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Does anyone know how to solve $$n^3 + 5n + 6 = 3\cdot 2^{1+n-k} $$ where n,k are natural numbers? I was told that there are prime number arguments that can be used but I am totally stuck. It is a subsection of a non-graded exercise sheet whose date is passed but I would still like to know how to do it. I also know that a pair of solutions are $n=7,k=1$ and $n=23,k=12$.

Many thanks.

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    $\begingroup$ factorize the polynomial and use GCD of factors... $\endgroup$ – MR_BD Feb 20 '16 at 16:07
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HINT:

$$n^3 + 5n + 6 = 3\cdot 2^{1+n-k} $$

We note that $-1$ is a root of the cubic equation $n^3 + 5n + 6 = 0$.

$$(n+1)(n^2-n+6) = 3\cdot 2^{1+n-k} $$ $$(n+1)\{n(n-1)+6\} = 3\cdot 2^{1+n-k} $$ $$n(n+1)(n-1)+6(n+1) = 3\cdot 2^{1+n-k} $$

A few more steps added:

$$\frac{n(n+1)(n-1)}{6}+\frac{(n+1)}{1} = 2^{n-k} $$ $$\frac{n(n+1)(n-1)}{3!}+\frac{(n+1)}{1!} = 2^{n-k} $$ $$\binom{n+1}{3}+\binom{n+1}{1} = 2^{(n+1)-k-1} $$ $$\binom{n}{3}+\binom{n}{1} = 2^{n-k-1} $$

See if this helps.

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  • $\begingroup$ It yields that $\displaystyle{m\choose3}+{m\choose1}$ is a power of two, where $m=n+1.$ $\endgroup$ – Lucian Feb 21 '16 at 0:40
  • $\begingroup$ Sorry, even after expanding the equation I am still lost, how do you know that it is a power of 2? $\endgroup$ – Rupert Gillyweed Feb 25 '16 at 19:28
  • $\begingroup$ @RupertGillyweed See my edit. I have updated my answer. $\endgroup$ – SchrodingersCat Feb 26 '16 at 12:10

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