Here when I was studying almost integers , I made the following conjecture -
Let $x$ be a natural number then For sufficiently large $n$ (Natural number) Let $$\Omega=(\sqrt x+\lfloor \sqrt x \rfloor)^n$$ then $\Omega$ is an almost integer . The value of $n$ depends upon the difference between the number $x$ and its nearest perfect square which is smaller than it.
Can anyone prove this conjecture.
Moreover, I can provide examples like
$(\sqrt 5+2)^{25}=4721424167835364.0000000000000002$
$(\sqrt 27+5 )^{15}=1338273218579200.000000000024486$

  • $\begingroup$ It would be more interesting if you constrained the notion of almost integer. The difference $|\Omega-\text{integer}|$ should be smaller than what? $\endgroup$ – Ian Mateus Oct 20 '13 at 13:02
  • $\begingroup$ @IanMateus it should be smaller than $0.0001$ $\endgroup$ – Shivam Patel Oct 20 '13 at 13:03
  • $\begingroup$ @ChristianBlatter , yes there are some corrections and I have made them $\endgroup$ – Shivam Patel Oct 20 '13 at 13:05
  • $\begingroup$ Well, your omega describes Pisot-Vijayaraghavan numbers (en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number) and thus, they all converge to integers (by definition). The Fibonacci numbers are examples. There are many others, in fact, an infinite number of higher order roots (e.g., cubics, derived from the Plastic constant, which is also a Pisot number) that behave like this. $\endgroup$ – Stefan Gruenwald Jul 25 '16 at 21:23
  • $\begingroup$ Perhaps $\big(\sqrt{x}+\lfloor{\sqrt{x}}\rfloor\big)^{x^2}$ is an almost integer? $\endgroup$ – Mr Pie May 27 '18 at 12:43

Well, we have that $$(\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n} + (-\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n} $$ is an integer (by the Binomial Theorem), but $(-\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n}\to 0$ if $\sqrt{x}$ was not already an integer.

  • $\begingroup$ Neat! This happens in the closed form for the Fibonacci numbers. $\endgroup$ – Ian Mateus Oct 20 '13 at 13:15

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