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A positive integer $n$ is called square-full or powerfull if $p^2|n$ for every factor $p$ of $n$ for every prime factor $p$.

I need to show that if $n$ is square-full show so it can be written in the form $n=a^2b^3$ for $a,b$ positive integer.

I looked in wikipedia but the proof was too complicated... is there any easy way to prove this?

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3 Answers 3

up vote 5 down vote accepted

Let $p^r$ be the exact power of $p$ dividing $n$. By hypothesis $r\geq2$.

If $r$ is even put $p^r$ under the '$A$' label. If $r$ is odd then $r\geq3$ so you can put $p^3$ under the '$B$' label and $p^{r-3}$ under the '$A$' label.

Do the above for all $p$ dividing $n$. When you are finished you have that $B$ is a product of cubes, thus itself a cube $B=b^3$, and $A$ is a product of primes with even exponents, thus a square $A=a^2$.

Obviously, $n=AB=a^2b^3$.

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Suppose $m \geq 2$ is an integer. Then $m = 2\alpha + 3\beta$, where $\alpha$ and $\beta$ are nonnegative integers. The claim follows from this fact, because for any prime $p$ dividing a square-full number $n$, the power of $p$ in the prime factorization of $n$ is $\geq 2$.

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Take all the primes $p_{\text{odd}}$ which divide $n$ an odd number of times (at least three each). Then let $$b= \prod p_{\text{odd}}$$ and $$a = \sqrt{\frac{n}{b^3}}.$$

All you need to show is that $a$ is a positive integer, as $b$ is by definition (noting that the empty product is $1$). Do this by looking first at primes which divide $n$ an odd number of times (at least three each so when you subtract $3$ you are left with a non-negative even exponent), and then at primes which divide $n$ an even number of times.

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