Proof of Theorem 1.1

We start with some basic properties of our sequences. For a complex number𝑧 we write𝑧 for its complex conjugate. Let𝜔̸= 1be a cubic root of1. Put can be proved, by induction for example, that the Binet formulas

𝐹𝑛 =𝛼^𝑛−𝛽^𝑛

The first formula in (3.1) is well known. The second one follows from the general theorem on linear recurrence sequences since the above polynomial is the charac-teristic polynomial of the Padovan sequence. Further, the inequalities

𝛼^𝑛−26𝐹𝑛 6𝛼^𝑛−1, 𝛾^𝑛−36𝑃𝑛6𝛾^𝑛−1 (3.2) also hold for all𝑛>1. These can be proved by induction. We note that

𝛾= 1.32471. . . , |𝛿|= 0.86883. . . , 𝑐1= 0.54511. . . , |𝑐2|= 0.28241. . . , and

𝛼= 1.61803. . . , |𝛽|= 0.61803. . .

Now we start with the study of our equation (1.3) in non-negative integers (𝑛, 𝑚, 𝑛1, 𝑚1) with (𝑛, 𝑚) ̸= (𝑛1, 𝑚1) where, as we have said, 𝑛, 𝑛1 ̸= 1,2,4, 𝑚, 𝑚1 ̸= 1. We note, if 𝑚 = 𝑚1 then 𝑃𝑛 = 𝑃𝑛1 which implies 𝑛 = 𝑛1, a con-tradiction. Thus, we assume that𝑚 > 𝑚1. Rewriting equation (1.3) as

𝑃𝑛−𝑃𝑛1=𝐹𝑚−𝐹𝑚1 (3.3)

we observe the right-hand is positive. So, the left-hand side is also positive and therefore,𝑛 > 𝑛1. Now, we compare both sides of (3.3) using (3.2). We have

where the inequality at the right-hand side is clear for both 𝑚1 = 0and 𝑚1̸= 0.

Thus,

(𝑛−8)log𝛾

log𝛼 6𝑚−1 and (𝑛−1)log𝛾

log𝛼 >𝑚−4. (3.4) Sincelog𝛾/log𝛼= 0.584357. . . we have that if𝑛6540then𝑚6318. A brute force search withMathematica in the range06𝑛1< 𝑛6540,06𝑚1< 𝑚6318, with our conventions, we obtained all solutions listed in Theorem 1.1.

From now on, we assume that𝑛 >540.Thus, from (3.4), we have that𝑚 >311 and also that𝑛 > 𝑚. From Binet’s formula (3.1), we rewrite our equation as

⃒⃒

⃒⃒𝑐1𝛾^𝑛−𝛼^𝑚

√5

⃒⃒

⃒⃒62|𝑐2||𝛿|^𝑛+ 1

√5+𝛾^𝑛1−1+𝛼^𝑚1−1<max{𝛾^𝑛1+6, 𝛼^𝑚1+4}. Dividing through by𝛼^𝑚/√

5 we get

⃒⃒

⃒√

5𝑐1𝛾^𝑛𝛼^−𝑚−1⃒⃒⃒<max{𝛾^{𝑛1−𝑛+16}, 𝛼^{𝑚1−𝑚+6}}, (3.5) where we have used𝛾^𝑛−86𝛼^𝑚−1,√

5< 𝛼𝛾²and√

5< 𝛼². LetΛbe the expression inside the absolute value in the left-hand side of (3.5). Observe thatΛ̸= 0. To see this, we consider the Q-automorphism 𝜎 of the Galois extension K := Q(𝛼, 𝛾, 𝛿) overQdefined by 𝜎(𝛾) :=𝛿, 𝜎(𝛿) :=𝛾 and𝜎(𝛼) :=𝛼. We note that𝜎(𝛿) =𝛿 and 𝜎(𝛽) =𝛽. IfΛ = 0then𝜎(Λ) = 0 and we get

𝛼^𝑚

√5 =𝜎(𝑐1𝛾^𝑛) =𝑐2𝛿^𝑛. Thus,

𝛼^𝑚

√5 =|𝑐2||𝛿|^𝑛 <1,

which is absurd since𝑚 >311. So,Λ̸= 0. We apply Matveev’s inequality toΛby taking

𝛼1=√

5𝑐1, 𝛼2=𝛾, 𝛼3=𝛼, 𝑏1= 1, 𝑏2=𝑛, 𝑏3=−𝑚.

