#A92 INTEGERS 25 (2025)
ON THE EXISTENCE OF {p, q}-DIOPHANTINE QUADRUPLES
AND QUINTUPLES
Volke Ziegle
Depa men o Ma hema ics, Uni e si y o Salzbu g, Salzbu g, Aus ia
[email p o ec ed]
Recei ed: 1/24/25, Accep ed: 9/25/25, Published: 11/5/25
Abs ac
Le Sbe a se o p imes. We call an m- uple (a1, . . . , am) o dis inc posi i e in ege s
S-Diophan ine i , o all i=j, he in ege s si,j := aiaj+ 1 ha e only p ime di iso s
coming om he se S, i.e., i all si,j a e S-uni s. In his pape we p o e ha o
S={p, q}, no S-Diophan ine quad uple exis s i p≡3 mod 4 o q≡3 mod 4.
We also p o e ha o S={p, q}no S-Diophan ine quin uple exis s.
1. In oduc ion
Le Abe a se o kdis inc , posi i e in ege s. Gy˝o y, S´a k¨ozy, and S ewa [2]
conside ed he p oduc
Π = Y
a,b∈A
a=b
ab + 1,
and ound lowe bounds o he numbe ωo p ime ac o s o Π in e ms o k. In
pa icula , hey showed ha ω≫log k. They also conjec u ed ha he la ges
p ime ac o o (ab + 1)(ac + 1)(bc + 1) ends o in ini y as max{a, b, c}→∞. A
weake o m, namely ha he la ges p ime ac o o
(ab + 1)(ac + 1)(bd + 1)(cd + 1)
ends o in ini y as max{a, b, c, d}→∞, was p o ed by S ewa and Tijdeman [5],
and he ull conjec u e was p o ed independen ly by Co aja and Zannie [1] and
He n´andez and Luca [3]. We wan o emphasize ha he esul s o S ewa and Ti-
jdeman [5] a e e ec i e, while he esul s o Co aja and Zannie [1] and He n´andez
and Luca [3] a e ine ec i e.
In o de o ob ain sha p lowe bounds o ω o small kwe in oduce he no ion o
S-Diophan ine m- uples. Le Sbe a se o p imes. We call an m- uple (a1, . . . , am)
DOI: 10.5281/zenodo.17535199
INTEGERS: 25 (2025) 2
o dis inc , posi i e in ege s S-Diophan ine i , o all i=j, he in ege s si,j :=
aiaj+ 1 ha e only p ime di iso s coming om he se S, i.e., i all si,j a e S-uni s.
We a e in e es ed in inding lowe bounds o min e ms o |S|. Indeed he esul
o Gy˝o y, S`a k¨ozy, and S ewa [2] implies ha o m≥exp(C|S|), whe e Cis
an e ec i ely compu able cons an , no S-Diophan ine m- uple exis s. This bound
seems o be a om he u h. Howe e , i seems o be a di icul p oblem o ind
sha p bounds o min e ms o |S|e en o small alues o |S|, as o |S|= 2 o
|S|= 3. Le us no e ha he case ha |S|= 3 has been s udied by he au ho in
[10]. In pa icula , all {p, q, }-Diophan ine quad uples wi h 2 ≤p<q< ≤100
ha e been ound.
Le us no e ha in he con ex o S-Diophan ine uples, he esul s o Co aja,
Zannie , He n´andez, and Luca [1, 3] imply ha o a ixed, ini e se o p imes Sonly
ini ely many S-Diophan ine iples exis . Mo eo e , he esul due o S ewa and
Tijdeman [5] implies ha o a ixed se o p imes Sall S-Diophan ine quad uples
can be e ec i ely compu ed. Howe e , he ollowing conjec u e is s ill open.
Conjec u e 1. Le S={p, q}be a se o p imes wi h p<q. Then no S-
Diophan ine quad uple exis s.
Conjec u e 1 has been con i med in he ollowing special cases:
•I p2∤qo dp(q)−1, q2∤po dq(p)−1, and q < pξholds o some ξ > 1, hen
he e exis s an e ec i ely compu able cons an C=C(ξ) such ha o all
such p imes p, q > C no S-Diophan ine quad uple exis s (see [6]).
•No S-Diophan ine quad uple exis s, i p≡q≡3 mod 4 (see [7]).
•No S-Diophan ine quad uple exis s, i p= 2 o p= 3 (see [9]).
•No S-Diophan ine quad uple exis s, i p≡3 mod 4 and
q(pq−1−1) ≥2,and p(qp−1−1) ≥max 2,log q
log p,
whe e p(x) and q(x) deno e he p-adic and q-adic alua ion o x, espec i ely
(see [9]).
