Nonexistence of Consecutive Powerful Triplets Around Cubes with Prime-Square Factors

#A103 INTEGERS 25 (2025)
NONEXISTENCE OF CONSECUTIVE POWERFUL TRIPLETS
AROUND CUBES WITH PRIME-SQUARE FACTORS
Jialai She
Phillips Academy, Ando e , Massachuse s
[email p o ec ed]
Recei ed: 7/9/25, Re ised: 8/31/25, Accep ed: 10/22/25, Published: 11/25/25
Abs ac
The E d˝os–Mollin–Walsh conjec u e, asse ing he nonexis ence o h ee consecu i e
powe ul in ege s, emains a celeb a ed open p oblem in numbe heo y. A na u al
line o inqui y, ollowing ecen wo k by Chan (2025), is o in es iga e po en ial
coun e examples cen e ed a ound pe ec cubes, which a e hemsel es powe ul.
This pape es ablishes a new non-exis ence esul o a amily o such in ege iple s
wi h dis inc s uc u al cons ain s, combining echniques om modula a i hme ic,
p-adic alua ion, Thue equa ions, and he heo y o ellip ic cu es.
1. In oduc ion
A posi i e in ege is called a powe ul numbe i each o i s p ime ac o s appea s
wi h an exponen o a leas wo. E e y powe ul numbe nadmi s a unique ep e-
sen a ion as
n=a2b3,(1)
whe e a, b ∈Zand bis squa e- ee (meaning ha i is no di isible by any pe ec
squa e o he han 1) (see [5] o example). The E d˝os–Mollin–Walsh conjec u e
[3, 9] asse s ha no h ee consecu i e in ege s a e all powe ul.
A na u al and in e es ing case a ises when he iple is cen e ed a a pe ec cube
x3, i sel always powe ul. Tha is, does he iple (x3−1, x3, x3+ 1) e e consis
en i ely o powe ul numbe s? Recen ly, Chan [2] esol ed a subcase by p o ing no
such iple s exis when
x3−1=p3y2, x3+1=q3z2,(2)
o p imes p, q and in ege s x, y, z > 0.
DOI: 10.5281/zenodo.17711516
INTEGERS: 25 (2025) 2
In his pape , we p o e he non-exis ence o a new amily o iple s in his se ing,
ex ending Chan’s esul while add essing dis inc cons ain s. Ou main esul is as
ollows.
Theo em 1. The e exis no consecu i e powe ul numbe s o he o m
x3−1=p2a3, x3, x3+1=q2b3.
whe e p, q a e p imes and a, b, x a e in ege s.
No ice ha he powe s in Theo em 1 di e om hose in (2). Fu he mo e, unlike
[2], we do no need o impose he es ic ion ha a iables x,a, and ba e posi i e.
This esul es ablishes he non-exis ence o a class o consecu i e powe ul iple s
no co e ed by p e ious li e a u e.
Co olla y 1. Fo any p imes p, q and any in ege s x, a wi h a= 0, he equa ion
x6−1=p2q2a3has no solu ion.
2. P oo o he Main Resul
Fi s , le us in oduce some lemmas.
Lemma 1. Le pbe a p ime and R, S, C be in ege s ha sa is y
R S =p2C3,
and se g= gcd(R, S). I g= 1, hen one o R, S is a pe ec cube and he o he is
p2 imes a pe ec cube. I gis a p ime, hen he e exis C1, C2∈Zsuch ha ei he
(R, S)=(gC3
1, gC3
2),{R, S}={gC3
1, g2p2C3
2},o {R, S}={gp2C3
1, g2C3
2}.
P oo . Fo g= 1, i p∤C, any p ime ac o =po Cdi ides exac ly one o R, S,
and i p|C, hen all p ac o s in p2C3mus be con ained en i ely wi hin ei he R
o C, bu no bo h. Assume gis a p ime. I g=p, hen bo h R/p and S/p a e
pe ec cubes. O he wise, he wo ac o s can be w i en as gC3
1and g2p2C3
2, o
gp2C3
1and g2C3
2, up o o de ing. The conclusion ollows.
Lemma 2. The Diophan ine equa ion u2+u+ 1 = 3 3has exac ly wo in ege
solu ions (u, )∈ {(−2,1),(1,1)}. The Diophan ine equa ion u2−u+ 1 = 3 3has
exac ly wo in ege solu ions (u, )∈ {(2,1),(−1,1)}.
