12

Is $(n,m) = (7,11)$ the only solution in positive integers to the equation

$2^n - n = m^2\text{?}\tag*{}$

It's not hard to verify by direct calculation that there are no other solutions for $n < 10000$ but that's no way to establish a general result. I'm aware that $n=7$ is also a solution to the Ramanujan-Nagell equation, $2^n - 7 = m^2$.

It also seems that $(n,m) = (5,3)$ may be the only solution in positive integers to the equation $2^n - n = m^3.$ So for a bigger challenge, one could consider all solutions in positive integers to the equation

$2^n - n = m^k.\tag*{}$

(Slight correction, thanks to Chickenmancer): there's also the trivial solution $(n,m) = (1,1)$ which solves both the original equation and the generalized one for all $k$.)

Servaes
  • 67,306
  • 8
  • 82
  • 171
Ted Hopp
  • 707
  • 3
    If $n$ is even, then this would make $2^{2n}$ (a square) be much closer to $m^2$ (another square) than two squares can be. Still thinking about the case where $n$ is odd though – Milo Moses Aug 02 '22 at 02:58
  • 2
    Similar: https://artofproblemsolving.com/community/c6h2567861p22022784 – insipidintegrator Aug 02 '22 at 03:09
  • Given @MiloMoses 's comment, for $k=2$, we can write $(2^n+m)(2^n-m) = n, n \equiv 0 \pmod 2$. Not certain how much that helps. – Eric Snyder Aug 03 '22 at 07:05
  • 2
    @EricSnyder - If $n$ is even and $k=2,$ then $m^2$ must be exactly $n$ less than $(2^{n/2})^2.$ However, the gap between consecutive squares is too large. We have $\left(2^{n/2}\right)^2−\left(2^{n/2}−1\right)^2=2^{(n+2)/2}−1 > n.$ A similar argument works, I think, whenever $\gcd(n,k)>1.$ – Ted Hopp Aug 03 '22 at 16:22
  • @Thissitehasbecomeadump. - When $x(2^{n/x}-1)^{x-1}$ is a decreasing function, that last inequality is in the wrong direction. So I think this argument is wrong (or at least incomplete). Also, throwing up a quick graph shows that $x(2^{n/x}-1)^{x-1}$ can increase for a bit after $x>1$. For instance, when $n=6,$ it is increasing past $x=3.$ (I don't know if it's ever increasing at $x=4$ for any $n\text{.)}$ – Ted Hopp Aug 04 '22 at 15:46
  • @Thissitehasbecomeadump. Yes, that's much better. – Ted Hopp Aug 05 '22 at 17:35
  • 2
    In the range $[2,10^5]$ , the only integers $n$ such that $2^n-n$ is a perfect power are $n=5$ and $n=7$. – Peter Aug 07 '22 at 10:42
  • Apparently this problem has been solved by investigating the distribution of squares around powers of $2$, here's the relevant link for the SE post https://math.stackexchange.com/a/4396749/1316192 – Curious Oct 14 '24 at 21:04

0 Answers0