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.
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.