A note on the second cuboid conjecture. Part I

0.17 MB·English

by  Ruslan Sharipov

A NOTE ON THE SECOND CUBOID
CONJECTURE. PART I.
Ruslan Sharipov
2
1
Abstract. The problem of finding perfect Euler cuboids or proving their non-
0
existence is an old unsolved problem in mathematics. The second cuboid conjec-
2
tureisoneofthethreepropositionssuggested asintermediatestages inprovingthe

n non-existence of perfect Euler cuboids. It is associated with a certain polynomial
a Diophantineequationoftheorder10. Inthispaperastructuraltheoremfortheso-
J lutions ofthis Diophantine equation isproved andsomeexamples of itsapplication

5 areconsidered.

]
T
N
1. Introduction.
.
h
Let’s denote throughQ (t) the following polynomialof the order10 depending
t pq
a on two integer parameters p and q:
m
10 2 2 2 2 8 8 2 6
[ Q (t)=t +(2q +p )(3q −2p )t +(q +10p q +
pq

4 4 6 2 8 6 2 2 8 2 6 4 4
1 +4p q −14p q +p )t −p q (p +10q p +4q p − (1.1)
v
6 2 8 4 6 6 2 2 2 2 2 10 10
−14q p +q )t −p q (2p +q )(3p −2q )t −q p .
9
2
2 Conjecture1.1(secondcuboidconjecture). Foranypositivecoprimeintegers
1 p6=q the polynomial Q (t) in (1.1) is irreducible in the ring Z[t].
pq
.
1
Thesecondcuboidconjecture1.1andthepolynomialQ (t)initwereintroduced
0 pq
in[1]. Theyareassociatedwiththe problemofconstructingaperfectEulercuboid
2
1 (see [2] and [3–37] for more details). Let’s write the equation
:
v
Q (t)=0. (1.2)
i pq
X
The equation (1.2) can be understood as a Diophantine equation of the order 10
r
a withtwo integerparametersp andq. The secondcuboidconjecture 1.1implies the
following theorem.
Theorem1.1. Foranypositivecoprimeintegersp6=qthepolynomialDiophantine
equation (1.2) has no integer solutions.
Note that a similar theoremassociatedwith the firstcuboid conjecture wasformu-
lated and proved in [38].
The theorem1.1isaweakerpropositionthantheconjecture 1.1itself. However,
even proving this proposition in the case of the second cuboid conjecture is rather
2000Mathematics Subject Classification. 11D41,11D72,12E05.
TypesetbyAMS-TEX
2 RUSLAN SHARIPOV
difficult. Below in section 4 we formulate and prove a structural theorem for the
solutions of the Diophantine equation (1.2), if any, and in section 5 we use it in
order to prove the theorem 1.1 for some particular values p and q.
2. The inversion symmetry and parity.
The polynomial Q (t) in (1.1) possesses some special properties. They are
pq
expressed by the following formulas which can be verified by direct calculations:
Q (p2q2/t)t10
Q (t)=− qp , Q (t)=Q (−t),
pq p10q10 pq pq
(2.1)
Q (p2q2/t)t10
Q (t)=− pq , Q (t)=Q (−t).
qp p10q10 qp qp
Note that in (2.1) we have two polynomials Q (t) and Q (t). The polynomial
pq qp
Q (t) is produced from (1.1) by exchanging parameters p and q:
qp
10 2 2 2 2 8 8 2 6
Q (t)=t +(2p +q )(3p −2q )t +(p +10q p +
qp
4 4 6 2 8 6 2 2 8 2 6 4 4
+4q p −14q p +q )t −q p (q +10p q +4p q + (2.2)
6 2 8 4 6 6 2 2 2 2 2 10 10
−14p q +p )t −q p (2q +p )(3q −2p )t −p q .
Two of the four symmetries in (2.1) contain the inversion of t. For this reason
they arecalledinversionsymmetries. Theothertwosymmetriesin(2.1)meanthat
the polynomials (1.1) and (2.2) are even with respect to their argument t.
