(PDF) Complete solution of the polynomial version of a problem of Diophantus

(PDF) Complete solution of the polynomial version of a problem of Diophantus
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Complete solution of the polynomial version of a problem of Diophantus
Andrej Dujella
2004, Journal of Number Theory
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Abstract
In this paper, we prove that if {a, b, c, d} is a set of four non-zero polynomials with integer coefficients, not all constant, such that the product of any two of its distinct elements plus 1 is a square of a polynomial with integer coefficients, then This settles the "strong" Diophantine quintuple conjecture for polynomials with integer coefficients.
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COMPLETE SOLUTION OF THE POLYNOMIAL VERSION
OF A PROBLEM OF DIOPHANTUS
ANDREJ DUJELLA∗ AND CLEMENS FUCHS‡
Abstract. In this paper, we prove that if {a, b, c, d} is a set of four
non-zero polynomials with integer coefficients, not all constant, such
that the product of any two of its distinct elements plus 1 is a square of
a polynomial with integer coefficients, then
(a + b − c − d)2 = 4(ab + 1)(cd + 1).
This settles the “strong” Diophantine quintuple conjecture for polynomials with integer coefficients.

1. Introduction
A set of m positive integers is called a Diophantine m-tuple if the product
of any two of its distinct elements increased by 1 is a perfect square (cf.
[3]). The first Diophantine quadruple, the set {1, 3, 8, 120}, was found by
Fermat. In 1969, Baker and Davenport [2] proved that the Fermat’s set
cannot be extended to a Diophantine quintuple. The “folklore” conjecture
is that there does not exist a Diophantine quintuple. Recently, the first
author proved that there does not exist a Diophantine sextuple and there
are only finitely many Diophantine quintuples (see [5]).
It was known already to Euler that every Diophantine pair {a, b} can be
extended to a Diophantine quadruple. Namely, if ab + 1 = r2 , then
{a, b, a + b + 2r, 4r(a + r)(b + r)}
is a Diophantine quadruple. A Diophantine triple of the form {a, b, a+b+2r}
is called a regular Diophantine triple. In 1979, Arkin, Hoggatt and Strauss
[1] proved that every Diophantine triple can be extended to a Diophantine
quadruple. More precisely, let ab + 1 = r2 , ac + 1 = s2 , bc + 1 = t2 , where
r, s, t are positive integers. Define
d± = a + b + c + 2abc ± 2rst.
Then
{a, b, c, d± }
The first author was supported by the Ministry of Science and Technology, Republic
of Croatia, grant 0037110.
‡The second author was supported by the Austrian Science Foundation FWF, grant
S8307-MAT.

ANDREJ DUJELLA AND CLEMENS FUCHS

are Diophantine quadruples (observe that d− < c). Indeed,
(1) ad± + 1 = (at ± rs)2 ,

bd± + 1 = (bs ± rt)2 ,

cd± + 1 = (cr ± st)2 .

Diophantine quadruples of this form are called regular Diophantine quadruples. Equivalently, {a, b, c, d} is regular, if and only if
(a + b − c − d)2 = 4(ab + 1)(cd + 1)
(see [9]). This is a quadratic equation in d with the roots d± .
There is even a stronger version of the “folklore” conjecture from above,
namely if we fix a Diophantine triple {a, b, c}, then there is a unique
positive integer d such that d > max{a, b, c} and {a, b, c, d} is a Diophantine
quadruple. This means that every Diophantine quadruple is regular. In the
mentioned result of the nonexistence of Diophantine sextuples, the author
proves this stronger conjecture for all triples satisfying some gap conditions.
A polynomial variant of the above problems was first studied by Jones
[10, 11].
Definition 1. A set {a1 , a2 , . . . , am } of m non-zero polynomials with integer
coefficients, which are not all constant, is called a polynomial Diophantine
m-tuple if for all 1 ≤ i < j ≤ m the following holds: ai · aj + 1 = b2ij , where
bij ∈ Z[X].
Observe that every polynomial pair can be extended to a polynomial
triple and that every polynomial triple can be extended to a polynomial
quadruple. In fact the relations from above are true after all, because they
are obtained by purely algebraic manipulations. Therefore, we define
Definition 2. A polynomial Diophantine quadruple {a, b, c, d} is called regular if
(a + b − c − d)2 = 4(ab + 1)(cd + 1),
or, equivalently, if d = d+ or d− , where
d± = a + b + c + 2abc ± 2rst
and r, s, t ∈ Z[X] are defined by
ab + 1 = r2 ,

ac + 1 = s2 ,

bc + 1 = t2 .

