 19: k≡ 0
(mod 6) i n= 3; mod 256: k≡1 (mod 4); mod
9: k≡ 1,5 (mod 6); mod 13: k≡ 9 (mod 12)
16 6
mod 128: n= 5; mod 9: kis odd i n= 3; mod 13:
kis e en i n= 3; mod 512: k≡1 (mod 8); mod
7: k≡ 1,5 (mod 6); mod 73: k≡ 9 (mod 24)
19 2 mod 32: k≡1 (mod 8); mod 17: k≡ 1 (mod 8)
20 2 mod 16: k≡1 (mod 4); mod 13: k≡ 1 (mod 4)
23 4 mod 32: n= 3; mod 64: k≡1 (mod 4); mod 7:
k≡3 (mod 6); mod 13: k≡ 9 (mod 12)
24 4 mod 32: n= 3; mod 128: k≡1 (mod 8); mod 17:
k≡ 1 (mod 8)
Table 1. Da a o he solu ion o equa ion (1) wi h x≡
0,3 (mod 4) and x < 25. I nis no indica ed hen n= 2.
[2] M. A. Benne , K. Gy˝o y, ´
A. Pin ´e , On he Diophan ine equa ion 1k+ 2k+
· · · +xk=yn, Compos. Ma h. 140 (2004), 1417–1431.
[3] P. Co aja, U. Zannie , Fini eness o odd pe ec powe s wi h ou nonze o
bina y digi s, Ann. Ins . Fou ie 63 (2013), 715–731.
[4] J. M. G au, A. M. Olle -Ma c´en, Abou he cong uence ∑n
k=1 k (n)≡0
(mod n), a Xi :1304.2678 1 [ma h.NT] 9 Ap 2013.
[5] J. M. G au, A. M. Olle -Ma c´en, J. Sondow, On he cong uence 1m+ 2m+
· · · +mm≡n(mod m)wi h n|m, a Xi :1309.7941 4 [ma h.NT] 2 Dec
2013.