3. Some prerequisites.
Assume that the polynomial Qpq(t) has an integer root t = A0. Since p 6= 0
and q 6= 0, we have A0 6= 0. Then due to the inversion symmetries in (2.1) the
polynomial Qqp(t) has an integer root t=B0, where
p2q2
B0 = . (3.1)
A0
Since p 6= 0 and q 6= 0, from (3.1) we derive B0 6= 0. Applying the parity symme-
try from (2.1), we conclude that the polynomial Q (t) has the other integer root
pq
t=−A0, while Qqp(t) has the other integer root t=−B0. As a result the polyno-
mials Q (t) and Q (t) split into factors
pq qp
2 2 2 2
Qpq(t)=(t −A0) C8(t), Qqp(t)=(t −B0) D8(t) (3.2)
with A0 > 0 and B0 > 0. Here C8(t) and D8(t) are eighth order polynomials
complementary to t2−A2 and t2−B2. Applying (2.1) to (3.2) we derive
0 0
C8(t)=C8(−t), D8(t)=D8(−t). (3.3)
Due to (3.3) the polynomials C8(t) and D8(t) are given by the formulas
8 6 4 2
C8(t)=t +C6t +C4t +C2t +C0,
(3.4)
8 6 4 2
D8(t)=t +D6t +D4t +D2t +D0.
A NOTE ON THE SECOND CUBOID CONJECTURE. PART I. 3
The coefficients of the polynomials (3.4) are integer numbers.
Now let’s apply the inversionsymmetries from (2.1) to (3.2). As a result we get
2 2 8 2 2 8
D8(p q /t)t C8(p q /t)t
C8(t)= p6q6A2 , D8(t)= p6q6B2 . (3.5)
0 0
Applying the symmetries (3.5) to (3.4), we derive a series of relationships for the
coefficients of the polynomials C8(t) and D8(t):
2 10 10 2 6 6
C0A0 =p q , C2A0 =p q D6,
2 2 2 2 2 2
C4A0 =p q D4, C6A0p q =D2,
2 6 6 2 10 10
A0p q =D0, D0B0 =p q , (3.6)
2 6 6 2 2 2
D2B0 =p q C6, D4B0 =p q C4,
2 2 2 2 6 6
D6B0p q =C2, B0p q =C0.
Theequations(3.6)areexcessive. Dueto(3.1)someofthemareequivalenttosome
others. For this reason we can eliminate excessive variables:
6 6 2 2 2 2
D0 =p q A0, D2 =C6p q A0,
(3.7)
6 6 2 2 2 2
C0 =p q B0, C2 =D6p q B0.
Substituting (3.7) into the formulas (3.4) for C8(t) and D8(t), we get
8 6 4 2 2 2 2 6 6 2
C8(t)=t +C6t +C4t +D6p q B0t +p q B0,
(3.8)
8 6 4 2 2 2 2 6 6 2
D8(t)=t +D6t +D4t +C6p q A0t +p q A0.
Unlike C0, D0, C2, and D2 in (3.7), the coefficients C4 and D4 in (3.8) are not
expressedthroughothercoefficients. However,theyarenotindependent. They are
related with each other by means of the equation
A0C4 =B0D4. (3.9)
The equation (3.9) is derived from (3.6) by means of the formula (3.1).
Havingderivedtheformulas(3.8),wesubstitutethembackintotherelationships
(3.2). As a result we derive the following formulas:
10 2 8 2 6 2 2 2
Qpq(t)=t +(C6−A0)t +(C4−A0C6)t +(D6p q B0−
(3.10)
2 4 2 2 2 4 4 2 2 2 6 6 2
−A0C4)t +q p B0(p q −A0D6)t −A0p q B0,
10 2 8 2 6 2 2 2
Qqp =t +(D6−B0)t +(D4−B0D6)t +(C6p q A0−
(3.11)
2 4 2 2 2 4 4 2 2 2 6 6 2
−B0D4)t +p q A0(p q −B0C6)t −A0p q B0.
