D(4)-Triples with Two Largest Elements in Common

Definition 1.

Let n ≠ 0 be an integer. We call a set of m distinct positive integers a D(n)-m-tuple, or m-tuple with the property D(n), if the product of any two of its distinct elements increased by n is a perfect square.

Conjecture 1.

Suppose that {a₁, b, c} and {a₂, b, c} are D(4)-triples with a₁ < a₂ < b < c. Then, {a₁, a₂, b, c} is a D(4)-quadruple.

Conjecture 2.

Suppose that {a₁, b, c, d} is a D(4)-quadruple with a₁ < b < c < d. Then, {a₂, b, c, d} is not a Diophantine quadruple for any integer a₂ with a₁ ≠ a₂ < b.

Theorem 1.1.

Suppose that {a, b, c} is a D(4)-triple, a ≠ 3. Then, {a+1, b, c} is not a D(4)-triple.

Proof of this theorem will be separated in two cases in Section 3, the first case will be proven by using the hypergeometric method and the second case by using linear forms in logarithms.

We further support the validity of conjectures by proving the next results.

Theorem 1.2.

If c < 0.25b³, then Conjecture 1 holds.

Corollary 1.

If either c < 0.25b³ or c ≥ 39247b⁴, then Conjecture 2 holds.

Lemma 2.1.

Let (z, x) and (z, y) be positive solutions of (2.3) and (2.4). Then there exist solutions (z₀, x₀) of (2.3) and (z₁, y₁) of (2.4) in the ranges

such that

where m and n are nonnegative integers.

Assumption 1.

At least one of {a, b′, b} and {a + 1, b′, b} is not a D(4)-triple for any b′ with 0 < b′ < b.

Since we are searching for intersections of sequences (2.5) and (2.6), we can describe the initial terms of sequences more precisely. We omit the proof since the result is proven similarly as [6: Lemma 2.3] by following cases from [11: Lemma 4].

Lemma 2.2.

If the equation v_m = w_n has a solution, then both m and n are even and z₀ = z₁ = 2ε, where ε ϵ {±1}.

Remark 1.

Assumption 1 is crucial for the proof of Lemma 2.2. If a = 3, then (2.1) would also have a fundamental solution (t₀, s₀) = (0, 1). In this case, for every b=bν=(sν2−4)/a , where s_ν is defined as in (2.2), there would exist b′ < b which arises from the sequence with this second fundamental solution.

From z₀ = z₁ = 2ε we have x₀ = y₁ = 2 so sequences (2.5) and (2.6) can be written in the form

Lemma 2.3.

Lemma 2.4.

If v_m = w_n has a solution, then n≤m≤32n+1 .

Proof. We follow the proof of [11: Lemma 5]. By using that b ≥ b₁ ≥ 96a ≥ 96, we get estimates for v_m and w_n which yield a double inequality

Also, (ab+4−1)3=(s−1)3>t2=(a+1)b+4 , so we conclude

Similarly, by observing

we get n − 1 ≤ m. Now the desired inequality follows from Lemma 2.2.

Lemma 2.5.

If v_m = w_n has a solution with m ≥ 2, then m > n.

Proof. We will prove by induction that v_n < w_n for n ≥ 2, so if v_m = w_n, then m > n for m ≥ 2.

Let n = 2. We first observe the case ε = 1, where x₀ = y₁ = 2 so v₂ = b(a + s) + 2 < b(a + 1 + t) + 2 = w₂. On the other hand, if ε = −1, first we observe that s + 1 < t, since s + 1 ≥ t is in a contradiction with b ≥ b₁ = 64a + 32. Using that inequality, we have v₂ = b(s – a) – 2 < b(t − 1 – a) − 2 = w₂.

Let us assume that v_n−1 < w_n−1 for some n ≥ 3. Since s ≤ t – 2 and sequence w_n is increasing, we have

which proves our statement.

Lemma 2.6.

If v_m = w_n has a solution with m > 0, then

Proof. Since v_m = w_n, from Lemma 2.3 we get that congruence

holds.