P oo . The co e o he p oo is o con e he o iginal equa ions in o he s anda d
o m o a Mo dell cu e h ough p ope ans o ma ions. Taking he second equa-
ion as an example, we i s mul iply he equa ion by 32and subs i u e u′= 3uand
′= 3 , which yields (u′)2−3u′+ 9 = ( ′)3.Mul iplying by 4 and comple ing he
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squa e on he le side allows o he subs i u ion u′′ = 2u′−3 and ′′ = ′, leading
o (u′′)2+ 27 = 4( ′′ )3.Finally, mul iplying bo h sides by 24and le ing x= 4 ′′
and y= 4u′′ educes he equa ion o he canonical Mo dell o m: y2=x3−432.
The in ege solu ions o his well-known cu e a e (x, y) = (12,±36) (see [4] o ex-
ample). Wo king backwa ds gi es he wo solu ions o (u, ). The o he equa ion
can be handled by subs i u ing u7→ −u.
Lemma 3. The in ege solu ions o u o he equa ion u2+u+ 1 = 3a e gi en
by u=−19,−1,0,18. Simila ly, he in ege solu ions o u o u2−u+ 1 = 3a e
−18,0,1,19.
P oo . The i s equa ion can be handled using he ans o ma ions and Mo dell
cu e echniques in Lemma 2, o di ec ly by ci ing he co olla y o [10]. The second
equa ion hen ollows by he subs i u ion u7→ −u.
Lemma 4. Fo any in ege x, we ha e gcdx−1, x2+x+ 1= gcd(x−1,3) and
gcdx+ 1, x2−x+ 1= gcd(x+ 1,3).
P oo . W i ing x2+x+ 1 = (x−1)(x+ 2) + 3 and x2−x+ 1 = (x+ 1)(x−2) + 3,
he Euclidean algo i hm on polynomials yields he conclusion.
Lemma 5. The Diophan ine equa ion u3− 3= 2 has he unique in ege pai
solu ion (1,−1),and he Diophan ine equa ion u3− 3= 1 has he in ege solu ions
(1,0) and (0,−1).
P oo . We may assume uand a e bo h nonze o. Conside u3− 3= 2. Since
u > , bo h ac o s u− and u2+u + 2a e posi i e in ege s. The e o e, we ha e
wo possibili ies:
u− = 2, u2+u + 2= 1,
o
u− = 1, u2+u + 2= 2.
Simple calcula ion yields he unique solu ion (1,−1). The o he equa ion can be
ea ed simila ly.
We a e now p epa ed o es ablish he main esul . Suppose o con adic ion
ha
x3−1=p2a3, x3+1=q2b3,(3)
wi h in ege s x, a, b and p imes p, q. Se g−= gcd(x−1, x2+x+ 1) and g+=
gcd(x+ 1, x2−x+ 1).By Lemma 4,
g−=(3x≡1 (mod 3),
1 o he wise, g+=(3x≡2 (mod 3),
1 o he wise.
INTEGERS: 25 (2025) 4
We p oceed by casewo k since he pai (g−, g+) can be (1,1), (1,3), o (3,1) only.
Case 1: (g−, g+) = (1,1),x≡0 (mod 3).By Lemma 1, we ha e he ollowing
possibili ies
(i) x−1=u3, x2+x+ 1 = p2 3,
(ii) x−1=p2u3, x2+x+ 1 = 3,and (a) x+1=s3, x2−x+ 1 = q2 3,
(b) x+1=q2s3, x2−x+ 1 = 3,
whe e u, , s, and a e in ege s. We explo e each subcase below.
•(i)+(a).We ob ain x−1 = u3, x+1 = s3, o s3−u3= 2.By Lemma 5, s= 1
and u=−1.Bu hen x= 0 and he e exis s no p ime psa is ying (3).
•(i)+(b).Applying Lemma 3 o x2−x+1 = 3yields x=−18,0,1,19. Howe e ,
none o hese alues sa is ies x+1=q2s3unde he gi en cons ain s ( o
example, aking x= 19 leads o q2s3= 20 and hus q= 2, bu no in ege s
exis s).
•(ii) + (a).This subcase can be ea ed simila ly by applying Lemma 3 o he
equa ion x2+x+ 1 = 3.
•(ii)+(b).In his subcase, x2−x+ 1 and x2+x+1 a e pe ec cubes. Lemma
3 o ces x= 0, bu hen no p ime qexis s sa is ying (3).
Case 2: (g−, g+) = (1,3),x≡2 (mod 3).Le p(n) deno e he p-adic alua ion
o n.