•No S-Diophan ine quad uple exis s, i p, q < 105(see [8]).
•No S-Diophan ine quad uple exis s i q=p+ 2 is a p ime and pis la ge
enough (see [4]).
Le us no e ha mos o he esul s lis ed abo e use esul s on lowe bounds o
linea o ms in complex and p-adic loga i hms. Howe e , in his pape we can p o e
by using only elemen a y me hods (mainly di isibili y p ope ies) he ollowing
heo em.
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Theo em 1. Assume ha S={p, q}is a se o p imes wi h a leas one o po q
cong uen o 3modulo 4. Then he e does no exis an S-Diophan ine quad uple.
In o he wo ds, o p o e Conjec u e 1 we only ha e o conside p imes wi h
p≡q≡1 mod 4. Al hough a comple e p oo o Conjec u e 1 seems o be ou o
each a he momen , a leas we can p o e ha no S-Diophan ine quin uple exis s
i |S|= 2.
Theo em 2. Le p<qbe p ime numbe s and S={p, q}. Then no S-Diophan ine
quin uple exis s.
In he nex sec ion we lis some use ul esul s conce ning S-Diophan ine m- uples,
which will be equen ly used in he p oo s o Theo ems 1 and 2. In Sec ion 3 we
p o e Theo em 1 and in Sec ion 4 we p o e Theo em 2.
2. P elimina ies
Le S={p, q}be a se o wo p imes wi h p<qand le (a, b, c, d) be a hypo he ical
S-Diophan ine quad uple. Then we w i e
ab + 1 =pα1qβ1, bc + 1 =pα4qβ4,
ac + 1 =pα2qβ2, bd + 1 =pα5qβ5,
ad + 1 =pα3qβ3, cd + 1 =pα6qβ6.
Mo eo e , we deno e by si he S-uni pαiqβi o i= 1,...,6. I we compu e abcd in
di e en ways, we ob ain
abcd = (ab)(cd)=(s1−1)(s6−1) = s1s6−s1−s6+ 1
= (ac)(bd)=(s2−1)(s5−1) = s2s5−s2−s5+ 1
= (ad)(bc)=(s3−1)(s4−1) = s3s4−s3−s4+ 1.
The e o e, we ob ain he h ee non-linea S-uni equa ions
s1s6−s1−s6=s2s5−s2−s5,(1)
s3s4−s3−s4=s2s5−s2−s5,(2)
s1s6−s1−s6=s3s4−s3−s4.(3)
We s a wi h he ollowing simple di isibili y condi ion which was p o ed in [6,
Lemma 2.1].
Lemma 1 ([6]).Le (a, b, c)be an S-Diophan ine iple, wi h a<b<c, hen s∤
wi h s=ac + 1 and =bc + 1.
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An immedia e consequence o Lemma 1 is ha we can exclude he ollowing
ela ions be ween exponen s:
α2=α4, α3=α5, α3=α6, α5=α6,
β2=β4, β3=β5, β3=β6, β5=β6.
On he o he hand, we ha e he ollowing lemma (c . [6, P oposi ion 1] o [8,
Lemma 2.1]), which one ob ains by aking p-adic and q-adic alua ions applied o
Equa ions (1), (2), and (3).
Lemma 2 ([6]).The smalles wo exponen s occu ing in each o he h ee quad uples
(α2, α3, α4, α5),(α1, α2, α5, α6), and (α1, α3, α4, α6)coincide. The same s a emen
also holds wi h α eplaced by β.
The ollowing lemma also p o es o be use ul and yields uppe bounds o a, b, c,
and d. A p oo o i can be ound in [6, Lemma 3].
Lemma 3 ([6]).We ha e
a|gcd s2−s1
gcd(s2, s1),s3−s1
gcd(s3, s1),s3−s2
gcd(s3, s2),
b|gcd s4−s1
gcd(s4, s1),s5−s1
gcd(s5, s1),s5−s4
gcd(s5, s4),
c|gcd s4−s2
gcd(s4, s2),s6−s2
gcd(s6, s2),s6−s4
gcd(s6, s4),
d|gcd s5−s3
gcd(s5, s3),s6−s3
gcd(s6, s3),s6−s5
gcd(s6, s5).
We also use he ollowing esul which is he con en o [7, Sec ion 4].
