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I saw a question here with the same title but it was not answered properly, or I just couldn't extract the solution from the comments/answers. This is an exercise from Beachy and Blair: Abstract algebra. A positive integer is called $\textbf{square-free}$ if it is a product of distinct primes. Prove that every positive integer can be written uniquely as a product of a square and a square-free integer.

Here is my try:

Let $n\in\mathbb{N}^+$, by the Fundamental theorem of arithmetic we can write $n$ uniquely as $$ n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\cdot\ldots\cdot p_n^{\alpha}. $$ with $p_n$ being distinct primes. Now $\alpha_i$ is either even or odd, lets rename the odd ones as $\beta_j$, then we have $$ \alpha_i=2k_i\qquad \beta_j=2l_j+1\qquad k_i,l_j\in\mathbb{Z}^+, \ \text{and} \ i+j=n. $$ Since multiplication is commutative we can rearrange the factors so all the even exponents come first so $$ n=p_1^{2k_1}\cdot p_2^{2k_2}\cdot\ldots\cdot p_i^{2k_i}\cdot p_{n-j+1}^{2l_{n-j+1}+1}\cdot p_{n-j+2}^{2l_{n-j+2}+1}\cdot \ldots\cdot p_n^{2l_n+1}= $$ $$ n=p_1^{2k_1}\cdot p_2^{2k_2}\cdot\ldots\cdot p_i^{2k_i}\cdot p_{n-j}^{2l_{n-j+1}}\cdot p_{n-j+1}\cdot p_{n-j+2}^{2l_{n-j+2}}\cdot p_{n-j+2}\cdot \ldots\cdot p_n^{2l_n}\cdot p_{n}. $$ Again by commutativity we can arrange so all the factors with exponent "$1$" get to the end of the expression so

$$ n=p_1^{2k_1}\cdot p_2^{2k_2}\cdot\ldots\cdot p_i^{2k_i}\cdot p_{n-j+1}^{2l_{n-j+1}}\cdot p_{n-j+2}^{2l_{n-j+2}}\cdot \ldots\cdot p_n^{2l_n}\cdot p_{n-j+1}\cdot p_{n-j+2}\cdot\ldots\cdot p_{n}. $$ finally $$ n=\underbrace{(p_1^{k_1}\cdot p_2^{k_2}\cdot\ldots\cdot p_i^{k_i}\cdot p_{n-j+1}^{l_{n-j+1}}\cdot p_{n-j+2}^{l_{n-j+2}}\cdot \ldots\cdot p_n^{l_n})^2}_{a^2}\cdot \underbrace{p_{n-j+1}\cdot p_{n-j+2}\cdot\ldots\cdot p_{n}}_{b}. $$ And we are done since this factorisation was unique from the beginning. Now one of my problems is that this doesn't look too nice. In my head it looks simpler ("bringing down" the ones from odd exponents) but the final product looks like a mess. Is there any easier way to describe this? (maybe I could have renamed the primes with odd exponents as $q$?) I mean if it is correct at all. And I have a further question, can I get any HINTS how to do this by induction? Thank you in advance!

Vinyl_cape_jawa
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  • I think what you have done is fine and concise. – SchrodingersCat Oct 07 '17 at 16:10
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    I was always wondering how do you guys find duplicate questions so fast, I mean I was trying to find this question here in MSE but I never came across this one (I mean the one you marked as being a duplicate of mine, or marked mine as being a duplicate of :)) am I this bad at searching for questions here? any hints on that. Actually it is not a "possible duplicate" mine is a duplicate, maybe the request in the end for hints about induction is making it different, but you get my point – Vinyl_cape_jawa Oct 07 '17 at 16:17
  • Aren't there some edge conditions here? Like, is $1$ supposed to be a "square-free integer"? (And also note the FToA applies to numbers $>1$.) Also, what is the correct factorization for squares, e.g. $4$? – geometrian Oct 07 '17 at 17:48
  • I think one has to be counted as both since 1=1$\cdot 1$ – Vinyl_cape_jawa Oct 07 '17 at 18:01

2 Answers2

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You could write the product as $$n=\prod p_i^{2r_i+\epsilon_i}$$ where $\epsilon_i$ is the remainder on dividing the exponent by $2$ (zero or one) so that $$n=\left(\prod p_i^{r_i}\right)^2\cdot\prod p_i^{\epsilon_i}$$This does not involve distinguishing between primes with even/odd exponent - the $\epsilon$s take care of that. You can also take the product over all primes, if you want to, because most of the time the exponent is zero (for all but a finite number of primes).

Mark Bennet
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Let $k$ be the largest square divisor of the given integer $n$. Then $n/k=a$ is square free (by our choice). In particular, $n=ak$.

Sri-Amirthan Theivendran
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