Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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! :)

share|improve this question
    
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

3 Answers 3

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.

share|improve this answer
1  
You could also use $n!$ for the exponent. –  lhf Mar 21 '13 at 18:28
6  
@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  
oeis.org/A003418 –  Charles Mar 21 '13 at 20:17

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

share|improve this answer
1  
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.

share|improve this answer
    
Thanks for your contribution! :) –  ANerd Mar 21 '13 at 18:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.