Lemma 4 ([7]).Le p, q be odd p imes and assume ha (a, b, c, d)is a {p, q}-
Diophan ine quad uple. Then he ollowing sys em o equa ions canno hold:
ab + 1 = qβ1, bc + 1 = pα4qβ4,
ac + 1 = pα2, bd + 1 = pα5,
ad + 1 = pα3qβ3, cd + 1 = qβ6.
(4)
3. Non-Exis ence o S-Diophan ine Quad uples
This sec ion is de o ed o he p oo o Theo em 1. A key esul o es ablish his
heo em is a esul ob ained by Ziegle [9, P oposi ion 5.1].
Lemma 5 ([9]).Assume ha p≡3 mod 4. Then one o he ou cases in Table 1
holds.
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Case The αexponen s The βexponen s
I 0 = α1=α6< α4=α5< α2< α3β1=β2=β3< β4< β5< β6
II 0 = α1=α6< α2=α5< α3, α4β3=β4< β1=β2< β5< β6
III 0 = α2=α5< α1=α3≤α4< α6β1=β6< β3=β4< β2< β5
IV 0 = α3=α4< α1=α2< α5< α6β1=β6< β2=β5< β3, β4
Table 1: Res ic ions o he exponen s
No e ha i necessa y we can exchange he oles o pand q o ensu e ha p≡3
mod 4. By he esul due o Szalay and he au ho [6] we may assume ha q≡1
mod 4, i.e., q≥5, and due o he au ho ’s esul [9] we may assume ha p= 3,
i.e., p≥7. In o de o p o e Theo em 1 we ha e o show ha he es ic ions om
Lemma 5, i.e., he cases om Table 1, yield con adic ions.
Be o e we p o e Theo em 1 we p o e wo lemmas i s . We s a wi h he ol-
lowing lemma.
Lemma 6. Assume ha Case III o Table 1 holds. Then we ha e d < 1.03·q2(β′−β)
and b < 1.03 ·qβ′−β.
P oo . Fi s , we no e ha due o Lemma 3 we ha e
d
s6−s4
gcd(s4, s6)=pα6qβ−pα′qβ′
pα′qβ=pα6−α′−qβ′−β
and, in pa icula , we ob ain d < pα6−α′.
Mo eo e , we ha e
d
b=s3−1
s1−1=pαqβ′−1
pαqβ−1=qβ′−β·1−p−αq−β′
1−p−αq−β.
Since we may assume ha α′> α > 0, β > 0, p≥7, and q≥5, we ge
qβ′−β<d
b<1.03 ·qβ′−β.
Also, no e ha s−1
3=p−αq−β′< s−1
1=p−αq−β.
Simila ly we ob ain
d
b=s6−1
s4−1=pα6qβ−1
pα′qβ′−1=pα6−α′qβ−β′·1−p−α6q−β
1−p−α′q−β′> pα6−α′qβ−β′.
Compa ing lowe and uppe bounds o d/b yields
pα6−α′qβ−β′<d
b<1.03 ·qβ′−β,
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and hence we ge pα6−α′<1.03 ·q2(β′−β). The e o e, we ob ain
d<pα6−α′<1.03 ·q2(β′−β).
Now, we ob ain he inequali y o bby he ollowing simple calcula ion:
b=d·b
d<1.03 ·q2(β′−β)
qβ′−β<1.03 ·qβ′−β.
Nex , we p o e he ollowing lemma.
Lemma 7. Assume ha Case IV o Table 1 holds. Then we ha e b < 1.042·qβ′−β.
P oo . We use Lemma 3 o ob ain
d
s6−s5
gcd(s6, s5)=pα6qβ−pα′qβ′
pα′qβ=pα6−α′−qβ′−β,
which implies d<pα6−α′. Fu he le us compu e
c
b=s2−1
s1−1=pαqβ′−1
pαqβ−1=qβ′−β·1−p−αq−β′
1−p−αq−β<1.03 ·qβ′−β.
The las inequali y comes om he ac ha we may assume ha α, β > 0 because
o Lemma 4. Also, no e ha we may assume ha p≥7 and q≥5.
On he o he hand we also ha e
c
b=s6−1
s5−1=pα6qβ−1
pα′qβ′−1=pα6−α′qβ−β′·1−p−α6q−β
1−p−α′q−β′> pα6−α′qβ−β′.
The las inequali y comes om he ac ha s−1
6=p−α6q−β< p−α′q−β′=s−1
5
holds. Combining he inequali ies o c/b we ge
pα6−α′<1.03 ·q2(β′−β).
The e o e, we ob ain om he p e iously ound uppe bound o d he uppe bound
d<pα6−α′<1.03 ·q2(β′−β).
