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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 for any odd natural number, \(a\), there exists a natural number, \(b\), such that \(2^b - 1\) is divisible by \(a\).