The result obtained by the first author already mentioned above about
the existence of only finitely many Diophantine quintuples implies that there
does not exist a polynomial Diophantine quintuple. In the present paper, we
will prove the “stronger” Diophantine quintuple conjecture for polynomials.
Namely, we have
Theorem 1. All polynomial Diophantine quadruples are regular.
This theorem has been proved by Jones in [11] in the case that a, b, c
are linear polynomials. Moreover, Jones has proved that there is no
polynomial Diophantine quadruple with four polynomials all having the

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

same positive degree (cf. [11, Corollary 1]). Other results related to polynomial versions of the above problem of Diophantus can be found in [6, 7, 8].
In the proof of Theorem 1, we follow the strategy from the paper of the
first author [5]. Namely, we first transform the problem into solving a system
of simultaneous Pellian equation, which reduces to finding intersections of
binary recurring sequences of polynomials. We will assume that we have an
irregular polynomial Diophantine quadruple {a, b, c, d} with minimal d. This
will lead, by using congruence relations and a gap principle, to a very precise
determination of the initial terms of the recurring sequences. From this we
will be able to prove our main result.
2. Reduction to intersections of recursive sequences
Let Z+ [X] denote the set of all polynomials with integer coefficients with positive leading coefficient. For a, b ∈ Z[X], a < b means
that b − a ∈ Z+ [X]. The usual fundamental properties of inequality hold
for this order. For a ∈ Z[X], we define |a| = a if a ≥ 0, and |a| = −a if a < 0.
If {a, b, c, d}, a < b < c < d is a Diophantine quadruple, then d is nonconstant. Assume now that a and b are constant polynomials. Considering
leading coefficients of ad + 1 and bd + 1 we conclude that ab is a perfect
square, contradicting the assertion that ab + 1 is also a perfect square.
Therefore, we proved that in a polynomial Diophantine quadruple there is
at most one constant polynomial. It is also clear that all leading coefficients
of the polynomials in a Diophantine m-tuple have the same sign. This
implies that there is no loss of generality in assuming that they are all
positive, i.e. that all polynomials are in Z+ [X].
Assume that {a, b, c, d}, where 0 < a < b < c < d, is an irregular
polynomial Diophantine quadruple with minimal d among all irregular
polynomial Diophantine quadruples. Under this assumption we will end up
with a contradiction, which implies that such a quadruple cannot exist.
Let r, s, t ∈ Z+ [X] be defined by
ab + 1 = r2 ,

ac + 1 = s2 ,

bc + 1 = t2 .

In this paper, the symbols r, s, t will always have this meaning. Moreover,
let
(2)

ad + 1 = x2 ,

bd + 1 = y 2 ,

cd + 1 = z 2 ,

with x, y, z ∈ Z+ [X]. Eliminating d from (2) we obtain the system
(3)

cx2 − az 2 = c − a,

(4)

cy 2 − bz 2 = c − b.

ANDREJ DUJELLA AND CLEMENS FUCHS

We will now describe the sets of solutions of equation (3) and (4). The
following lemma is an analogue of the result proved in [4] for the classical
case of Pellian equations in integers. A similar lemma for polynomials was
also proved in [6].
Let
deg a = A, deg b = B and deg c = C.
The letters A, B, C will have this meaning for the rest of the paper.
Lemma 1. If (z, x) and (z, y), with x, y, z ∈ Z+ [X], are polynomial solutions
of (3) and (4) respectively, then there exist z0 , x0 ∈ Z[X] and z1 , y1 ∈ Z[X]
with
(i) (z0 , x0 ) and (z1 , y1 ) are solutions of (3) and (4) respectively,
(ii) the following inequalities are satisfied:
(5)

deg x0 ≤

A+C
< deg s,

deg z0 ≤

3C − A
< C,

(6)

deg y1 ≤

B+C
< deg t,

deg z1 ≤

3C − B
< C,

and
x0 , |z0 |, y1 , |z1 | > 0,

(7)

and there exist integers m, n ≥ 0 such that
(8)
z a + x c = (z0 a + x0 c)(s + ac)m ,
z b + y c = (z1 b + y1 c)(t + bc)n ,
√ √
where this means that the coefficients of a, b and c respectively on both
sides are equal.
(9)

Proof. The proof of the statements follow from [8, Lemma 4].

In that way, our problem reduces to solving equations of the form
vm = wn ,
where vm and wn are binary recursive sequences defined by
(10)

v0 = z0 ,

v1 = sz0 + cx0 ,

vm+2 = 2svm+1 − vm ,

for some solution (z0 , x0 ) of (3) with (5), and
(11)

w0 = z1 ,

w1 = tz1 + cy1 ,

wn+2 = 2twn+1 − wn ,

for some initial values (z1 , y1 ) as above.
We will need information on the degrees of these sequences and we collect
these in the lemma below.

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

Lemma 2. Let (vm ), (un ) be the sequences from above. Then
A+C
B+C
deg wn = deg w1 + (n − 1)
Proof. The proof runs by induction on m, n respectively and follows easily
from (10) and (11).
deg vm = deg v1 + (m − 1)

