The Fermat equation is solved in integral two by two matrices of determinant one as well as in finite order integral three by three matrices.

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...