Suppose to the contrary that m ≤ 0.4672(a + 1)^−1/2b^1/2. Observe that

and

since (a + 1)b ≥ (a + 1)b₁ ≥ 192. This implies that congruence (2.10) is an equality. After multiplying with tn + sm, we get

Since m > n, both sides of equation are positive. Also, by using Lemmas 2.2 and 2.4 on equation (2.11), we get

must hold. Then

Since a ≥ 1 and b ≥ b₁ = 64a + 32 ≥ 96, we get

a contradiction.

Lemma 2.7.

If m ≥ 1, then z=vm>(s+ab2)m .

The next lemma can be proven by following the idea of [6: Lemma 4.1].

Lemma 2.8.

If v_m = w_n has a solution with m ≥ 2, then

Theorem 3.1.

Let a be a positive integer and N a multiple of a(a + 1). Assume that N ≥ 270a(a + 1)². Then the numbers θ1=1+4a/N and θ2=1+4(a+1)/N satisfy

for all integers p₁, p₂, q with q > 0, where

Lemma 3.1.

(cf. [11: Lemma 14]). Let N = a(a + 1)b and let θ₁, θ₂ be as in Theorem 3.1. Then, all positive solutions to the system of Pellian equations (2.3) and (2.4) satisfy

Proof of Theorem 1.1 in the case b ≥ b₂

Let us assume that b ≥ b₂ = 1024a³ + 1536a² + 704a + 96 and that Assumption 1 holds. We can apply Theorem 3.1 with N = a(a + 1)b, p₁ = (a + 1)sx, p₂ = aty and q = a(a + 1)z and Lemma 3.1 to show that the next inequality holds

Inserting expression for λ and approximating λ < 2 on the right hand side of the previous inequality yields

Combining Lemmas 2.6 and 2.7 with inequality (3.1) implies

Since b ≥ b₂ ≥ 1024a²(a + 1) > 1024a³ and the right-hand side of the inequality is decreasing in b, we get an inequality

If a > 5 (i.e. a ≥ 6), applying a + 1 < 1.2a to the previous inequality and solving for a returns a ≤ 5, a contradiction. On the other hand, for a ≤ 5, we can insert values for a and b₃ in the inequality (3.2) and see it cannot hold, which means it remains to consider only the pairs (a, b₂), 1 ≤ a ≤ 5.

Since b achieves its maximum for a = 5, we can use it in the inequality from the proof of [10: Theorem 1]

to get m < 4.3 · 10¹⁹. Also, from Lemma 2.6, we have m>0.4672b2/(a+1)>0.4672⋅32a>14 . Now we can use the Baker-Davenport reduction method on the remaining pairs as described in [8], and each case returned the bound m < 7, which is a contradiction.

Lemma 3.2.

If v_m = w_n has a solution with m ≥ 2 and b = b₁ = 64a + 32, then

where ν = m – n.

Theorem 3.2.

([15: Corollary 2]). Assume that α₁ and α₂ are real, positive and multiplicatively independent algebraic numbers in a field K of degree D. Set

where b₁ and b₂ are positive integers. Let A₁ and A₂ be real numbers greater than one such that

Set

Then,

Proof of Theorem 1.1 in the case b = b₁

Let us denote ν = m – n and rewrite Λ as follows

Let

Multiplicative independence of α₁ and α₂ over ℚ(ab,(a+1)/b) can be verified similarly as in [14: Lemma 19], so the linear form in logarithms Λ with these parameters satisfies the hypothesis of Theorem 3.2.

We have

By observing conjugates of γ whose absolute values are greater than one (see [6]) and noting that the leading coefficient of the minimal polynomial of γ is divisor of a²(b – a – 1)², we estimate

This implies

where we have used that ab ≥ 96 and ν ≥ 2. Since γ ≤ 2, we also have

hence we may take

Now

By using b ≥ 96, we can estimate that β < 1.43α, hence log α + log β < log(1.43α²) < 2.16 log α. From Theorem 3.2 and (2.14) it follows

If log1.16m2(ν+2)log α>5.25 , the inequality (3.3) implies that

In the other case, when log1.16m2(ν+2)log α≤5.25 the same inequality holds. By combining Lemma 3.2 and (3.4), we obtain

Therefore it remains to verify only finitely many D(4)-pairs {a, b₁}, 1 ≤ a ≤ 18072. Three steps of the Baker-Davenport reduction method ended with the bound m < 2, which is a contradiction.