3. Gap principle and congruence relations
As we have seen, the polynomials d0 = d+ and d− have the property that
ad0 + 1, bd0 + 1, cd0 + 1 are perfect squares. We repeat this construction in
the following lemma which was proved e.g. in [7].
Lemma 3. Let {a, b, c} be a polynomial Diophantine triple and let ab + 1 =
r2 , ac + 1 = s2 , bc + 1 = t2 . Then for
d± = a + b + c + 2abc ± 2rst,
we have
ad± + 1 = u2 , bd± + 1 = v 2 , cd± + 1 = w2
with u = at ± rs, v = bs ± rt, w = cr ± st. Furthermore, it holds
c = a + b + d± + 2(abd± ∓ ruv).
Let us remark that an easy computation shows that
d+ · d− = (c − a − b − 2r)(c − a − b + 2r).
The trivial observation that if d− 6= 0, then d− ≥ 1 leads to the very useful
gap principle, which was already proved by Jones in [11].
Lemma 4. If {a, b, c} is a polynomial Diophantine triple and a < b < c,
then c = a + b + 2r or c ≥ 2abd− + 1, where d− is defined as above and
d− 6= 0.
Proof. This was shown for example in the proof of Lemma 3 in [7] and
follows easily from Lemma 3.
Observe that from the gap principle it follows that we either have
C ≥ A + B,
i.e. C is larger than A, B, since A ≥ 0 and B > 0, or c = a + b + 2r holds.
We will use this fact several times later on.
Let us consider the sequences (vm ) and (wn ) modulo 2c. From (10) and
(11) it is easily seen (by induction) that
(12)
(13)

v2m ≡ z0 (mod 2c),
w2n ≡ z1 (mod 2c),

v2m+1 ≡ sz0 + cx0 (mod 2c),
w2n+1 ≡ tz1 + cy1 (mod 2c).

We will deduce later very precise information on the initial terms z0 and z1 .

ANDREJ DUJELLA AND CLEMENS FUCHS

As a consequence of Lemma 1 and the relations (12) and (13), we obtain
the following lemma.
Lemma 5. We have:
1) If the equation v2m = w2n has a solution, then z0 = z1 .
2) If the equation v2m+1 = w2n has a solution, then cx0 − s|z0 | = |z1 |.
3) If the equation v2m = w2n+1 has a solution, then cy1 − t|z1 | = |z0 |.
4) If the equation v2m+1 = w2n+1 has a solution, then cx0 − x|z0 | =
cy1 − t|z1 |.
Proof. We split the proof according to the statements of the lemma.
1) From Lemma 1 and equation (12) we have |z0 −z1 | < 2c and z0 ≡ z1 (mod
2c), which implies z0 = z1 .
2) Observe that
(c + s)(cx0 − s|z0 |) ≤ (cx0 + s|z0 |)(cx0 − s|z0 |) =
= c2 x20 − s2 z02 = c2 − ac − z02 ≤ c2 − s2 ,
thus cx0 − s|z0 | < c. On the other hand we have
c(c − a)
3c(c − a)
> 0,
since deg z0 ≤ 3C
2 which follows from (5) of Lemma 1 and deg c(c − a) = 2C.
This last equation follows trivially if C > A and otherwise we have (by the
gap principle) that c = a+b+2r and therefore deg c−a = deg b = C. Hence,
c2 − ac − z02 ≥ c2 − ac −

0 < cx0 − s|z0 | < c.
By (12) we have sz0 + cx0 ≡ z1 (mod 2c). Thus we conclude that if z0 > 0
then z1 = sz0 − cx0 , and if z0 < 0, then z1 = sz0 + cx0 .
3) As in 2), we find that
0 < cy1 − t|z1 | < c,
which implies that if z1 > 0, then z0 = tz1 − cy1 , and if z1 < 0, then
z0 = tz1 + cy1 .
4) We have already proved that
0 < cx0 − s|z0 | < c,

0 < cy1 − t|z1 | < c.

Hence, we have two possibilities: if z0 > 0 then also z1 > 0 and
sz0 − cx0 = tz1 − cy1 , and if z0 < 0 then z1 < 0 and sz0 + cx0 = tz1 + cy1 . 
In the following lemma we will consider the sequences (vm ) and (wn )
modulo 4c2 .
Lemma 6. We have:
1) v2m ≡ z0 + 2c(az0 m2 + sx0 m) (mod 8c2 )
2) v2m+1 ≡ sz0 + c [2asz0 m(m + 1) + x0 (2m + 1)] (mod 4c2 )
3) w2n ≡ z1 + 2c(bz1 n2 + ty1 n) (mod 8c2 )
4) w2n+1 ≡ tz1 + c [2btz1 n(n + 1) + y1 (2n + 1)] (mod 4c2 )