Fi s , applying he Li ing- he-Exponen lemma (see, e.g., Theo em 1.37 o [7])
o 3(x3+ 1) gi es 3(x+ 1) + 1 = 3(x3+ 1) = 3(x+1)+ 3(x2−x+ 1),o
3(x2−x+ 1) = 1.(4)
We spli in o wo subcases based on q.
•q= 3 : Fo = 3 as an a bi a y p ime ac o o x2−x+ 1,since g+= 3, we
ha e
(x2−x+ 1) = ((x+ 1)(x2−x+ 1)) = (9b3)=3 (b).
Combined wi h (4), x2−x+ 1 can be w i en as 3 Qn
i=1 3αi
i,o 3s3 o some
s∈Z. Applying Lemma 2 gi es x=−1 o 2.Ye nei he yields a alid
solu ion o (3) wi h p ime p.
•q= 3 : He e, we ha e 3(x3+ 1) = 3(q2b3)=3 3(b). Combined wi h (4),
3(x3+ 1) ≥3, om which i ollows ha
3(x+ 1) = 3((x3−1)/(x2−x+ 1)) ≥3−1 = 2,
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and hus
x≡ −1 (mod 9).(5)
Nex , applying Lemma 1 o (x−1)(x2+x+1) = p2a3, we ha e wo possibili ies:
(i) x−1 = u3, x2+x+1=p2 3,and (ii) x−1=p2u3, x2+x+1= 3.
Combining (i) and (5) yields u3≡ −2 (mod 9), which is impossible. Fo (ii),
(5) and Lemma 3 o ce x=−19 o −1.Nei he yields a alid solu ion o (3)
wi h p ime p.
Case 3: (g−, g+) = (3,1),x≡1 (mod 3).This case is analogous o he p e i-
ous one, wi h x eplaced by −x. The esul ollows om a combina ion o p-adic
analysis and Lemmas 1, 2, and 3.
3. P oo o he Co olla y
The p oo o Co olla y 1 combines he a gumen s o ou main heo em wi h ha
o Co olla y 1 o [2]. We begin wi h a ew p epa a o y lemmas.
Lemma 6. I 3|x−1, hen 3(x2+x+1) = 1; i 3|x+1, hen 3(x2−x+1) = 1.
P oo . The a gumen is analogous o he one used in he main heo em’s p oo .
Lemma 7. The only in ege solu ions o he Diophan ine equa ion u3−2 3= 1
a e (1,0) and (−1,−1).
P oo . Bo h (1,0) and (−1,−1) sa is y u3−2 3= 1. By he Delone–Nagell heo em
(see, e.g., [1, Theo em V, §72]), he e is a mos one solu ion in addi ion o (1,0).
The conclusion ollows.
Al hough he ollowing esul is likely known, we we e unable o loca e a speci ic
e e ence o hese pa ame e s. Fo he sake o comple eness, we p o ide a b ie ,
sel -con ained p oo based on Lemma 2.
Lemma 8. Fo d∈ {4,18,36}, he Diophan ine equa ion u3−d 3= 1 has he
unique in ege solu ion (1,0).
P oo . Fi s , o d= 4, we conside he equa ion u3−4 3= 1. Reducing his
modulo 9 implies ha u3≡1 (mod 9) and 3≡0 (mod 9), and so u≡1 (mod 3)
and ≡0 (mod 3). Le = 3 0 o some in ege 0. The equa ion becomes
(u−1)(u2+u+ 1) = 4 ·33· 3
0.

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Lemma 6 gi es 3(u2+u+1) = 1, om which i ollows ha gcd(u−1, u2+u+1) = 3.
We can he e o e se u−1=3aand u2+u+ 1 = 3bwi h gcd(a, b) = 1, and ob ain
ab = 12 3
0. Since u2+u+ 1 is odd, bmus also be odd. As 3(3b)=1,3∤b. I
ollows ha b=s3, and u2+u+ 1 = 3s3. By Lemma 2, u= 1,−2. No ing umus
be odd, he conclusion ollows.
Simila ly, o d= 18, he equa ion u3−18 3= 1 aken modulo 9 gi es u3≡1
(mod 9), which implies u≡1 (mod 3). By Lemma 6, 3(u2+u+ 1) = 1, and hus
gcd(u−1, u2+u+1) = 3. Se ing u−1 = 3aand u2+u+1 = 3bwi h gcd(a, b)=1
esul s in ab = 2 3. Since bmus be odd and is cop ime o a, we ha e b=s3,
which yields u2+u+ 1 = 3s3. The unique in ege solu ion (1,0) again ollows
by Lemma 2. The case o d= 36 ollows om an iden ical a gumen , applying a
modulo 9 educ ion along wi h Lemma 6 and Lemma 2.
