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I am trying to prove that for every integer $n \ge 1$, there exists uniquely determined $a > 0$ and $b > 0$ such that $n = a^2 b$, where $b$ is squarefree.

I am trying to prove this using the properties of divisibility and GCD only. Is it possible?

Let me assume that $n = a^2 b = a'^2b'$ where $a \ne a'$ and $b \ne b$'. Can we show a contradiction now?

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    It is probably worth mentioning that the question to which the previous comments refer is [Show that every n can be written uniquely in the form n=ab, with a square-free and b a perfect square](http://math.stackexchange.com/questions/21282/show-that-every-n-can-be-written-uniquely-in-the-form-n-ab-with-a-squa). (See [revision history](http://math.stackexchange.com/posts/139737/revisions).)2015-06-12

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For existence: given $n\geq1$, there is a maximal-for-divisibility integer $a$ such that $a^2$ divides $n$ and then $n=a^2b$ for some $b$. If $b$ is divisible by the square of a positive integer $c$, then $n$ is divisible by $(ac)^2$, then the maximality of $a$ implies that $ac=a$, that is that $c=1$: this means that $b$ is squarefree.

For uniqueness: suppose now that we also have $n=c^2d$ with $d$ squarefree. The maximality of $a$ implies that $c\mid a$, so there is an $e$ such that $a=ce$, and then from $c^2e^2b=a^2b=c^2d$ we get $e^2b=d$: since $d$ was supposed to be square free, $e=1$. It follows that $a=c$ and $b=d$.

It should be noted that proving there exists a maximal-for-divisibility $a$ such that $a^2\mid n$ depends on the basic properties of the divisibility relation and the GDC. All the rest is more or less tech-free.

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    @MarianoSuárez-Álvarez Then the conclusion fo4 $c\mid a$ requires some addiitonal step with properties of the divisor lattice of $n$, e.g., that $a^2\mid n$ and $c^2\mid n$ implies $\operatorname{lcm}(a,c)^2\mid n$. If instead of "the set of divisors of $n$", we consider the set $S=\{1,2,3,4,6,9\}$, then $S$ contains *two* maximal-for-divisibility squares, namely $2^2$ and $3^2$, but $2\nmid 3$.2017-04-03
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Hint $\ $ If $\rm\: a^2 d =n = b^2 c\:$ for squarefree $\rm\:d,c\:$ then $\rm a\:|\:b\:|\:a\:\Rightarrow\:a=b,\:$ since, by your prior question, for $\rm\: z\:$ squarefree, $\rm\ x^2\:|\:y^2 z\:\Rightarrow\: x\:|\:y,\:$ which we apply twice above, in both directions.

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    @LoneLearner But my proof in the linked thread doesn't use that. Rather, it uses only that $\rm\:c\:$ is squarefree (necessarily true when its cofactor $\rm\:b^2\:$ is a maximal square divisor of $\rm\:n = b^2 c,\:$ else \rm\:d^2\:|\:c,\ d > 1\: $\Rightarrow$ \rm\:(bd)^2\:|\:n,\ bd > b,\: contra maximality of $\rm\:b).$2012-05-02
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For existence, let $a$ be the largest integer, in the usual ordering, such that $a^2$ divides $n$. If $n=a^2q$, then $q$ must be square-free.

For uniqueness, call a positive integer bad if it has two different decompositions $a^2 c$ and $b^2 d$, where $c$ and $d$ are square-free, and $a$ and $b$ are positive. If there are bad positive integers, let $M$ be the smallest bad one.

If $a$ and $b$ are not relatively prime, we can produce a bad positive integer smaller than $M$. So $a$ and $b$ are relatively prime.

We show that $a^2$ and $b^2$ are relatively prime. There are various approaches. One I like is that there exist integers $x$ and $y$ such that $ax+by=1$. Cube both sides. We get $a^2(ax^3+3x^2by)+b^2(3axy^2+by^3)=1,$ which says that $a^2$ and $b^2$ are relatively prime.

Since $a^2c=b^2d$ and $a^2$ and $b^2$ are relatively prime, we have $a^2\mid d$. This contradicts the fact that $d$ is square-free, unless $a=1$. Similarly, $b=1$, and therefore $M$ cannot be bad.