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Good night, everyone!

According to a celebrated result by J. H. Cohn, the only perfect squares in the Fibonacci sequence are $F_{0}=0$, $F_{1}=F_{2}=1$, and $F_{12}=144$. It is also known that the only perfect squares in the Lucas numbers are $L_{1}=1$ and $L_{3}=4$.

Do you know other (interesting) examples of Lucas sequences with few perfect squares among their terms? I am specially interested in examples in which the determination of the squares is more straightforward than Cohn's proof of the aforementioned theorem.

Thanks for your attention.

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    $\begingroup$ You may like to check my paper On Integral Points on Biquadratic Curves and Near-Multiples of Squares in Lucas Sequences although it's more of computational rather than analytical nature. $\endgroup$ Sep 26, 2019 at 2:25
  • $\begingroup$ I don't know what you consider interesting. $2^n-1$ and $2^n+1$ are Lucas sequences with very easy proofs of rarity of squares. $\endgroup$ Sep 26, 2019 at 7:42
  • $\begingroup$ The main branch of the Collatz conjecture $1,5,21,85,\ldots=\frac{4^n-1}3$ is a Lucas sequence which factors into $\frac{(2^n-1)(2^n+1)}{3}$ and since these differ from a perfect square by $1$ I can't see this sequence containing many squares. $\endgroup$ Sep 9, 2020 at 9:54

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I will give a few references which could be considered as some answer to this question. Though the matter of "straight forward" is up to interpretation. Furthermore, I am far from any kind of expert on these sequences. I have written a paper on (some combinatorial aspects of) Lucas sequences and remember some references from when I was reading up on them.

First there are a lot of second order linear reccurences which have few (= finitely many) perfect square terms. In Perfect powers in second order linear recurrences by Pethö a special case of the main theorem says that for a second order linear recurrence $G_n = AG_{n-1} - BG_{n-2}$ there are only finitely many perfect square terms provided some conditions. The conditions are that $A \neq 0$, $|G_0| + |G_1| \neq 0$, $\gcd(A,B) = 1$, $A^2 \neq iB$ for $i \in \{1,2,3,4\}$, and $D$ is not a perfect square if $BC = 0$. Here $C = G_1^2 - AG_0G_1 + BG_1^2$ and $D = A^2 - 4B$. This paper gives some bounds on where perfect squares can arise by Baker’s method, but since Lucas sequences are special one may want to know more.

In The Square Terms in Lucas Sequences by Ribenboim and McDaniel is in shown more precisely where perfect square terms can arise in Lucas sequences under some conditions. Under the conditions that $P$ and $Q$ are odd, $\gcd(P,Q) = 1$, and $D = P^2 - 4Q > 0$ it is shown that

  • if $V_n(P,Q)$ is a perfect square, then $n \in \{1,3,5\}$.
  • if $U_n(P,Q)$ is a perfect square, then $n \in \{0,1,2,3,6,12\}$.

Here the notation matches with the usual notation which is given the OEIS link "Lucas sequences" in the question. We see this result agrees with what is known for $F_n$ and $L_n$. Again I am not an expert, but I think exactly where squares occur depends on particular choices of $P$ and $Q$ and I am not familiar with particular of methods used in various cases. You can find more papers like Lucas sequences whose 12th or 9th term is a square by Bremner and Tzanakis. Looking at these papers with their references + google scholar gives more information on this problem.

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One of the most known Lucas sequences, after the Fibonacci and the Lucas numbers, is the sequence of Pell numbers:
$P_n=\begin{cases}0&\mbox{if }n=0;\\1&\mbox{if }n=1;\\2P_{n-1}+P_{n-2}&\mbox{otherwise.}\end{cases}$

The sequence of Pell numbers starts with $0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378$. It is known that the only squares in this sequence are 1 and 169. Furthemore, there are no higher powers of integers.

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