Comparing the formula (3.10) with (1.1) and comparing the formula (3.11) with
(2.2), we derive ten equations for the coefficients of the polynomials (3.8). Two of
them are equivalent to the equation (3.1) written as
2 2
A0B0 =p q . (3.12)
4 RUSLAN SHARIPOV
The other eight equations are written as follows:
2 2 2 2 2
C6−A0 =(2q +p )(3q −2p ),
(3.13)
2 2 2 2 2
D6−B0 =(2p +q )(3p −2q ),
2 8 6 2 4 4 2 6 8
C4−A0C6 =p −14p q +4p q +10p q +q ,
(3.14)
2 8 6 2 4 4 2 6 8
D4−B0D6 =q −14q p +4q p +10q p +p ,
2 2 2 2 2 2 8 6 2 4 4 2 6 8
A0C4−D6p q B0 =p q (q −14q p +4q p +10q p +p ),
(3.15)
2 2 2 2 2 2 8 6 2 4 4 2 6 8
B0D4−C6p q A0 =p q (p −14p q +4p q +10p q +q ),
2 2 4 4 4 4 2 2 2 2
B0(A0D6−p q )=p q (2p +q )(3p −2q ),
(3.16)
2 2 4 4 4 4 2 2 2 2
A0(B0C6−p q )=p q (2q +p )(3q −2p ).
The equations (3.12), (3.13), (3.14), (3.15), (3.16) are excessive. Indeed, the equa-
tions (3.16) follow from (3.12) and (3.13). Similarly, the equations (3.15) can be
derivedfrom(3.14)withtheuseof(3.9)and(3.12). Asfortheequations(3.13)and
(3.14), when complemented with the equations (3.9) and (3.12), they constitute a
system of Diophantine equations with respect to C4, D4, C6, D6, A0 and B0. The
results of the above calculations are summarized in the following lemma.
Lemma 3.1. For p6=0 and q 6=0 the polynomial Q (t) in (1.1) has integer roots
pq
if and only if the system of Diophantine equations (3.9), (3.12), (3.13), and (3.14)
is solvable with respect totheinteger variables C4, D4, C6, D6, A0 >0 andB0 >0.
4. The structural theorem.
Below we continue studying the equations (3.9), (3.12), (3.13), (3.14) implicitly
assumingp6=q tobetwopositivecoprimeintegernumbers. Letp1, ... , pm bethe
prime factors of p and let q1, ... , qn be the prime factors of q:
p=pα1 ·...·pαm, q =qβ1 ·...·qβn. (4.1)
1 m 1 n
Usuallythemultiplicitiesα1, ... , αm andβ1, ... , βnin(4.1)arepositivenumbers.
However, in order to cover two special cases p = 1 and q = 1 we assume them to
be non-negative numbers. Since p and q are assumed coprime, i.e.
gcd(p,q)=1, (4.2)
the prime factors p1, ... , pm and q1, ... , qn in (4.1) are distinct, i.e. pi 6=qj.
Lemma 4.1. For any solution of the Diophantine equations (3.9), (3.12), (3.13),
and (3.14) with A0 > 0 and B0 > 0 if A0 6= 1, each prime factor r of A0 is a
prime factor of p or a prime factor of q, i.e. r =p or r =q .
i j
Lemma 4.2. For any solution of the Diophantine equations (3.9), (3.12), (3.13),
and (3.14) with A0 > 0 and B0 > 0 if B0 6= 1, each prime factor r of B0 is a
prime factor of p or a prime factor of q, i.e. r =p or r =q .
i j
The lemmas 4.1 and 4.2 are immediate from (4.2) and (3.12). Due to the lem-
A NOTE ON THE SECOND CUBOID CONJECTURE. PART I. 5
mas 4.1 and 4.2 we can write the following expansions for A0 and B0:
A0 =p1µ1 ·...·pmµmq1ν1 ·...·qnνn,
(4.3)
B0 =p1η1 ·...·pmηmq1τ1 ·...·qnτn.