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

Proof. The proof runs by induction and can be done totally in the same
way as in [4, Lemma 4].
4. Precise determination of initial terms
From the estimates of initial terms and congruence condition modulo 2c,
it follows that if the equation vm = wn has a solution, then there exists a
solution with m = 0 or 1 (see Lemma 5). But, that small solution induces
d0 < c such that ad0 + 1, bd0 + 1 and cd0 + 1 are perfect squares. From our
minimality assumption it follows that d0 = 0 or d0 = d− . This conclusion
leads to very precise determination of initial terms:
Lemma 7. We have:
1) If v2m = w2n has a solution, then either
1.1) z0 = z1 = ±1, or
1.2) z0 = z1 = ±(cr − st).
2) If v2m+1 = w2n has a solution, then z0 = ±t and z1 = ±(st − cr).
3) If v2m = w2n+1 has a solution, then z0 = ±(cr − st) and z1 = ∓s.
4) If v2m+1 = w2n+1 has a solution, then z0 = ±t and z1 = ±s.
Proof. 1) From Lemma 5 we have z0 = z1 . Define d0 = (z02 −1)/c ∈ Z+ [X].
Then we have
cd0 + 1 = z02 .
We have already seen that d0 = 0 or d− . If d0 = 0 then z0 = ±1. If d0 = d−
then
z02 = cd− + 1 = (cr − st)2 ,
(see equation (1)) and z0 = ±(cr − st).
2) By Lemma 5, if z1 > 0 then z0 < 0 and we define z 0 = z1 = cx0 + sz0 ,
and if z1 < 0 then z0 > 0 and we define z 0 = −z1 = cx0 − sz0 . Thus z 0 > 0.
Define d0 = (z 02 − 1)/c ∈ Z+ [X]. We have again
cd0 + 1 = z 02 .
Now we show that d0 = 0 is impossible. This is clearly true if z 0 > 1. In the
case that z 0 = 1, it follows that
c(c − a) − z02 = c2 − ac − z02 = (cx0 + s|z0 |)z 0
and therefore

3C
2C = deg (cx0 + s|z0 |)z 0 ≤
which is a contradiction. Here we have again used that deg c(c − a) = 2C;
this is clearly true if C > A and otherwise we have c − a = b + 2r, i.e.
deg(c − a) = deg b = C.
This means that d0 = d− and as above z 0 = ±(cr − st). Thus z1 =
±(cr − st) and
|z1 | = cr − st = cx0 − s|z0 |.
Observe that we have cr − st > 0 since c2 r2 = abc2 + c2 > abc2 + ac +
bc + 1 = (ac + 1)(bc + 1) = s2 t2 . The fact that c2 > bc + ac + 1 follows

ANDREJ DUJELLA AND CLEMENS FUCHS

easily from the gap principle (Lemma 4). The last equation can be rewritten
as c(x0 − r) = s(|z0 | − t). From gcd(c, s) = 1 (which is a consequence of
ac + 1 = s2 ) it follows that |z0 | ≡ t (mod c), and since |z0 | < c, t < c (which
follows from Lemma 1 and bc+1 = t2 ), we conclude that |z0 | = t and x0 = r.
3) In analogy to above, let z 0 = z0 = cy1 + tz1 if z0 > 0, and z 0 = −z0 =
cy1 − tz1 if z0 < 0, and define d0 = (z 02 − 1)/c. Then
cd0 + 1 = z 02
and 0 < d0 < c. Thus d = d− , which implies |z 0 | = cr − st. Hence,
|z0 | = cr − st and c(y1 − r) = t(|z1 | − s). Since gcd(t, c) = 1 we have |z1 | ≡ s
(mod c), which implies |z1 | = s.
4) Let z 0 = cx0 − s|z0 | = cy1 − t|z1 | and d0 = (z 02 − 1)/c. We have d0 = d−
and therefore |z 0 | = cr − st. We have already shown that this fact implies
|z0 | = t and |z1 | = s.
Let a = α2 δX A + · · · , b = β 2 δX B + · · · , c = γ 2 δX C + · · · .
We will consider several subcases:
B < C,

A < B = C,

A = B = C.

These cases correspond to the different types of standard Diophantine triples
(see Definition 1 in [5]). Note that if B = C, then c = a + b + 2r. Indeed, by
our gap principle (Lemma 4) we have that if c 6= a + b + 2r, then c ≥ 2abd− ,
where d− is defined in Definition 1 and d− 6= 0. Assume that B = C. Then a
and d− are both constant polynomials. From this it follows that B = C > 0
(because there can be at most one constant polynomial in the Diophantine
triple) and d− = µ2 δ because the leading coefficient of bd− + 1 is a perfect
square. But then we have that ad− + 1 and ad− are both perfect squares, a
contradiction.
In the next section we will turn our discussion to the remaining cases.
Before we do this, we collect some technical information in the following
lemmata.
Lemma 8. Assume that B < C. Then, we have
deg(cr − st) = C −

A+B

Moreover, if z0 = cr − st then
deg(cx0 − sz0 ) =

B+C

deg(cy1 − tz1 ) =

A+C

and if z1 = cr − st then

Proof. First, we conclude from
(cr − st)(cr + st) = c2 r2 − s2 t2 = c2 − ac − bc − 1

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

that
A+B
= 2C
and therefore the first part of the lemma follows. Now, observe
deg(cr − st) + C +

C −B
C −A
cy12 = b(cr − st)2 + c − b =⇒ deg y1 =

cx20 = a(cr − st)2 + c − a =⇒ deg x0 =

Thus, we have
deg sz0 = deg cx0 =

3C − B

By using the equation
(cx0 − sz0 )(cx0 + sz0 ) = c2 x20 − s2 z02 =
= ca(cr − st)2 + c2 − ac − s2 (cr − st)2 = c2 − ac − (cr − st)2 ,
and by observing that C > A, we get
deg(cx0 − sz0 ) =

