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### Fibonacci numbers and Fermat's last theorem

Acta Arithmetica

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 ${F}_{p-\left(5/p\right)}\equiv 0\left(modp\right)$ for any odd prime p, where (-) denotes the Legendre symbol. In 1960 D. D. Wall [13] asked whether $p²|{F}_{p-\left(5/p\right)}$ 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 ${F}_{p-\left(5/p\right)}/p$ and a criterion for the relation $p|{F}_{\left(p-1\right)/4}$ (if p ≡ 1 (mod 4), where p ≠ 5 is an odd prime. We also prove that the affirmative answer to...

Acta Arithmetica

### Finite coverings of groups

Fundamenta Mathematicae

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

### On sums of binomial coefficients modulo p²

Colloquium Mathematicae

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

### The tangent function and power residues modulo primes

Czechoslovak Mathematical Journal

Let $p$ be an odd prime, and let $a$ be an integer not divisible by $p$. When $m$ is a positive integer with $p\equiv 1\phantom{\rule{4.44443pt}{0ex}}\left(mod\phantom{\rule{0.277778em}{0ex}}2m\right)$ and $2$ is an $m$th power residue modulo $p$, we determine the value of the product ${\prod }_{k\in {R}_{m}\left(p\right)}\left(1+tan\left(\pi ak/p\right)\right)$, where ${R}_{m}\left(p\right)=\left\{0 In particular, if $p={x}^{2}+64{y}^{2}$ with $x,y\in ℤ$, then $\prod _{k\in {R}_{4}\left(p\right)}\left(1+tan\pi \frac{ak}{p}\right)={\left(-1\right)}^{y}{\left(-2\right)}^{\left(p-1\right)/8}.$

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

Acta Arithmetica

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