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I know there are no integers for which the value is specifically a perfect square for the first 73 integers because I plugged the function $\sqrt{\lfloor{e^{\lfloor{x}\rfloor}}\rfloor}-\lfloor{\sqrt{\lfloor{e^{\lfloor{x}\rfloor}\rfloor}}}\rfloor$ into Desmos. There are no zeroes up to that point, and this expression would be zero (as the two terms would be the same) if this satisfied the condition.

Desmos has trouble calculating this above 73, so I couldn't check higher than that. I wonder if there is any way to prove that any such integers exist, or whether they are unlikely/likely to exist. I guess that as n gets higher, perfect powers and values of e^n get more spaced out, so it gets less and less likely for them to land on the same integer, however I do not know how to find the probability of this or prove it.

volcanrb
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    Wait. The question in the title makes sense, but the expanded one doesn't. Am I guessing correctly that you just assumed that if $\lfloor{e^n}\rfloor$ was to be a perfect square then the root could only be $e^{\lfloor{n}\rfloor}$? – Arnaud Mortier Feb 11 '18 at 22:56
  • Sorry about that, I just edited it to correct the equation. I was missing a floor sign within the square root. – volcanrb Feb 11 '18 at 22:57
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    Okay it makes more sense now! – Arnaud Mortier Feb 11 '18 at 22:57
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    The title mentions "perfect power" but then you consider only square roots. Do you mean perfect square? – Arnaud Mortier Feb 11 '18 at 22:59
  • I only tested it for square roots, but to expand the scope of the question I decided to include any perfect powers. – volcanrb Feb 11 '18 at 23:00
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    Not sure if that helps but if $\lfloor e^n \rfloor = a^k$ for some integers $a, k \gt 1$ then $e^n \leq a^k \lt e^n + 1$ so $e^{n/k} \leq a \lt e^{n/k}(1 + e^{-n})^{1/k}$. So we need $e^{n/k}$ close to an integer with about $e^{-n}/k$ accuracy. So this is about finding rational powers of $e$ that are very close to integers, or in other point of view, good approximations of $e$ of the form $e = a^r$ for integer $a$ and rational $r$. – Tob Ernack Feb 12 '18 at 01:17

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From the heuristic point of view, say you want $$0<e^n-N^2<1$$ for some integers $n$ and $N$.

It follows from the inequality $$\sqrt{N^2+1}-\sqrt{N^2}<\frac{1}{2N}$$ that such an $n$ would have to satisfy $$|\sqrt{e^n}-N|<\frac{1}{2N}\sim \frac{1}{2\sqrt{e^n}}$$

Therefore you want $\sqrt{e^n}$ to be close to an integer, but how close is actually more and more demanding as $n$ increases. This is not in favour of a positive heuristical argument.

Arnaud Mortier
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