B+C

The last part can be obtained analogously.
Lemma 9. Let e = 2rst − 2cr + c. Then, we have
(i) deg e = C − A − B < C, if C > A + 2B,
(ii) deg e ≤ B = C − A − B < C, if C = A + 2B, and
(iii) deg e = B, if C < A + 2B.
Proof. Let U = e(st + cr). We have U = 2acr + 2bcr + 2r + cst − c2 r and
by the previous lemma, deg U ≤ 2C − A+B
since deg(st − cr) = C − A+B
(observe that for C = A + 2B the polynomials 2bcr and cst − c2 r < 0 have
the same degree). Hence, deg e ≤ C − A − B < C.
For C > A + 2B we get deg e = C − A − B.
If we assume C < A + 2B, then the dominant summand in U is 2bcr and
therefore deg U = B + C + A+B
2 , which in turn implies deg e = B.
Lemma 10. Let {a, b, c} with a < b < c be a polynomial Diophantine triple
and assume that B < C = A + 2B. Then
{a, b, d− , c} = {a, b, a + b ± 2r, 4r(r ± a)(b ± r)}.
Moreover, in this case we have e = ∓2r.
Proof. First, by using the equation
d+ · d− = (c − a − b − 2r)(c − a − b + 2r),
which yields deg c = deg(abd− ), we get deg d− = B. Since B < C we have
three possibilities, namely 0 < a < b < d− < c, 0 < a < d− < b < c or
0 < d− < a < b < c. Observe that d− 6= 0, since d− = 0 implies c = a+b±2r
and thus B = C, a contradiction. Moreover, d− 6= a, b, since this would lead
to a2 + 1 = r2 , b2 + 1 = s2 respectively, which is also a contradiction.

10

ANDREJ DUJELLA AND CLEMENS FUCHS

Now, since A ≤ B = deg d− , we apply Lemma 3 to the triple {a, b, d− }.
We have that e+ = c. But now, by the equation
e+ · e− = c · e− = (d− − a − b − 2r)(d− − a − b + 2r),
we get, since C = A + 2B ≥ 2B, that deg e− ≤ 0. Assume that e− 6= 0.
By observing that ae− + 1, be− + 1 are squares, we conclude by comparing
coefficients that e− = ψ 2 δ. Moreover, we have A = 0 and therefore a = α2 δ,
which yields a contradiction. Therefore we conclude e− = 0 and thus that
the triple {a, b, d− } is regular, i.e. d− = a + b ± 2r. Now, we use once more
that c can be recalculated by a, b and d− . We have (cf. Lemma 3)
c = a + b + d− + 2(abd− + ruv),
where u2 = ad− + 1 and v 2 = bd− + 1. From above it follows that u =
r ± a, v = b ± r and
c = a + b + a + b ± 2r + 2ab(a + b ± 2r) + 2ruv =
= 2(a + b)r2 ± 2r(2ab + 1) + 2ruv = 2r(ar + ab ± ab ± r2 ) + 2ruv =
= 2r(r ± a)(b ± r) + 2ruv = 4ruv.
It implies that
c = 4r(r ± a)(b ± r).
In this case we have
s = 2r2 ± 2ar − 1,
t = 2br ± (2r2 − 1).
Now, let e = 2rst − 2cr2 + c. Direct computation shows that e = ∓2r.

5. Proof of the Theorem
We conclude the proof of our theorem by showing our conjecture for
all solutions, which come from intersections of the recurring sequences
obtained with the initial values described in Lemma 7.
Case 1.1) v2m = w2n , z0 = z1 , |z0 | = 1.
Observe that we get by equation (3) and (4), x20 = y02 = 1 and as they are
positive by Lemma 1 we conclude x0 = y0 = 1. Therefore, by Lemma 2
A+C
deg vm = C + (m − 1)
B+C
deg wn = C + (n − 1)
if B < C or z0 = 1 and
A+C
deg vm = C + (m − 1)
B+C
A+B
deg wn =
+ (n − 1)

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

11

if B = C and z0 = −1. The only fact one has to be aware of is that by
(c − s)(c + s) = c2 − s2 = c2 − ac − 1
we can conclude that deg(c − s) = C and analogously we get deg(c − t) = C
if B = C, since
if A ≤ B < C, while deg(c − s) = B and deg(c − t) = A+B
in this case we have c = a + b + 2r.
As in [5, formula (29)], we get from Lemma 6
(14)

±am2 + sm ≡ ±bn2 + tn

(mod 4c).