The multiplicities µ1, ... , µm, ν1, ... , νn, η1, ... , ηm, and τ1, ... , τn in the ex-
pansions (4.3) obey the following relationships:
µ +η =2α , ν +τ =2β . (4.4)
i i i i i i
Theformulas(4.4)areeasilyderivedbysubstituting theexpansions(4.1)and(4.3)
into the equation (3.12).
Now let’s return back to the equations (3.13) and (3.14). It is easy to see that
the equations (3.13) can be explicitly resolved with respect to C6 and D6:
2 2 2 2 2
C6 =A0+(2q +p )(3q −2p ),
(4.5)
2 2 2 2 2
D6 =B0 +(2p +q )(3p −2q ).
Uponsubstituting(4.5)intotheequations(3.14)wecanexplicitlyresolvetheequa-
tions (3.14) with respect to the variables C4 and D4:
4 2 2 2 2 2
C4 =A0+(2q +p )(3q −2p )A0+
(4.6)
8 6 2 4 4 2 6 8
+p −14p q +4p q +10p q +q ,
4 2 2 2 2 2
D4 =B0 +(2p +q )(3p −2q )B0+
(4.7)
8 6 2 4 4 2 6 8
+q −14q p +4q p +10q p +p .
And finally, we can substitute (4.6) and (4.7) into the equation (3.9). As a result
we derive the following equation for the variables A0 and B0:
4 2 2 2 2 2
A0(cid:0)A0+(2q +p )(3q −2p )A0+
8 6 2 4 4 2 6 8
+p −14p q +4p q +10p q +q =
(cid:1)
(4.8)
4 2 2 2 2 2
= B0(cid:0)B0 +(2p +q )(3p −2q )B0+
8 6 2 4 4 2 6 8
+q −14q p +4q p +10q p +p .
(cid:1)
Summarizing these calculations, we can formulate the following lemma.
Lemma4.3. ThesystemoffourDiophantineequations(3.9),(3.12),(3.13),(3.14)
is equivalent to the system of two Diophantine equations (3.12) and (4.8).
Lemma 4.4. For any solution of the Diophantine equations (3.9), (3.12), (3.13),
and(3.14)withA0 >0andB0 >0if µi >0andηi >0in(4.3), thenµi =ηi =αi.
Proof. The proof is by contradiction. Assume that µ > 0, η > 0, and µ 6= η .
i i i i
Then from (4.6) and (4.7), applying (4.1), (4.2), (4.3), and (4.4), we derive
8 8
C4 ≡q (mod pi), D4 ≡q (mod pi). (4.9)
6 RUSLAN SHARIPOV
Moreover,the formula (4.3) yields the relationships
A0 =A′0piµi, B0 =B0′ piηi, (4.10)
where A′ 6≡0(mod p ) and B′ 6≡0(mod p ). Our assumption µ 6=η means that
0 i 0 i i i
µ >η or µ <η . If µ >η , then substituting (4.10) into (3.9), we derive
i i i i i i
A′0C4piµi−ηi =B0′ D4. (4.11)
Due to (4.10), (4.9) and (4.2) the left hand side of (4.11) is zero modulo p , while
i
the right hand side of (4.11) is nonzero modulo p , which is contradictory.
i
Similarly, if µ <η , substituting (4.10) into (3.9), we derive
i i
A′0C4 =B0′ D4piηi−µi. (4.12)
In this case the left hand side of (4.12) is nonzero modulo p , while the right hand
i
side of (4.12) is zero modulo p , which is also contradictory.
i
The contradictions obtained prove that µ = η . The equalities µ = α and
i i i i
η =α are immediate from µ =η due to (4.4). The lemma 4.4 is proved. (cid:3)
i i i i
Lemma 4.5. For any solution of the Diophantine equations (3.9), (3.12), (3.13),
and (3.14) with A0 >0 and B0 >0 if νi >0 and τi >0 in (4.3), then νi =τi =βi.
