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Given the recurrence $$a_{n+2} = 14a_{n+1} - a_n - 6$$ with $a_1=1$ and $a_2=8$, how do I prove that none of the $a_n$'s apart from $a_1$ is a perfect square. This is not a homework problem, rather part of a small side project.

PS: I am aware that we can obtain a closed form solution for the recurrence, however that doesn't seem to help me answer the question, i.e., it is equivalent to proving that $\dfrac{(2+\sqrt3)^{2k-1}-(2-\sqrt3)^{2k-1}}{4\sqrt3} + \dfrac12$ is not a perfect square except for $k=1$.

EDIT: Elementary quadratic residue based arguments for small primes enable me to obtain that $n \equiv a \pmod b$, but doesn't seem to help me in ruling out all possible cases except $n=1$.

EDIT: To give the problem more context, I was trying to prove that the Mordell's equation $y^2=x^3+1$ has the solutions only as $(-1,0)$, $(0,\pm1)$ and $(2,\pm3)$. After some algebra, you need to find integer solutions to $2m^3-3m^2+1 = n^2$, which in turn results in the above claim.

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  • $\begingroup$ This recurrence can be solved (using Z transform is one method). Have you tried doing that? $\endgroup$
    – Paul
    Apr 9 '15 at 21:23
  • $\begingroup$ @Paul But how does that help me in identifying the squares. $\endgroup$
    – Adhvaitha
    Apr 9 '15 at 21:24
  • $\begingroup$ Don't know - let's see! $\endgroup$
    – Paul
    Apr 9 '15 at 21:25
  • $\begingroup$ @Mercy Yes, So? We want to show that none of these are perfect squares except for $a_1=1$. $\endgroup$
    – Adhvaitha
    Apr 9 '15 at 21:34
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    $\begingroup$ If it's you who come up with this sequences, it's not very easy to find squares in non trivial "sequences" (without congruence relations). For example there is only three squares in Fibonacci numbers and the proof is not very easy, and we don't have any strong methods to solve this problems other then congruences $\endgroup$
    – Elaqqad
    Apr 9 '15 at 21:35
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The recursion preserves the quadratic form $$ Q(x_n, x_{n+1}) = {x_{{n}}}^{2}-14\,x_{{n+1}}x_{{n}}+{x_{{n+1}}}^{2}+6\,x_{{n}}+6\,x_{{n +1}} $$ In this case $Q(1,8) = 7$. Now the solutions of $Q(s,t) = 7$ are $t = 7s - 3 \pm 4 \sqrt{3s^2 - 3s + 1}$. If $s$ and $t$ are integers, $3s^2 - 3 s + 1$ must be a square. That is, $3 a_n^2 - 3 a_n + 1$ is a square. With $a_n = x^2$, we are led to consider the curve $$ 3 x^4 - 3 x^2 + 1 - y^2 = 0$$ which Maple says has genus $1$, so it's elliptic.

Well, are there integer points other than $[0,\pm 1]$ and $[\pm 1, \pm 1]$ on this curve?

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  • $\begingroup$ Thanks for this. Though I was actually trying to prove that there are no other integer solutions to the above curve, using the fact that there are no non-trivial squares in this recurrence. $\endgroup$
    – Adhvaitha
    Apr 9 '15 at 22:54
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    $\begingroup$ Oddly enough, I was about to post an answer with the exact same equation ($y^2=3x^4-3x^2+1$) based on perusing the OEIS entry oeis.org/A001922 that I linked to in a comment beneath the OP. $\endgroup$ Apr 9 '15 at 22:58
  • $\begingroup$ Could you let me know how you came up with the quadratic form or in general, how do I obtain the preserved QF given a recurrence. $\endgroup$
    – Adhvaitha
    Apr 9 '15 at 23:48
  • $\begingroup$ Once you guess that there is such a form, let $Q(s,t) = c_{11} s^2 + c_{12} s t + c_{22} t^2 + c_1 s + c_2 t$, expand out $Q(s,t) - Q(t, 14 t - s -6)$ and solve linear equations for the coefficients. $\endgroup$ Apr 10 '15 at 0:27

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