Assume that m, n 6= 0 (because m = n = 0 leads to the trivial solution
d = 0).
a) Let B < C. Then (14) implies that ±am2 + sm = ±bn2 + tn. This
further implies A+C
= B+C
and A = B. But this means that m = n (by
comparing degrees in v2m = w2n ), ±m(a − b) = t − s and by the equation
m2 (a − b)(t + s) = ±m(t − s)(t + s) = ±mc(b − a)
we get ±m(t + s) = c. The comparison of the degrees gives B+C
= C, a
contradiction.
b) Let A < B = C. Then c = a + b + 2r, which yields s = a + r and
t = b + r. Therefore (14) implies
a(±m2 ± n2 + n + m) = r(∓2n2 − n − m).
Since a and r have different degrees, both sides of this equation are equal
to 0. This implies (m, n) = (0, 0), which yields d = 0, or (m, n) = (1, 1),
which yields d = 4r(a + r)(b + r) = a + b + c + 2abc + 2rst = d+ since
z0 = −1, x0 = 1, thus v1 = c − s = b + r and finally
z = v2 = w2 = 2sv1 − v0 = 2(a + r)(b + r) + 1,
leading to the d as claimed above.
c) Let A = B = C. Then m = n and c = a + b + 2r. Therefore (14)
becomes
(±m2 + m)(b − a) ≡ 0 (mod c).
This implies (for ±m2 + m 6= 0, otherwise we would have d = 0 or d = d+ ,
namely for m = 0 or m = 1, respectively, with the same arguments as above)
that there exist integers k, p, q (k 6= 0) such that
pb − qa = 2kr.
We may assume that p 6= 0, since otherwise from q 2 a2 = 4k 2 (ab + 1) we
would obtain that a and b are constant polynomials, a contradiction. With
v = pq + 2k 2 , we obtain
(15)

p2 b2 − 2vab + a2 q 2 = 4k 2 .

We have v 2 − p2 q 2 = (2k(m2 ± m))2 . Since the discriminant of the quadratic
polynomial f (x) = p2 x2 − 2vx + a2 is a perfect square, we can factorize the
left hand side of (15). With l = 2k(m2 ± m) we have
(16)

(p2 b − va − la)(p2 b − va + la) = 4k 2 p2 .

12

ANDREJ DUJELLA AND CLEMENS FUCHS

We conclude that the both factors on the left hand side of (16) are constant.
Since l 6= 0, we obtain that a is constant. But then (16) implies that b is
also a constant, and we obtained a contradiction as before.
This finishes the proof in the case 1.1).
In all other cases we may assume that B < C. Indeed, if B = C, then
c = a + b + 2r < 4b and it implies cr − st = 1. But the case z0 = z1 =
±1 was already handled in case 1.1). In all remaining cases we obtain a
contradiction. If z0 = ±t, then deg t = deg(b+r) = C > deg z0 , contradicting
Lemma 1. If z1 = ±s, then from z12 = ac + 1 and deg z1 ≤ C/2 we conclude
that A = 0. We also have y1 = r. Consider now the relation
2(t − 1)y12 ≤ b(c − b)
(see the proof of Lemma 1 from [4]). Since deg(2(t−1)r2 ) = 2B and deg(b(c−
b)) = deg(b(a + 2r)) = 23 B, we obtain a contradiction again.
Now, we can consider the remaining cases.
Case 1.2) v2m = w2n , z0 = z1 = ±(cr − st).
First let us assume that we have A = B. From Lemmas 2 and 8 we get
that
( 3C−B )
A+C
deg vm =
+ (m − 1)
B+C
( 3C−A )
B+C
deg wn =
+ (n − 1)
A+C

where the first case in each of these formulae correspond to the first possible
sign, i.e. to z0 = z1 = cr − st and the second case to the minus sign, i.e. to
z0 = z1 = −(cr − st). We will carry on using this notation below.
Now comparing degrees, which means to consider deg vn = deg wn , implies
m = n. Moreover, from Lemma 6 we get by observing that x0 = rs − at and
y1 = rt − bs (cf. [7, p. 28]) that
∓astm(m ± 1) + rm ≡ ∓bstn(n ± 1) + rn

(mod c)

and by multiplying with 2st we obtain
(17)

∓2 [am(m ± 1) − bn(n ± 1)] ≡ 2rst(n − m)

(mod c).

Since m = n we conclude
∓2m(m ± 1)(a − b) = 0,
which can only hold for (m, n) = (0, 0), which leads to d = d− < c, or
(m, n) = (1, 1). The last case leads to the only solutions which is allowed,
namely d = d+ = a + b + c + 2abc + 2rst, since z = v2 = w2 = 2sv1 − v0 =
cr + st.

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

13

Thus, we may assume that A < B < C. From Lemmas 1 and 8 we
conclude that C ≤ 2A + B. Using the equation
d+ · d− = (c − a − b − 2r)(c − a − b + 2r),
which yields deg c = deg(abd− ), we therefore get deg d− ≤ A.
Now we are intended to show that the leading coefficients of b and e are
equal. We have
e = 2rst − 2cr2 + c and d− = a + b + c + 2abc − 2rst.
From this it follows
d− = a + b + c + 2abc − (e + 2cr2 − c) = a + b − e
and thus the conclusion follows since deg d− ≤ A < B = deg e.
Now from (17) we conclude
∓2[am(m ± 1) − bn(n ± 1)] = e(n − m)
and by comparing the leading coefficients we get
±2n(n ± 1) = n − m
or
m = n ∓ 2n(n ± 1) =

−n(2n + 1)
n(2n − 1)

Both cases can hold with m = n = 0, leading to d = d− . The first case can
only hold in this situation. From deg v2n = deg w2n we conclude
m=n

B+C
A+C

and therefore we get in the second case that
(18)

2n − 1 =

B+C
2C
= 2.
A+C

Therefore, we have n < 2 and the only remaining possibilities are (m, n) =
(0, 0) and (m, n) = (1, 1), which lead to d = d− and d = d+ .
This finishes the proof in the case 1.2).
Case 2) v2m+1 = w2n , z0 = ±t, z1 = ±(st − cr).
We apply Lemma 1 and Lemma 8 and conclude
3C − A
B+C
A+B
3C − B
deg(±(st − cr)) = C −
deg(±t) =

=⇒ C ≥ A + 2B,
=⇒ C ≤ 2A + B.

