What should be assumed about the integral polynomials $f₁(x),...,f_{k}(x)$ in order that the solvability of the congruence $f₁(x)f₂(x) ⋯ f_{k}(x) ≡ 0 (mod p)$ for sufficiently large primes p implies the solvability of the equation $f₁(x)f₂(x) ⋯ f_{k}(x) = 0$ in integers x? We provide some explicit characterizations for the cases when $f_j(x)$ are binomials or have cyclic splitting fields.
Consider a recurrence sequence $(x_{k})_{k∈ℤ}$ of integers satisfying $x_{k+n} = a_{n-1}x_{k+n-1} + ... + a₁x_{k+1} + a₀x_{k}$, where $a₀,a₁,...,a_{n-1} ∈ ℤ$ are fixed and a₀ ∈ {-1,1}. Assume that $x_{k} > 0$ for all sufficiently large k. If there exists k₀∈ ℤ such that $x_{k₀} < 0$ then for each negative integer -D there exist infinitely many rational primes q such that $q|x_{k}$ for some k ∈ ℕ and (-D/q) = -1.
7
Dostęp do pełnego tekstu na zewnętrznej witrynie WWW
The product of consecutive integers cannot be a power (after Erdős and Selfridge), but products of disjoint blocks of consecutive integers can be powers. Even if the blocks have a fixed length l ≥ 4 there are many solutions. We give the bound for the smallest solution and an estimate for the number of solutions below x.
8
Dostęp do pełnego tekstu na zewnętrznej witrynie WWW