The lemma 4.5 is analogous to the lemma 4.4. Its proof is similar to the above
proof of the lemma 4.4.
Now,relyingonthelemmas4.4and4.5,wedefinethefollowingintegernumbers:
ap =Ypiαi, bp =Ypiαi, cp =Ypiαi, (4.13)
ηi=0 µi=0 µi>0
ηi>0
aq =Yqiβi, bq =Yqiβi, cq =Yqiβi. (4.14)
τi=0 νi=0 νi>0
τi>0
Note that the numbers (4.13) and (4.14) are pairwise mutually coprime, i.e.
gcd(a ,b )=1, gcd(a ,c )=1, gcd(a ,a )=1,
p p p p p q
gcd(a ,b )=1, gcd(a ,c )=1, gcd(b ,c )=1,
p q p q p p
gcd(b ,a )=1, gcd(b ,b )=1, gcd(b ,c )=1, (4.15)
p q p q p q
gcd(c ,a )=1, gcd(c ,b )=1, gcd(c ,c )=1,
p q p q p q
gcd(a ,b )=1, gcd(a ,c )=1, gcd(b ,c )=1.
q q q q q q
If µ = 0, then η = 2α , and if η = 0, then µ = 2α . Similarly, if ν = 0, then
i i i i i i i
τ = 2β and if τ = 0, then ν = 2β . These implications are derived from (4.4).
i i i i i
The lemmas 4.4 and 4.5 say that if µ > 0 and η > 0, then µ = η = α , and if
i i i i i
ν >0 and τ >0, then ν =τ =β . As a result from (4.13) and (4.14) we derive
i i i i i
2 2 2 2
A0 =apcpaqcq, B0 =bpcpbqcq,
(4.16)
p=a b c , q =a b c .
p p p q q q
A NOTE ON THE SECOND CUBOID CONJECTURE. PART I. 7
The result expressed by the formulas (4.15) and (4.16) is rather important. For
this reason it is formulated as a lemma.
Lemma 4.6. For any solution of the Diophantine equations (3.9), (3.12), (3.13),
and (3.14)with A0 >0 and B0 >0 there are six positive pairwise mutually coprime
integer numbers ap, bp, cp, aq, bq, cq such that the numbers A0, B0, p, q are
expressed through them by means of the formulas (4.16).
Let’s substitute (4.16) into the equation (4.10). As a result we obtain the fol-
lowing equation with respect to the numbers a , b , c , a , b , c :
p p p q q q
10 10 4 4 6 10 2 6 4 8 8 4 4 2 2
a a c c +6a a c c b −a a c c b b −
p q p q p q p q q p q p q q p
10 6 6 2 4 6 6 4 4 4 4 10 2 8 8
−2a a c c b +4a a b c b c +a a b c +
p q p q p p q p p q q p q p p
2 10 8 8 4 8 2 2 6 6 8 4 6 6 2 2
+a a b c +10a a b c b c −14a a b c b c =
p q q q p q p p q q p q p p q q
(4.17)
10 10 4 4 10 6 6 2 4 8 8 4 4 2 2
=b b c c +6b b c c a −b b c c a a −
p q p q p q p q p p q p q q p
6 10 2 6 4 6 6 4 4 4 4 2 10 8 8
−2b b c c a +4b b a c a c +b b a c +
p q p q q p q p p q q p q q q
10 2 8 8 8 4 6 6 2 2 4 8 2 2 6 6
+b b a c +10b b a c a c −14b b a c a c .
p q p p p q p p q q p q p p q q
Lemma4.7. Foragivenpairofpositivecoprimeintegernumbersp6=q thesystem
of Diophantine equations (3.9), (3.12), (3.13), and (3.14) is resolvable if and only
if there are six positive integer numbers a , b , c , a , b , c obeying the equation
p p p q q q
(4.17),obeyingthecoprimalityconditions(4.15),andsuchthatpandqareexpressed
through them by means of the formulas p=a b c and q =a b c .
p p p q q q
The lemma 4.7 follows from the lemmas 4.3 and 4.6 due to the calculations in
deriving the equation (4.17) from the equation (4.8). Combining the lemmas 3.1
and 4.7, now we derive the following structural theorem for the solutions of the
Diophantine equation (1.2).
