Is there a number which is a perfect square, cube, fourth power and so on?
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! :)
$\endgroup$ 23 Answers
$\begingroup$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.
$\endgroup$ 4 $\begingroup$This may be trivial, but 0 and 1 are solutions.
$\endgroup$ 2 $\begingroup$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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