We now p o e Co olla y 1 by con adic ion. The equa ion in he co olla y can
be w i en as
x3−1x3+ 1=p2q2a3.(6)
Since gcd(x3−1, x3+ 1) |x3+ 1 −(x3−1) = 2, we ha e gcd(x3−1, x3+ 1) = 1 o
2.
Fo gcd(x3−1, x3+ 1) = 1, he e a e wo possibili ies o he ac o s on he
le -hand side o (6). Assume p2q2di ides one o he ac o s. This implies ha he
o he ac o mus be a pe ec cube. I x3+1=a3
1o x3−1=a3
1, hen xhas
no alid solu ion by Lemma 5. The e o e, p2di ides one ac o and q2di ides he
o he ; wi hou loss o gene ali y, assume x3−1 = p2a3
1and x3+ 1 = q2a3
2. Bu his
con adic s ou main heo em.
Nex , conside gcd(x3−1, x3+ 1) = 2. I one o he p imes is 2, say p= 2, hen
we ha e wo possible sys ems o equa ions:
(x3−1=2q2u3
x3+ 1 = 2 3,o (x3−1=2u3
x3+ 1 = 2q2 3.(7)
By Lemma 7, x=±1, which iola es a= 0.
Fo he emainde o he p oo , assume p= 2, q= 2, and (6) becomes (x3−
1)(x3+ 1) = 8p2q2b3. One possibili y is ha p2q2di ides one o he ac o s on he
le . This leads o ou subcases, each o which yields a con adic ion by applying
ei he Lemma 7 o Lemma 8:
•x3−1=2p2q2u3, x3+ 1 = 4 3. By Lemma 8, x=−1, iola ing a= 0.
•x3−1=4p2q2u3, x3+ 1 = 2 3. By Lemma 7, x=±1, iola ing a= 0.
•x3−1=2u3, x3+ 1 = 4p2q2 3. By Lemma 7, x=±1, iola ing a= 0.
•x3−1=4u3, x3+ 1 = 2p2q2 3. By Lemma 8, x= 1, iola ing a= 0.
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The e o e, p2mus di ide one o x3−1 and x3+ 1, and q2mus di ide he o he .
I emains o s udy he sys em
x3−1=2p2u3, x3+ 1 = 4q2 3.(8)
Indeed, when he ac o s o 2 and 4 a e swapped, x3−1 = 4p2u3and x3+1 = 2q2 3
educe o he o m in (8) unde he change o a iables x7→ −x, u 7→ − , 7→
−u, p 7→ q, and q7→ p. The equa ions in (8) esemble hose in (3), bu a e dis inc :
o example, x3−1 he e is no necessa ily a powe ul numbe . Ne e heless, he
same p oo s a egy used in he p e ious sec ion can be applied.
Recall he de ini ions o g−and g+in he p oo o he main heo em. We p oceed
in a simila manne .
Case 1: (g−, g+) = (1,1).The ac o iza ion o he i s equa ion in (8) yields ou
possibili ies: (i) x−1 = 2 3
1, x2+x+ 1 = p2 3
2, (ii) x−1=2p2 3
1, x2+x+ 1 = 3
2,
(iii) x−1 = 3
1, x2+x+ 1 = 2p2 3
2, and (i ) x−1 = p2 3
1, x2+x+ 1 = 2 3
2,
and simila ly, he second equa ion gi es: (a) x+ 1 = 4 3
3, x2−x+ 1 = q2 3
4, (b)
x+ 1 = 4q2 3
3, x2−x+1= 3
4, (c) x+ 1 = 3
3, x2−x+ 1 = 4q2 3
4, and (d)
x+1=q2 3
3, x2−x+ 1 = 4 3
4, whe e ia e in ege s.
Since x2±x+1 = x(x±1)+1 is always odd, (iii), (i ), (c), and (d) a e impossible.
Among he emaining combina ions, hose in ol ing (ii) o (b) can be excluded by
Lemma 3. Fo example, unde (i) + (b),we ha e xodd, and Lemma 3 o ces
x∈ {1,19};x= 1 gi es a= 0 (a con adic ion), while x= 19 yields x+ 1 = 20, so
no p ime qexis s sa is ying (8).
Thus only (i) + (a) emains. In his case, 2 3
3− 3
1= 1.By Lemma 7, 1=±1,
hence x∈ {3,−1}, bu nei he alue p oduces a p ime psa is ying (8).
Case 2: (g−, g+) = (3,1).I p= 3, Lemma 8 implies x= 1, iola ing a= 0.
