A sequence $$a_1, a_2, a_3, \ldots$$ of real numbers satisfies 

$$\frac{a_{n+3} - a_{n+2}}{a_n - a_{n+1}} = \frac{a_{n+3} + a_{n+2}}{a_n + a_{n+1}}$$

for all $$n \ge 1$$. Suppose $$a_{55} = 6$$, $$a_{66} = 2$$ and $$a_{77} = 1$$. Let $$N$$ denote the sum $$a_1^2 + a_2^2 + \cdots + a_{2026}^2$$. What is the sum of the digits of $$N$$?

Cross-multiplying and cancelling gives $$a_{n+1}a_{n+3} = a_n a_{n+2}$$, so all products $$a_n a_{n+2}$$ are equal and no term can be zero. Cancelling $$a_{n+2}$$ in $$a_n a_{n+2} = a_{n+2} a_{n+4}$$ gives $$a_{n+4} = a_n$$, so the sequence has period 4, and the given indices force $$a_3 = 6$$, $$a_2 = 2$$, $$a_1 = 1$$, hence $$a_4 = 3$$ and the repeating block $$1, 2, 6, 3$$. Each block contributes $$1 + 4 + 36 + 9 = 50$$, and since $$2026 = 4 \times 506 + 2$$ we get $$N = 506 \times 50 + 1 + 4 = 25305$$, whose digit sum is 15.