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Is $\sqrt1*\sqrt2*\dots*\sqrt n$ ever an integer?

Note: Deusovi has already given a nice and simple answer. I still wish to welcome more answers as I want to see if there are other ways to approach this question .

Source: I came up with the above question myself after seeing the following question :

Is $\sqrt1+\sqrt2+\dots+\sqrt n$ ever an integer?

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Yes: $\sqrt{1}$ is an integer.

After that, no. If you've multiplied up to $n$, Bertrand's postulate states that there's always a new prime somewhere from $\frac12n$ to $n$. So there's always at least one prime that doesn't have a partner inside the square root: in other words, the number $1×2×3×\cdots$ cannot be a perfect square.

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  • $\begingroup$ Is there a simpler answer to the question ? Using logarithms or something else ? $\endgroup$
    – Hemant Agarwal
    Commented Jun 16, 2021 at 21:57
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    $\begingroup$ @HemantAgarwal I'm not sure how much simpler you can get than this. Other than the invocation of Bertrand's postulate (which is pretty intuitive, IMO), the solution is just simple properties of square roots. $\endgroup$
    – Deusovi ♦
    Commented Jun 16, 2021 at 22:01
  • $\begingroup$ yes..yours is a much simpler answer than I initially thought of it to be .. But, does this comprehensively prove that these products cannot be an integer ? What I mean is is that even though, the square root of the prime number(s) which cannot be paired up, will be a fraction, don't we also need to prove that the products of all these fractions cannot result in an integer ? $\endgroup$
    – Hemant Agarwal
    Commented Jun 16, 2021 at 22:21
  • $\begingroup$ @HemantAgarwal Deusovi's implied and almost-entirely-written-out proof by contradiction: In order for the square root to be a perfect square, there must be no prime numbers under the radical (i.e, from 1 to n). However, Bertrand's Postulate says there's at least one between 0.5n and n (which covers the "n=2" case). Therefore, there is always a prime number under the radical, which means the number cannot be a perfect square, and therefore the radical's product cannot be an integer $\endgroup$
    – Punintended
    Commented Jun 16, 2021 at 22:49
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    $\begingroup$ @HemantAgarwal The product of square roots is the square root of the product, at least for nonnegative numbers. And a square-root-of-something is only an integer if the thing under the root is a square number. (It might be clearer with an example: $√3×√5×√7$ is not an integer, because it's equal to $\sqrt{3×5×7}$, and $3×5×7$ is not a square number.) $\endgroup$
    – Deusovi ♦
    Commented Jun 16, 2021 at 23:03
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Possibly, depending on what sequence you use

Whilst the default assumption is that $1, 2, ... n$ would be filled in with an arithmetic sequence where each member is exactly 1 higher than the previous, there could be other contexts in which a different sequence is implied, such as a geometric sequence $1, 2, 4, 8, ..., n$ (with $n$ implicitly a power of 2), or a fibonacci sequence $1, 2, 3, 5, 8, ..., n$, or anything many other sequences mentioned in the OEIS or elsewhere. Typically when another sequence is used, more members are given (as in these examples) so that "this is not a simple counting sequence" is made clear, but the syntax is imprecise to start with, so there aren't clear-cut rules on that.

With plenty of sequences to choose whose members in appropriate context could be summarised as $1, 2, ... n$, there are many ways this question could be interpreted to allow for a this being an integer.

As a simple example:

$\sqrt1*\sqrt2*\sqrt4*\sqrt8 = \sqrt{64} = 8$, which is an integer, and indeed half of the members of this continued sequence are also integers - at least if you extend it in the same way I'm assuming you will!

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