Consider a ‘square root’ function defined by f:k -> √(1+6k), k∈R, k≥0. It is easily shown that f(2n(3n+1)) = 6n+1, n≥0, and f(2n(3n-1)) = 6n-1, n≥1/3. Clearly then f(2n(3n+1)) = 6n+1 and f(2n(3n-1)) = 6n-1, n∈ω. The set of primes ≥ 5 ⊂ {6n+1: n∈ω} ∪ {6n-1: n∈ω} and so f() has an infinite number of ‘integer’ points including all of those […]