Assume p= 3. The case condi ion equi es x≡1 (mod 3), and applying Lemma
6 gi es 3(x2+x+ 1) = 1. Combining his wi h he pa i y a gumen om Case
1, i su ices o analyze he ac o iza ions: (i) x−1 = 18 3
1, x2+x+ 1 = 3p2 3
2,
o (ii) x−1 = 18p2 3
1, x2+x+ 1 = 3 3
2 o he i s equa ion in (8), and (a)
x+ 1 = 4 3
3, x2−x+1=q2 3
4, o (b) x+ 1 = 4q2 3
3, x2−x+1= 3
4 o he second
equa ion. By a simila line o easoning, applying Lemma 2 and Lemma 3 elimina es
all bu one non i ial combina ion, (i) + (a), om which i ollows ha 2 3
3−9 3
1= 1.
Reducing he equa ion modulo 9 gi es 2 3
3≡1 (mod 9), a con adic ion.
Case 3: (g−, g+) = (1,3).I q= 3, Lemma 8 implies x=−1, iola ing a= 0.
Assume q= 3. Applying Lemma 6 gi es 3(x2−x+ 1) = 1. Combining he 3-adic
alua ion and he pa i y a gumen as in Case 2, he i s equa ion in (8) leads o wo
possibili ies: (i) x−1=2 3
1, x2+x+1 = p2 3
2, o (ii) x−1=2p2 3
1, x2+x+1 = 3
2, and
he second equa ion leads o wo possibili ies: (a) x+ 1 = 36 3
3, x2−x+ 1 = 3q2 3
4,
o (b) x+1 = 36q2 3
3, x2−x+1 = 3 3
4. Applying Lemma 2 and Lemma 3 lea es only
one non i ial case, x−1=2 3
1, x2+x+1 = p2 3
2, x+1 = 36 3
3, and x2−x+1 = 3q 3
4.
This combina ion implies 18 3
3− 3
1= 1. Then Lemma 8 gi es 1=−1, and hus
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x=−1, iola ing a= 0.
The p oo is comple e.
4. Conclusion
We ha e p o ed ha no h ee consecu i e in ege s cen e ed a a pe ec cube can
all be powe ul unde he s uc u al cons ain s s udied he e. This esul ex ends
ecen ad ances on he E d˝os–Mollin–Walsh conjec u e by elimina ing a no able
amily o po en ial coun e examples h ough modula and ellip ic cu e me hods.
One na u al ex ension is o examine whe he simila non-exis ence holds when
cen e ed a ound highe powe s o o he special in ege s, and wha u he con-
s ain s migh elimina e all consecu i e powe ul numbe s en i ely. A ela ed and
mo e undamen al ques ion ha a ose du ing ou esea ch is he ollowing: we con-
jec u e ha o e e y in ege x > 1 and e e y in ege n>2, he numbe xn−1 is no
powe ul. This is a s onge claim han Mih˘ailescu’s Theo em ( o me ly Ca alan’s
Conjec u e) [8]. A p oo o his gene al asse ion would ha e signi ican implica-
ions; he speci ic case o n= 3 would esol e he ques ion add essed in Theo em
1 and p o ide deepe insigh in o he s uc u e o powe ul numbe s.
Acknowledgemen . The au ho is g a e ul o Tudo Popescu o b inging Chan’s
wo k o his a en ion and o p oposing he key conjec u e add essed in his pape .
In an ea lie d a [6], he au ho es ablished he esul o he co olla y wi h 2xin
place o x; he au ho is g a e ul o D . Tsz Ho Chan o his insigh ul sugges ion
o in es iga e emo ing he ac o o 2, which was accomplished in his e ision.
The au ho also hanks he anonymous e e ee o independen ly sugges ing his
same imp o emen , p o iding a help ul p oo ske ch, and o e ing many s ylis ic
sugges ions ha imp o ed he exposi ion o he pape . The conjec u e s a ed in he
inal sec ion was i s aised by he au ho in he pe sonal communica ion wi h D .
Chan.
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1964.
[2] T.H. Chan, A no e on h ee consecu i e powe ul numbe s, In ege s 25 (2025), #A7, 7 pp.
[3] P. E d˝os, P oblems and esul s on consecu i e in ege s, Publ. Ma h. Deb ecen 23 (1976),
271-282.
INTEGERS: 25 (2025) 9
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[9] R.A. Mollin and P.G. Walsh, On powe ul numbe s, In . J. Ma h. Ma h. Sci. 9(1986), 801-806.
[10] N. Tzanakis, The Diophan ine equa ion x3
−3xy2
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