Theorem 4.1. For a given pair of positive coprime integer numbers p 6= q the
Diophantine equation (1.2) is resolvable with respect to the variable t if and only
if there are six positive integer numbers a , b , c , a , b , c obeying the equation
p p p q q q
(4.17),obeyingthecoprimalityconditions(4.15),andsuchthatpandqareexpressed
through them by means of the formulas p = a b c and q = a b c . Under these
p p p q q q
conditions the equation (1.2) has at least two solutions given by the formulas
2 2 2 2
t=a c a c , t=−a c a c . (4.18)
p p q q p p q q
The structuraltheorem4.1is the mainresultofthis paper. The formulas(4.18)
in this theorem are immediate from (4.16).
5. Some applications of the structural theorem.
Let’s choose q =1 and assume that p is some prime number. Then p and q are
coprime, i.e. the relationship (4.2) is fulfilled. Applying the formula q = a b c
q q q
from the theorem 4.1 to this case, we get
a =1 b =1 c =1. (5.1)
q q q
8 RUSLAN SHARIPOV
Similarly, applying the formula p = a b c from the theorem 4.1 and taking into
p p p
account that p is prime, we derive three options
a =p, b =1, c =1; (5.2)
p p p
a =1, b =p, c =1; (5.3)
p p p
a =1, b =1, c =p. (5.4)
p p p
Substituting (5.1) and (5.2) into (4.17), we obtain the following equation for p:
2 8
16p −16p =0. (5.5)
The left hand side of the equation (5.5) factorizes as
2 2 2
−16p (p−1)(p+1)(p +p+1)(p −p+1)=0. (5.6)
Therefore the only integer solutions of the equation (5.6) are
p=−1, p=1. (5.7)
The first ofthem is negative. The secondone is positive, but p=1 contradicts the
inequality p6=q in the theorem 4.1 since q =1 in our present case.
The second option is given by the formulas (5.3). Substituting (5.1) and (5.3)
into (4.17), we obtain another equation for p:
10 8 6 4 2
−8p −8p −16p +16p +8p +8=0. (5.8)
The left hand side of the equation (5.8) factorizes as follows:
8 6 4 2
−8(p−1)(p+1)(p +2p +4p +2p +1)=0. (5.9)
Again the only integer solutions of the equation (5.9) are given by the formulas
(5.7). The first of these two solutions is negative, while the second contradicts the
inequality p6=q in the theorem 4.1.
And finally, the third option is given by the formulas (5.4). Substituting (5.1)
and (5.4) into (4.17), we obtain the third equation for p:
6 2
−32p +32p =0. (5.10)
The left hand side of the equation (5.10) factorizes as follows:
2 2
−32p (p−1)(p+1)(p +1)=0. (5.11)
The integer solutions of the equation (5.11) are given by the formula (5.7). The
first of them is negative, while the second one contradicts the inequality p 6= q.
Thus, none of the equations (5.6), (5.9), and (5.11) has a solution suitable for the
theorem 4.1. Applying this theorem, we can formulate the following result.
Theorem 5.1. For q = 1 and for any positive prime integer p the polynomial
Diophantine equation (1.2) has no integer solutions.
A NOTE ON THE SECOND CUBOID CONJECTURE. PART I. 9
Exchanging p and q in the theorem 5.1, we obtain another theorem.
Theorem 5.2. For p = 1 and for any positive prime integer q the polynomial
Diophantine equation (1.2) has no integer solutions.
The proof of the theorem 5.2 is similar to the above proof of the theorem 5.1.
The theorems 5.1 and 5.2 prove the theorem 1.1 in two special cases where p is
prime and q =1 and where q is prime and p= 1. Other examples of applying the
structural theorem 4.1 will be given in a separate paper.
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