Thus, 𝐵 = 𝑛. Further, ℎ(𝛼2) = log𝛾/3, ℎ(𝛼3) = log𝛼/2. For 𝛼1 we use the properties of the height to conclude

ℎ(𝛼1)6log𝛾+ 7 log 2.

So we take𝐴1= 30.8,𝐴2= 0.57,𝐴3= 1.45. From Matveev’s inequality we obtain log|Λ|>−𝐶(1 + log𝑛)·30.8·0.57·1.45>−3.66336×10¹⁴(1 + log𝑛), which, compared with (3.5) we obtain

min{(𝑛−𝑛1) log𝛾,(𝑚−𝑚1) log𝛼}63.66337×10¹⁴(1 + log𝑛).

Now we study each one of these two possibilities.

Case 1. min{(𝑛−𝑛1) log𝛾,(𝑚−𝑚1) log𝛼}= (𝑛−𝑛1) log𝛾.

In this case, using Binet’s formulas (3.1), we rewrite our equation as

⃒⃒ LetΛ1be the expression inside the absolute value in the left-hand side of (3.6). We note that Λ1 ̸= 0. For if not, we apply the above𝜎 to it and we have𝜎(Λ1) = 0.

Thus,

𝛼^𝑚

√5 =|𝜎(𝑐1)(𝛿^𝑛−𝛿^𝑛1)|62|𝑐2|<1,

which is absurd since𝑚 >311. We apply Matveev’s inequality toΛ1 and for this we take

𝛼1=√

5𝑐1(𝛾^𝑛−𝑛1−1), 𝛼2=𝛾, 𝛼3=𝛼, 𝑏1= 1, 𝑏2=𝑛1, 𝑏3=−𝑚.

We have𝐵 =𝑛. The heights of𝛼2 and 𝛼3 are already calculated. For 𝛼1 we use the height properties and we get

ℎ(𝛼1)6 3.66338×10¹⁴(1 + log𝑛)

3 .

Thus, we can take 𝐴1 = 7.32676×10¹⁴(1 + log𝑛) and 𝐴2, 𝐴3 as above. From Matveev’s inequality we obtain

log|Λ1|>−𝐶(1 + log𝑛)·(7.32676×10¹⁴(1 + log𝑛))·0.57·1.45, which compared with (3.6) gives

(𝑚−𝑚1) log𝛼 <8.71446×10²⁷(1 + log𝑛)². Case 2. min{(𝑛−𝑛1) log𝛾,(𝑚−𝑚1) log𝛼}= (𝑚−𝑚1) log𝛼.

To this case, we rewrite our equation as

⃒⃒

where the left-hand side inequality holds since 𝑚 > 311, which is absurd. So, Λ2̸= 0 and we apply Matveev’s inequality to it. To do this, we take

𝛼1=𝛼^𝑚−𝑚1−1

√5𝑐1

, 𝛼2=𝛾, 𝛼3=𝛼, 𝑏1= 1, 𝑏2=−𝑛, 𝑏3=𝑚1.

Thus,𝐵 =𝑛. The heights of𝛼2and𝛼3are already calculated. From the properties of the height for𝛼1we obtain

ℎ(𝛼1)6 3.66338×10¹⁴(1 + log𝑛)

2 .

Thus, we can take 𝐴1 = 1.09901×10¹⁵(1 + log𝑛) and 𝐴2, 𝐴3 as above. Hence, from Matveev’s inequality we obtain

log|Λ2|>−𝐶(1 + log𝑛)·(︀

1.09901×10¹⁵(1 + log𝑛))︀

·0.57·1.45, which compared with (3.7) we get

(𝑛−𝑛1) log𝛾 <1.30717×10²⁸(1 + log𝑛)². So, from the conclusion of the two cases we have that

max{(𝑛−𝑛1) log𝛾,(𝑚−𝑚1) log 2}<1.30717×10²⁸(1 + log𝑛)². Now we get a bound on𝑛. To do this we rewrite our equation as

⃒⃒ inside the absolute value in the left-hand side of (3.8). As above, if Λ3 = 0 we apply the above𝜎 and we obtain𝜎(Λ3) = 0. Then

1< 𝛼^𝑚−1(𝛼−1)

√5 6𝛼^𝑚−𝛼^𝑚1

√5 =|𝑐2(𝛿^𝑛−𝛿^𝑛1)|62|𝑐2|< 2 3,

and as above, we get a contradiction. Thus, Λ3 ̸= 0 and we apply Matveev’s inequality to it. To do this, we take

𝛼1=√

5𝑐1 𝛾^𝑛−𝑛1−1

𝛼^𝑚−𝑚1−1, 𝛼2=𝛾, 𝛼3=𝛼, 𝑏1= 1, 𝑏2=𝑛1, 𝑏3=−𝑚1.