Nex , we no e ha
ad =qβ3−1 = qβ3(1 −q−β3)≥0.992 ·qβ3,
since β3≥3. Also, no e ha qβ3=s3> s2=pαqβ′. These wo inequali ies
oge he wi h he uppe bound o dyields
a≥0.992 ·qβ3
d>0.992 ·qβ3+2β−2β′
1.03 >0.96 ·pαq2β−β′.
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Hence, we ob ain
b=s1−1
a<pαqβ
0.96 ·pαq2β−β′= 1.042 ·qβ′−β.
Now, we s a wi h he p oo o Theo em 1.
P oo o Theo em 1. We conside he ou cases as shown in Table 1.
Case I o Table 1. Due o Lemma 3 we ha e ha
d
s5−s3
gcd(s5, s3)=pα5qβ5−pα3qβ3
pα5qβ3=qβ5−β3−pα3−α5.
This implies d<qβ5−β3. Thus we ob ain ha
b+ 1 = s5−1
d+ 1 >s5
d=pα5qβ3> qβ3.
We also ha e β3=β1and α1= 0, which yields
b+ 1 > qβ3=qβ1=s1=ab + 1 ≥b+ 1,
which is a con adic ion.
Case II o Table 1. In his case we ge
cd + 1
bd + 1 ·b
c−1
=
pα6qβ6·b−pα5qβ5·c
pα5qβ5·c
=
N
cpα5
=
c−b
bcd +c
<1
bd + 1 =1
pα5qβ5,
whe e Nis some in ege unequal o 0. No e ha N= 0 would imply b=c, which
is excluded. This inequali y yields
1≤ |N|<c
qβ5
and he e o e c > qβ5. By a simila a gumen , we also ob ain a lowe bound o b.
Namely, we conside
bd + 1
ad + 1 ·a
b−1
=
pα5qβ5·a−pα3qβ3·b
pα3qβ3·b
=
N
bpα3−α5
=
b−a
abd +b
<1
ad + 1 =1
pα3qβ3,
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and deduce ha
1≤ |N|<b
pα5qβ3.
Hence, we ha e b > pα5qβ3. Combining he lowe bounds o band cwe ob ain
pα5qβ5< pα5qβ3+β5< bc < bd + 1 = pα5qβ5,
which is an ob ious con adic ion.
Case III o Table 1. Le us w i e α=α1=α3,α′=α4,β=β1=β6, and
β′=β3=β4. Wi h his no a ion we ha e
ab + 1 = s1=pαqβ, ac + 1 = s2=qβ2, ad + 1 = s3=pαqβ′,
bc + 1 = s4=pα′qβ′, bd + 1 = s5=qβ5, cd + 1 = s6=pα6qβ.
Le us no e ha , due o Lemma 4, α, β = 0, ha is, we ha e 0 < α ≤α′and
0< β < β′.
Nex , we no e ha by Lemma 3 we ha e ha
c
s4−s2
gcd(s4, s2)=pα′qβ′−qβ2
qβ′=pα′−qβ2−β′,
which implies c < pα′. Toge he wi h he bound o b, which we ob ained in
Lemma 6, we ind
pα′qβ′=bc + 1 <1.03 ·qβ′−βpα′+ 1 < pα′qβ′,
which is a con adic ion.
Case IV o Table 1. Le us w i e α=α1=α2,α′=α5,β=β1=β6, and
β′=β2=β5. Wi h his no a ion we ha e
ab + 1 = s1=pαqβ, ac + 1 = s2=pαqβ′, ad + 1 = s3=qβ3,
bc + 1 = s4=qβ4, bd + 1 = s5=pα′qβ′, cd + 1 = s6=pα6qβ.
Again we ha e α, β = 0 due o Lemma 4, ha is, we ha e 0 < α ≤α′and 0 < β < β′.
Nex , le us ind an uppe bound o c. Indeed we ind
c
s4−s2
gcd(s4, s2)=qβ4−pαqβ′
qβ′=qβ4−β′−pα,
which yields c < qβ4−β′. Wi h his uppe bound o ccombined wi h he uppe
bound o bob ained in Lemma 7 we ob ain
qβ4=bc + 1 <1.042 ·qβ′−βqβ4−β′+ 1 = 1.042 ·qβ4−β+ 1.
This inequali y canno hold unless β= 0. Howe e , we can exclude he case ha
β= 0 due o Lemma 4.
