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Prove or disprove that for a fixed $n \in N$, there exists at least one prime among the integers of the form $2^{k}n+2^k-1$ for an arbitrary $k \in N$.

tesgoe
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  • Who says????????????? – Will Jagy Jul 20 '14 at 21:08
  • Will Jagy, are you asking about the source of this question? If yes, the source: http://mishabucko.wordpress.com/2014/05/09/quotes-that-i-think-of-while-eating-and-learning-to-learn-about-thinking/ – tesgoe Jul 20 '14 at 21:10
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    True for $n=k=1$ – frabala Jul 20 '14 at 21:34
  • Do you want a prove of this theorem/statement? Please don't just state something. Also, please explain what you have tried yourself to prove this, so we can help you better. For a proof, I would start to assume all numbers in the row $(2^k(n+1)-1)_{k\in \mathbb N}$ are prime and try to derive a contradiction. Or, even better, just give a factorisation of $2^k(n+1)-1$ for some $k\in \mathbb N$. – Ragnar Jul 20 '14 at 22:29
  • Ragnar, I added "prove or disprove" at the beginning of the problem. – tesgoe Jul 21 '14 at 06:41
  • Swap out $n+1$ for $n$, and the quantity is the same as $2^kn-1$. So you are asking if every odd $n$ can be paired with a power of $2$ coefficient to get one less than a prime? – 2'5 9'2 Jul 21 '14 at 06:56
  • I feel like the "for an arbitrary $k\in N$" is unnecessary at best, and confusing at worst. If it should be kept to clarify that the exponents under consideration on the $2$ are natural numbers, then the phrase could be replaced with "with $k\in\mathbb{N}$". – 2'5 9'2 Jul 21 '14 at 07:02

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There exist infinitely many values of $n$ for which $2^k n+2^k-1 = 2^k(n-1)-1$ is composite for every positive integer $k$. The integers $n-1$ with this property are called Riesel numbers (related to Sierpinski numbers). The smallest known Riesel number is $n-1= 509$,$203$. Proving that this is a Riesel number is related to covering systems of congruences.

Greg Martin
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