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as the title asks, is there an integer which is a perfect square, cube, fourth power, fifth power, etc until, well, it's a tenth power per say? Are there integers that are squares, cubes, and so on until it is a... Say, 100th power?

I was wondering because I saw this olympiad problem which asked for a square root of a number times a cube root of the same number, in which case I thought the best way to solve this would be to think of a number that is both a square and a cube and then work out the answer manually.

A proof or explanation, or any general useful contribution, will be greatly appreciated. Thanks! :)

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What does this have to do with linear algebra? – Chris Eagle Mar 21 '13 at 18:22
Lol I just couldn't think of a suitable tag :P – ANerd Mar 21 '13 at 18:35
up vote 15 down vote accepted

Yes. In general, $x^{\mathrm{lcm}(1,2,\ldots,n)}$ is a perfect first, second, etc. to $n$th power for any integer $x$. This is because the exponent on $x$ is divisible by $1,2,\ldots,n$ (and is in fact the smallest exponent which is). Furthermore, this describes all integers which are perfect first through $n$th powers.

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You could also use $n!$ for the exponent. – lhf Mar 21 '13 at 18:28
@lhf Yes, but I wanted to give the smallest one possible, and emphasize divisibility. – Alex Becker Mar 21 '13 at 18:32
Thanks! Your answer was nice, simple and understandable :) – ANerd Mar 21 '13 at 18:36
1 – Charles Mar 21 '13 at 20:17

This may be trivial, but 0 and 1 are solutions.

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Haha, nice one! Does that mean 0 is also a solution? – ANerd Mar 21 '13 at 18:36
Oops, I just caught that. My bad. – Daniel Geisler Mar 21 '13 at 18:42

Any number $a$ gives $a^{2 \cdot 3 \cdot 2 \cdot 5 \cdot 3 \cdot 7 \cdot 2 \cdot 3 \cdot 5} = a^{37800}$ which is a square, cube, ..., 10-th power.

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Thanks for your contribution! :) – ANerd Mar 21 '13 at 18:37

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