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4. Non-Exis ence o S-Diophan ine Quin uples
This sec ion is de o ed o he p oo o Theo em 2. The e o e, le (a, b, c, d, e) be a
{p, q}-Diophan ine quin uple wi h a<b<c<d<e. Then we w i e
ab + 1 = s1=pα1qβ1, cd + 1 = s6=pα6qβ6,
ac + 1 = s2=pα2qβ2, ae + 1 = s7=pα7qβ7,
ad + 1 = s3=pα3qβ3, be + 1 = s8=pα8qβ8,
bc + 1 = s4=pα4qβ4, ce + 1 = s9=pα9qβ9,
bd + 1 = s5=pα5qβ5, de + 1 = s10 =pα10 qβ10 .
Be o e we s a ou p oo o Theo em 2 we p o e es ic ions o he exponen s
αiand βi o an S-Diophan ine quad uple (a, b, c, d).
P oposi ion 1. Assume ha (a, b, c, d)is a {p, q}-Diophan ine quad uple. Then
one o he 14 cases lis ed in Table 2 holds.
Case exponen s Case exponen s
Iα1=α2=α3< α4, α6< α5VIII α1=α4=α5< α2, α6< α3
β1=β4=β5< β2, β6< β3β1=β2=β3< β4, β6< β5
II α1=α2=α3< α4< α5< α6IX α1=α6< α4=α5< α2< α3
β1=β6< β4=β5< β2< β3β1=β2=β3< β4< β5< β6
III α1=α6< α2=α3< α4< α5Xα1=α4=α5< α2< α3< α6
β1=β4=β5< β2< β3< β6β1=β6< β2=β3< β4< β5
IV α2=α5< α1=α3< α4< α6XI α1=α6< α3=α4< α2< α5
β1=β6< β3=β4< β2< β5β2=β5< β1=β3< β4< β6
Vα2=α5< α1=α4< α3< α6XII α1=α6< α3=α4< α2< α5
β1=β6< β3=β4< β2< β5β2=β5< β1=β4< β3< β6
VI α1=α6< α2=α5< α3, α4XIII α3=α4< α1=α2< α5< α6
β3=β4< β1=β2< β5< β6β1=β6< β2=β5< β3, β4
VII α2=α5< α3=α4< α1< α6XIV α1=α6< α3=α4< α2< α5
β1=β6< β3=β4< β2< β5β2=β5< β3=β4< β1< β6
Table 2: Res ic ions on he exponen s o he S-Diophan ine quad uple (a, b, c, d)
Le us no e ha he Cases I-VII o Table 2 a e he same cases as he Cases
VIII-XIV o Table 2 bu wi h he oles o he α’s and β’s exchanged. This is easily
explained by exchanging he oles o pand q.
P oo o P oposi ion 1. We conside S-uni Equa ion (2) in iew o Lemma 2 and
ob ain ha one o he ollowing ou cases holds:
•α2=α3< α4, α5,
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Re e ences
[1] P. Co aja and U. Zannie , On he g ea es p ime ac o o (ab + 1)(ac + 1), P oc. Ame .
Ma h. Soc. 131 (6) (2003), 1705–1709.
[2] K. Gy˝o y, A. S´a k¨ozy, and C. L. S ewa , On he numbe o p ime ac o s o in ege s o he
o m ab + 1, Ac a A i h. 74 (4) (1996), 365–385.
[3] S. He n´andez and F. Luca, On he la ges p ime ac o o (ab + 1)(ac + 1)(bc + 1), Bol. Soc.
Ma . Mexicana (3) 9(2) (2003), 235–244.
[4] F. Luca, Diophan ine S-quad uples wi h wo p imes which a e win, Ac a Ma h. Hunga . 159
(2) (2019), 589–602.
[5] C. L. S ewa and R. Tijdeman, On he g ea es p ime ac o o (ab + 1)(ac + 1)(bc + 1), Ac a
A i h. 79 (1) (1997), 93–101.
[6] L. Szalay and V. Ziegle , On an S-uni a ian o Diophan ine m- uples, Publ. Ma h. Deb ecen
83 (1-2) (2013), 97–121.
[7] L. Szalay and V. Ziegle , S-Diophan ine quad uples wi h wo p imes cong uen o 3 modulo
4, In ege s 13 (2013), #A80, 9 pp.
[8] L. Szalay and V. Ziegle , S-Diophan ine quad uples wi h S={2, q},In . J. Numbe Theo y
11 (3) (2015), 849–868.
[9] V. Ziegle , On he exis ence o S-Diophan ine quad uples, Glas. Ma ., III. Se . 54 (2) (2019),
279–319.
[10] V. Ziegle , Finding all S-diophan ine quad uples o a ixed se o p imes S,Mona sh. Ma h.
196 (3) (2021), 617–641.