But A + 2B ≤ C ≤ 2A + B, implies that A = B and C = A + 2B = 3A.

14

ANDREJ DUJELLA AND CLEMENS FUCHS

We get from (3) that x0 = r. Therefore, for the degrees of vm and wn we
have by Lemma 2 and Lemma 8:
A+C
A+B
deg vm = C ±
+ (m − 1)
( A+C2 )
B+C
deg wn =
+ (n − 1)
3C−A

By comparing degrees in vm = wn we get
4A + 2A(m − 1) = 2A + 2A(n − 1) =⇒ n = m + 1, or
2A + 2A(m − 1) = 4A + 2A(n − 1) =⇒ m = n + 1.
Now we use Lemma 6 and get
sz0 + c[2asz0 m(m + 1) + x0 (2m + 1)] ≡ z1 + 2c(bz1 n2 + ty1 n) (mod 4c2 ).
Observe again that y1 = rt − bs > 0 and that x0 = r. By diving through c
and multiplying with 2st we conclude
(19)

∓[2am(m + 1) − 2bn(n ∓ 1)] ≡ 2rst(m − n + δ)

(mod 4c),

where δ = 1 or 0. Applying Lemma 10, we see that 2rst ≡ e = ∓2r (mod c).
Observe that A = B = deg r. Thus, we get
2am(m + 1) − 2bn(n ∓ 1) = 2r(m − n + δ)
or
2m(m + 1)(a − b) = 0 or
2(n + 1)(n + 2)a − 2n(n + 1)b = 2r.
The first case implies either a = b, a contradiction, or m = 0 leading to
d = 0. In the second case we define the integers
p = 2(n + 1)(n + 2), q = 2n(n + 1)
and receive the same contradiction as in the case 1.1).
This finishes the proof in the case 2).
Case 3) v2m = w2n+1 , z0 = ±(cr − st), z1 = ∓s.
We start again by applying Lemma 1 and Lemma 8 and we conclude
2A + B ≤ C ≤ A + 2B.
As above (compare also with equation (48) in [5]) we conclude from
Lemma 6
(20)

∓2[am(m ± 1) − bn(n + 1)] = e(n − m + δ),

where δ = 0 or 1.
First, let us assume that C = A + 2B. Hence, by Lemma 10 we get
e = ∓2r. By comparing degrees in (20), we conclude that either m = n = 0,
which leads to a contradiction, or A = B. We get from (4) that y1 = r.

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

15

Therefore, for the degrees of vm and wn we have by Lemma 2 and Lemma
8:
( 3C−B )
A+C
deg vm =
+ (m − 1)
B+C

A+B
B+C
deg wn = C ∓
+ (n − 1)
By comparing the coefficients in vm = wn we get similarly as before
n = m + 1, or m = n + 1.
As in case 2) we derive by inserting this relation in (20) that
0 > am(m + 1) − b(m + 1)(m + 2) = r > 0,
which is a contradiction, or
2[a(n + 1)n − bn(n + 1)] = 0,
which implies n = 0 or a = b and thus also a contradiction.
We are left with the case C < A + 2B. Since we have C ≥ 2A + B, we
conclude that A < B must hold. Moreover, we derive from the equation
d+ · d− = (c − a − b − 2r)(c − a − b + 2r)
that deg d− < B. Since by Lemma 9 we have deg e = B we get as in case 1.2)
that b and e have the same leading coefficients. By comparing the leading
coefficients in (20) we get
±2n(n + 1) = n − m + δ
or
m = n ∓ 2n(n + 1) + δ =

−n(2n + 1)
(2n + 1)(n + 1)

As in case 1.2) the first case is only possible for (m, n) = (0, 0) leading to
d = d− . By comparing the degrees of v2m and w2n+1 we get in the second
case that m(A + C) = n(B + C) and thus
2n + 1 =

B+C
2C
= 2,
n+1
A+C

a contradiction.
Therefore the proof in case 3) is also finished.
Case 4) v2m+1 = w2n+1 , |z0 | = t, |z1 | = s.
Our starting point is the following relation (compare with equation (38)
in [5]), which is a consequence of Lemma 6:
(21) ±2astm(m + 1) + r(2m + 1) ≡ ±2bstn(n + 1) + r(2n + 1)
where we again have divided through c.

(mod 4c),

16

ANDREJ DUJELLA AND CLEMENS FUCHS

First, we again get from (3) and (4) that x0 = y1 = r. Therefore, for the
degrees of vm and wn we have:
A+B
A+C
deg vm = C ±
+ (m − 1)
A+B
B+C
deg wn = C ±
+ (n − 1)
From the comparison of degrees we have (A + C)(m − 1) = (B + C)(n − 1).
If A = B, then m = n and (21) becomes m(m + 1)(a − b) ≡ 0 (mod 2c), a
contradiction. Thus A < B.
We compare the degree of |z0 | = t with the estimate for deg z0 in Lemma
1. Therefore, we can conclude that C ≥ A + 2B.
As in [5], (21) can be rearranged in the following form
(22)

±2[am(m + 1) − bn(n + 1)] ≡ 2rst(n − m)

(mod 2c).

