Look at this formula found by Euler: \(n^2 +n +41\). It has a remarkable property: for every integer number from \(1\) to \(21\) it always produces prime numbers. For example, for \(n=3\) it is \(53\), a prime. For \(n=20\) it is \(461\), also a prime, and for \(n=21\) it is \(503\), prime as well. Could it be that this formula produces a prime number for any natural \(n\)?
Prove that there is a power of \(3\) that ends in \(001\). You can take the following fact as given: if the product \(a\times b\) of two numbers is divisible by another number \(c\), but \(a\) and \(c\) share no prime factors (we say that \(a\) and \(c\) are coprime) then \(b\) must be divisible by \(c\).
Prove that if \((m, 10) = 1\), then there is a repeated unit \(E_n\) that is divisible by \(m\). Will there be infinitely many repeated units?
Prove that for any odd natural number, \(a\), there exists a natural number, \(b\), such that \(2^b - 1\) is divisible by \(a\).