The $$n^{\text{th}}$$ spread polynomial, denoted by $S_n(s)$, is recursively defined by $S_0(s) \equiv 0$, $S_1(s) \equiv s$, and

$S_n(s) \equiv 2(1-2s) S_{n-1}(s) - S_{n-2}(s) + 2s,$

for $n \geq 2$. We will, for the context of this note, restrict our discussion to the finite field $\mathbb{F}_p$, where $p$ is an odd prime. As an example, we see that in $\mathbb{F}_3$,

$S_{0}(0) = 0 \quad s_{0}(1) = 0 \quad s_{0}(2) = 0$

$S_{1}(0) = 0 \quad s_{1}(1) = 1 \quad s_{1}(2) = 2$

$S_{2}(0) = 0 \quad s_{2}(1) = 0 \quad s_{2}(2) = 1$

$S_{3}(0) = 0 \quad s_{3}(1) = 1 \quad s_{3}(2) = 0$

$S_{4}(0) = 0 \quad s_{4}(1) = 0 \quad s_{4}(2) = 2$

$S_{5}(0) = 0 \quad s_{5}(1) = 1 \quad s_{5}(2) = 1$

$S_{6}(0) = 0 \quad s_{6}(1) = 0 \quad s_{6}(2) = 0$

In fact, for any positive integer $n$ we will have that $S_n(s) = S_{n+6}(s)$. Thus, a natural question to be asked is whether this phenomenon is also seen for arbitrary finite fields; as it turns out, it does and in fact we have $S_n(s) = S_{n+k}(s)$ in $\mathbb{F}_p$, where $k \equiv \frac{1}{2}(p-1)(p+1)$. How do we prove that this is true? Are we able to discern other patterns within particular values of $s$? If so, how do we ascertain this mathematically?

Note by A Former Brilliant Member
2 years, 7 months ago

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