This is just an equivalent formulation:

Find squares in the linear recurrence sequence (besides first two terms):

9, 25, 481, 14961, 500889, 16973353, ...

which can be described equivalently by

• recurrence $t_n = 41 t_{n-1} - 246 t_{n-2} + 246 t_{n-3} - 41 t_{n-4} + t_{n-5}$ for $n\geq 5$;
• generating function $$\frac{9 - 344x + 1670x^2 - 824x^3 + 33x^4}{(1-x)(1-6x+x^2)(1-34x+x^2)}$$
• explicit formula $$t_n = \frac38 \big((1+\sqrt2)^{4n} + (1-\sqrt2)^{4n}\big) + \frac{4-\sqrt2}2 (1+\sqrt2)^{2n} + \frac{4+\sqrt2}2 (1-\sqrt2)^{2n} + \frac{17}4.$$

Suppose that $t_n=z^2$ for some integer $z$, then values of $a,b,c$ are given by $$\begin{cases} a = \frac{(1+\sqrt2)^{2n} - (1-\sqrt2)^{2n}}{4\sqrt2}; \\ b = \frac{(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n} + 2}{4}; \\ c = \frac{z-1}2. \end{cases}$$

I believe it a hard question to determine squares in linear recurrences like that. It is more or less solved only for recurrences of the second order (like Fibonacci numbers).

This is just an equivalent formulation:

Find squares in the linear recurrence sequence (besides first two terms):

$$9, 25, 481, 14961, 500889, 16973353, \dotsc$$

which can be described equivalently by

• recurrence $t_n = 41 t_{n-1} - 246 t_{n-2} + 246 t_{n-3} - 41 t_{n-4} + t_{n-5}$ for $n\geq 5$;
• generating function $$\frac{9 - 344x + 1670x^2 - 824x^3 + 33x^4}{(1-x)(1-6x+x^2)(1-34x+x^2)}$$
• explicit formula $$t_n = \frac38 \big((1+\sqrt2)^{4n} + (1-\sqrt2)^{4n}\big) + \frac{4-\sqrt2}2 (1+\sqrt2)^{2n} + \frac{4+\sqrt2}2 (1-\sqrt2)^{2n} + \frac{17}4.$$

Suppose that $t_n=z^2$ for some integer $z$, then values of $a,b,c$ are given by $$\begin{cases} a = \frac{(1+\sqrt2)^{2n} - (1-\sqrt2)^{2n}}{4\sqrt2}; \\ b = \frac{(1+\sqrt2)^{2n} + (1-\sqrt2)^{2n} + 2}{4}; \\ c = \frac{z-1}2. \end{cases}$$

I believe it is a hard question to determine squares in linear recurrences like that. It is more or less solved only for recurrences of the second order (like Fibonacci numbers).