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Let Fₙ be the Fibonacci sequence defined by F₀=0, F₁=1, $F_{n+1}=Fₙ+F_{n-1}\left(n\ge 1\right)$. It is well known that $F_{p-(5/p)}\equiv 0\left(modp\right)$ for any odd prime p, where (-) denotes the Legendre symbol. In 1960 D. D. Wall [13] asked whether $p²|F_{p-(5/p)}$ is always impossible; up to now this is still open. In this paper the sum ${\sum }_{k\equiv r\left(mod10\right)}\left(\genfrac{}{}{0pt}{}{n}{k}\right)$ is expressed in terms of Fibonacci numbers. As applications we obtain a new formula for the Fibonacci quotient $F_{p-(5/p)}/p$ and a criterion for the relation $p|F_{(p-1)/4}$ (if p ≡ 1 (mod 4), where p ≠ 5 is an odd prime. We also prove that the affirmative answer to...

Let p be an odd prime and let a be a positive integer. In this paper we investigate the sum $\genfrac{}{}{0pt}{}{{\sum }_{k=0}^{p^a-1}(h{p}^a-1}{\genfrac{}{}{0pt}{}{k\left)\right(2k}{k)/m^k\left(modp²\right)}}$, where h and m are p-adic integers with m ≢ 0 (mod p). For example, we show that if h ≢ 0 (mod p) and $p^a>3$, then $\genfrac{}{}{0pt}{}{{\sum }_{k=0}^{p^a-1}(h{p}^a-1}{\genfrac{}{}{0pt}{}{k\left)\right(2k}{{k)(-h/2)}^k\equiv ((1-2h)/\left(p^a\right))(1+h({(4-2/h)}^{p-1}-1))\left(modp²\right)}}$, where (·/·) denotes the Jacobi symbol. Here is another remarkable congruence: If $p^a>3$ then $\genfrac{}{}{0pt}{}{{\sum }_{k=0}^{p^a-1}(p^a-1}{\genfrac{}{}{0pt}{}{k\left)\right(2k}{{k)(-1)}^k\equiv 3^{p-1}(p^a/3)\left(modp²\right)}}$.

Let $p$ be an odd prime, and let $a$ be an integer not divisible by $p$. When $m$ is a positive integer with $p\equiv 1(mod2m)$ and $2$ is an $m$th power residue modulo $p$, we determine the value of the product ${\prod }_{k\in R_m\left(p\right)}(1+tan(\pi ak/p))$, where $$R_m\left(p\right)=\{0<k<p:k\in \mathbb{Z}\text{is}\text{an}m\text{th}\text{power}\text{residue}\text{modulo}p\}.$$ In particular, if $p=x^2+64y^2$ with $x,y\in \mathbb{Z}$, then $$\prod _{k\in R_4\left(p\right)}\left(1+tan\pi \frac{ak}{p}\right)={(-1)}^y{(-2)}^{(p-1)/8}.$$