Hence,𝐵 =𝑛. The height of𝛼2 and𝛼3 have already been calculated. For𝛼1 we use the properties of the height to conclude that

ℎ(𝛼1) 6 log𝛾+ (𝑛−𝑛1)log𝛾

3 + (𝑚−𝑚1)log𝛼

2 + 9 log 2

< 6.53586×10²⁸(1 + log𝑛)²

6 .

Thus, we can take𝐴1= 6.53586×10²⁸(1 + log𝑛)²and𝐴2,𝐴3 as above. From Matveev’s inequality we get

log|Λ3|>−𝐶·(︀

(1 + log𝑛)·6.53586×10²⁸(1 + log𝑛)²)︀

·0.57·1.45, which compared with (3.8) yields𝑛 <2.2116×10⁴³(log𝑛)³. Thus, from Lemma 2.3 we obtain

𝑛 <1.75894×10⁵⁰. (3.9)

Now we reduce this upper bound on𝑛. To do this, letΓbe defined as Γ =𝑛log𝛾−𝑚log𝛼+ log(︁√

5𝑐1

)︁,

and we go to (3.5). Assume that min{𝑛−𝑛1, 𝑚−𝑚1} > 20. Observe that 𝑒^Γ−1 = Λ̸= 0. ThereforeΓ̸= 0. IfΓ>0, then

0<Γ< 𝑒^Γ−1 =|Λ|<max{𝛾^{𝑛1−𝑛+16}, 𝛼^{𝑚1−𝑚+6}}.

IfΓ<0, we then have1−𝑒^Γ =|𝑒^Γ−1|=|Λ|<1/2. Thus,𝑒^|Γ|<2and we get 0<|Γ|< 𝑒^|Γ|−1 =𝑒^|Γ||Λ|<2 max{𝛾^{𝑛1−𝑛+16}, 𝛼^{𝑚1−𝑚+6}}.

So, in both cases we have

0<|Γ|<2 max{𝛾^{𝑛1−𝑛+16}, 𝛼^{𝑚1−𝑚+6}}. Dividing through log𝛼we get

0<|𝑛𝜏−𝑚+𝜇|<max {︂ 374

𝛾^𝑛−𝑛1, 75 𝛼^𝑚−𝑚1

}︂

, where

𝜏:= log𝛾

log𝛼, 𝜇:= log(︀√

5𝑐1)︀

log𝛼 .

We apply Lemma 2.2. To do this we take𝑀:= 1.75894×10⁵⁰ which is the upper bound on𝑛by (3.9). With the help ofMathematicawe found that the convergent

𝑝111

𝑞111 = 10550181102903844192795827490150215250922708545039517997 18054337085897707605265391296915471978898809258369491754

of 𝜏 satisfies that𝑞111 >6𝑀 and that 𝜀:=‖𝑞111𝜇‖ −𝑀‖𝑞111𝜏‖ = 0.450294>0.

Thus, by Lemma 2.2 with𝐴:= 374,𝐵:=𝛾or𝐴:= 75,𝐵:=𝛼, we get that either 𝑛−𝑛16476 or 𝑚−𝑚16275.

Now we study each one of these two cases. We first assume that𝑛−𝑛16476 and𝑚−𝑚1>20. In this case, we consider

Γ1=𝑛1log𝛾−𝑚log𝛼+ log(√

5𝑐1(𝛾^𝑛−𝑛1−1))

and we go to (3.6). We see that𝑒^Γ1−1 = Λ1̸= 0. Thus,Γ1̸= 0and, with a similar argument as the previous one we obtain

0<|Γ1|< 2𝛼⁶ 𝛼^𝑚−𝑚1. Dividing through log𝛼we get

0<|𝑛1𝜏−𝑚+𝜇|< 75 𝛼^𝑚−𝑚1, where𝜏 is the same one as above and

𝜇:=log(√

5𝑐1(𝛾^𝑛−𝑛1−1))

log𝛼 .

We note that 𝑛1 > 0, since otherwise we would have 𝑛 6476 which contradicts 𝑛 >540. Thus, we can apply Lemma 2.2. Consider

𝜇𝑘 :=log(√

5𝑐1(𝛾^𝑘−1))

log𝛼 , 𝑘= 1,2, . . . ,476.

