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Let $(a_n)$ a linear recurrence sequence

$$a_{n+d}= \sum_{k=1}^d c_k a_{n+d-k}$$

for all $n\ge 0$, ($c_1$, $\ldots$, $c_d$ are fixed integers).

Assume that $a_n$ is a perfect square for all $n\ge 0$.

Then there exists an integral linear recurrence sequence $(b_n)$ such that $a_n = b_n^2$

Notes: There are questions where $a_n$ is a given integral recurrence sequence and it is required to show that $a_n$ is a perfect square for all $n$. One approach is to produce a sequence $b_n$ such that $b_n^2 = a_n$. Moreover, it appears that $(b_n)$ is also a recurrence sequence.

The result is true if $a_n = P(n)$ where $P$ is a polynomial ( classic result).

Any feedback would be appreciated!

$\bf{Added:}$ Unsurprisingly, this turns out to be a particular case of an old problem, the "Hadamard root problem". I figure it might not originate with Hadamard, rather it has to do with "Hadamard square". Knowing that the square of a linear recurrence sequence ( the Hadamard square) is again a recurrence sequence, the question is going the other way. Links to original sources in van de Poorten et al. p. 69.

orangeskid
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  • The only proof I know for the classical case of $a_n = P(n)$ is the number-theoretic one (e.g., https://math.stackexchange.com/a/306685/1048496), however it does not have any chances to be applicable to the general linear recurrences (probably when all the roots of characteristic polynomial are integer?) . Are there other proofs? – Pavel Gubkin Feb 09 '23 at 09:07
  • @Pavel Gubkin: If $(b_n)_{n\ge 0}$ is a sequence of integers and $(\Delta^k b)_n \to 0$ as $n \to \infty$ then $(\Delta^k b)_n = 0$ for $n >>0$ ( $\Delta$ is a finite difference operator). Now apply this for $b_n = \sqrt{a_n}$. – orangeskid Feb 09 '23 at 17:50

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