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Let the integer sequence $n_k$, ($k\ge 0$) be defined as $$ n_0=1$$ $$n_1=64$$ $$ n_k=38 n_{k-1}-n_{k-2}-90$$ How can one find the squares in such a sequence?

Besides $ n_0=1^2, n_1=8^2$, we also have $n_3=298^2$, and is that all?

COMMENT: the problem comes from this question. At the end, it is asked whether the set of solutions $x$ to the equation: $$2 (x^2 - 4) (x^2 - 1) = 45 (y^2 - 1)$$ is $$\{1, 2, 4, 8, 11, 23, 298\}$$

This is equivalent to solving $$X^2 - 90 y^2 = -81$$ with $$X=2x^2-5$$ then $x^2$ (a square) belongs to one of the three sequences defined by the recurrence $$ n_k=38 n_{k-1}-n_{k-2}-90$$ with three initial conditions $$ n_0=1,\ \ n_1=64=8^2$$ $$ n_0=4=2^2,\ \ n_1=121=11^2$$ $$ n_0=16=4^2,\ \ n_1=529=23^2$$

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René Gy
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  • What is the original problem that gives you this sequence? Your numbers are typical of either all the $x$ values or all the $y$ values in some $ax^2 + bxy+ cy^2 = t$ for fixed target $t,$ and $b^2 - 4ac > 0$ but not a square. – Will Jagy Apr 18 '16 at 17:32
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    @Will Jagy : Indeed, it comes from such a quadratic equation. Actually, it comes from this question at the end. I will make a comment in the above question. – René Gy Apr 18 '16 at 18:35

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