With the help of Mathematica we found that the denominator of the 111-th con-vergent above of 𝜏 is such that 𝑞111 > 6𝑀 and 𝜀𝑘 > 0.00129842 > 0 for all 𝑘 = 1,2, . . . ,476. Thus, by Lemma 2.2 with 𝐴 := 75, 𝐵 := 𝛼 we obtain that the maximum value of log(𝑞111·75/𝜀𝑘)/log𝛼, 𝑘 = 1,2, . . . ,476, is less than 287.

Therefore𝑚−𝑚16287.

In a similar way we study the other case. Assume that 𝑚−𝑚1 6 275 and 𝑛−𝑛1>20. In this case we consider

Γ2=𝑛log𝛾−𝑚1log𝛼+ log (︃ √

5𝑐1

𝛼^𝑚−𝑚1−1 )︃

and we go to (3.7). Observe that1−𝑒^−Γ2 = Λ2 ̸= 0. Hence,Γ2 ̸= 0and, with an argument as above we conclude that

0<|Γ2|< 2𝛾⁷ 𝛾^𝑛−𝑛1,

Dividing through bylog𝛼we get

0<|𝑛𝜏−𝑚1+𝜇|< 30 𝛾^𝑛−𝑛1. where𝜏 is as above and

𝜇:= log(︀√

5𝑐1/(𝛼^𝑚−𝑚1−1))︀

log𝛼 .

We note that 𝑚1 > 0. Indeed, for if not, we get 𝑚 6 275 which contradicts 𝑚 >311. Thus, we can apply Lemma 2.2 again. Consider

𝜇ℓ:= log(︀√

5𝑐1/(𝛼^ℓ−1))︀

log𝛼 , ℓ= 1, . . . ,275.

Again, with Mathematica we quickly found that the same 111-th convergent of 𝜏 satisfies 𝑞111 > 6𝑀 and 𝜀ℓ >0.000693865 >0 for allℓ = 1, . . . ,257. Thus, from Lemma 2.2 with 𝐴 := 30, 𝐵 := 𝛾 we obtain that the maximum value of log(𝑞111·30/𝜖ℓ)/log𝛾,ℓ= 1, . . . ,257is6490. Hence,𝑛−𝑛16490.

Summarizing what we have done, we first got that either 𝑛−𝑛1 6 476 or 𝑚−𝑚1 6 257. Assuming the first one we obtained that 𝑚−𝑚1 6 287, and assuming the second one we obtained 𝑛−𝑛1 6490. So, altogether we have that 𝑛−𝑛16490,𝑚−𝑚16287. It remains to study this case.

Consider

Γ3=𝑛1log𝛾−𝑚1log𝛼+ log (︂√

5𝑐1

𝛾^𝑛−𝑛1−1 𝛼^𝑚−𝑚1−1

)︂

,

and we go to (3.8). Note that 𝑒^Γ3−1 = Λ3̸= 0. Thus,Γ3̸= 0and since𝑛 >540 with an argument as before we get

0<|Γ3|< 2𝛾¹⁶ 𝛾^𝑛 . Dividing through bylog𝛼we obtain

𝑜 <|𝑛1𝜏−𝑚1−𝜇|< 374 𝛾^𝑛 , where𝜏 is as above and

𝜇:= log(︀√

5𝑐1(𝛾^𝑛−𝑛1−1/𝛼^𝑚−𝑚1−1))︀

log𝛼 .

As above we note that 𝑛1 and 𝑚1 are positives. We apply Lemma 2.2 again.

Consider

𝜇𝑘,𝑙:= log(︀√

5𝑐1(︀

𝛾^𝑘−1/𝛼^ℓ−1)︀)︀

log𝛼 , 𝑘= 1, . . . ,490 ℓ= 1, . . . ,287.

With Mathematica we find that the same 111-th convergent above of 𝜏 works again. That is, 𝑞111 > 6𝑀 and 𝜀𝑘,ℓ ≥5.28933⁻⁸ > 0 for all𝑘 = 1, . . . ,490 and ℓ = 1, . . . ,287. Thus, by Lemma 2.2 with 𝐴 := 374 and 𝐵 := 𝛾 we obtain that the maximum value of log(𝑞111374/𝜀𝑘,ℓ)/log𝛾, 𝑘= 1, . . . ,490 andℓ = 1, . . . ,287, is 6533. Thus, 𝑛6533 which contradicts our assumption on 𝑛. This completes the proof of Theorem 1.1.

Acknowledgements. We thank the anonymous referee for valuable comments.

The second author thanks Juan Manuel Pérez Díaz for helpful advice and kind support. He also thanks Lidia González García for valuable bibliography support.

This paper started during a visit of the third author to the Universidad Autónoma de Zacatecas, in February 2018. He thanks this Institution for their hospitality.

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