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Working on isometric paths in hypercubes, I came up with the following simple, yet (imo) interesting problem. For what natural numbers $n$ exists a natural number $t$ such that $n^2+n+2=2^t$? The first few terms are $n=0,1,2,5,90$, and these are all below one million. Does someone have any idea how to approach this problem? I basically only want to know whether there exist infinitely many $n$ or not (maybe even that 0, 1, 2, 5, and 90 are the only possible ones).

Thanks,

Sacha

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  • $\begingroup$ Beyond ($n=1,t=2$), $t$ will always be odd, because $n^2 < n^2+n+2 < (n+1)^2$. $\endgroup$ – Colonel Panic Aug 22 '12 at 15:05
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    $\begingroup$ It is easily equivalent to a beautiful relation $$ \sum_{i=1}^{n} i = \sum\limits_{j=1}^{t-1}2^j $$ $\endgroup$ – Ilya Aug 22 '12 at 15:07
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That there are no more positive solutions was proved by Nagell, settling a conjecture of Ramanujan. The problem is discussed in this paper, and in this Wikipedia article.

To see that these solve the problem, a small preliminary transformation of your equation is useful. Rewrite it as $4n^2+4n+8=2^k$, and then as $(2n+1)^2+7=2^k$. We have arrived at the Ramanujan-Nagell equation.

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  • $\begingroup$ Thanks! Reading the Wikipedia article, I just realized I forgot the simplest of all solutions: $n=0$. $\endgroup$ – Sacha Aug 23 '12 at 7:07
  • $\begingroup$ @Sacha: And there area few negatives, which presumably you are not interested in. That's why in the answer I wrote there are no more positive. $\endgroup$ – André Nicolas Aug 23 '12 at 7:10
  • $\begingroup$ Ah yep, thanks. I just rephrased the question accordingly. $\endgroup$ – Sacha Aug 23 '12 at 7:17
  • $\begingroup$ @AndréNicolas : So basically there are only few solutions as in wikipedia article given right ? $\endgroup$ – Theorem Aug 23 '12 at 7:20

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