Let e = 2rst − 2cr2 + c. We have 2rst ≡ e (mod c) and deg e < C by Lemma
9. Thus (22) implies
(23)

±2[am(m + 1) − bn(n + 1)] = e(n − m).

If C > A + 2B, then deg e = C − A − B (by Lemma 9) and the comparison
of degrees in (23) gives C = A + 2B, a contradiction.
Hence, we have C = A+2B. But, by Lemma 10 we conclude that e = ∓2r.
Hence, deg e = A+B
< B, which is a contradiction to equation (23).
This finishes the proof in the case 4).
Altogether the statement of Theorem 1 follows.

References
[1] J. Arkin, V. E. Hoggatt and E. G. Strauss, On Euler’s solution of a problem of
Diophantus, Fibonacci Quart. 17 (1979), 333-339.
[2] A. Baker and H. Davenport, The equations 3x2 − 2 = y 2 and 8x2 − 7 = z 2 , Quart.
J. Math. Oxford Ser. (2) 20 (1969), 129-137.
[3] Diophantus of Alexandria, Arithmetics and the Book of Polygonal Numbers, (I.
G. Bashmakova, Ed.) (Nauka, 1974) (in Russian), 85-86, 215-217.
[4] A. Dujella, An absolute bound for the size of Diophantine m-tuples, J. Number
Theory 89 (2001), 126-150.
[5] A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew.
Math., to appear.
[6] A. Dujella and C. Fuchs, A polynomial variant of a problem of Diophantus and
Euler, Rocky Mount. J. Math., to appear.
[7] A. Dujella, C. Fuchs and R. F. Tichy, Diophantine m-tuples for linear polynomials, Period. Math. Hungar. 45 (2002), 21-33.
[8] A. Dujella and F. Luca, On a problem of Diophantus with polynomials, preprint.
[9] P. Gibbs, A generalised Stern-Brocot tree from regular Diophantine quadruples,
preprint, math.NT/9903035.
[10] B. W. Jones, A variation of a problem of Davenport and Diophantus, Quart. J.
Math. Oxford Ser.(2) 27 (1976), 349-353.
[11] B. W. Jones, A second variation of a problem of Davenport and Diophantus, Fibonacci Quart. 15 (1977), 323-330.

POLYNOMIAL VERSION OF A PROBLEM OF DIOPHANTUS

Andrej Dujella
Department of Mathematics
University of Zagreb
Bijenička cesta 30
10000 Zagreb, Croatia
e-mail:
[email protected]
Clemens Fuchs
Institut für Mathematik
TU Graz
Steyrergasse 30
A-8010 Graz, Austria
e-mail:
[email protected]
17
References (11)
J. Arkin, V. E. Hoggatt and E. G. Strauss, On Euler's solution of a problem of Diophantus, Fibonacci Quart. 17 (1979), 333-339.
A. Baker and H. Davenport, The equations 3x 2 -2 = y 2 and 8x 2 -7 = z 2 , Quart. J. Math. Oxford Ser. (2) 20 (1969), 129-137.
Diophantus of Alexandria, Arithmetics and the Book of Polygonal Numbers, (I. G. Bashmakova, Ed.) (Nauka, 1974) (in Russian), 85-86, 215-217.
A. Dujella, An absolute bound for the size of Diophantine m-tuples, J. Number Theory 89 (2001), 126-150.
A. Dujella, There are only finitely many Diophantine quintuples, J. Reine Angew. Math., to appear.
A. Dujella and C. Fuchs, A polynomial variant of a problem of Diophantus and Euler, Rocky Mount. J. Math., to appear.
A. Dujella, C. Fuchs and R. F. Tichy, Diophantine m-tuples for linear polyno- mials, Period. Math. Hungar. 45 (2002), 21-33.
A. Dujella and F. Luca, On a problem of Diophantus with polynomials, preprint.
P. Gibbs, A generalised Stern-Brocot tree from regular Diophantine quadruples, preprint, math.NT/9903035.
B. W. Jones, A variation of a problem of Davenport and Diophantus, Quart. J. Math. Oxford Ser.(2) 27 (1976), 349-353.
B. W. Jones, A second variation of a problem of Davenport and Diophantus, Fi- bonacci Quart. 15 (1977), 323-330.
July 08, 2025
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Andrej Dujella
International Journal of Number Theory, 2008
In this paper, we prove that there does not exist a set of 11 polynomials with coefficients in a field of characteristic 0 with the property that the product of any two distinct elements plus 1 is a perfect square. Moreover, we prove that there does not exist a set of 5 polynomials with the property that the product of any two distinct elements plus 1 is a perfect kth power with k ≥ 7. Combining these results, we get an absolute upper bound for the size of a set with the property that the product of any two elements plus 1 is a pure power.
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