Every integer n>1 is the product of a square-free integer?

[Math100]

Homework Statement

An integer is said to be square-free if it is not divisible by the square of any integer greater than $1$. Prove the following:
(b) Every integer $n>1$ is the product of a square-free integer and a perfect square.
[Hint: If $n=p_{1}^{k_{1}} p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}$ is the canonical factorization of $n$, then write $k_{i}=2q_{i}+r_{i}$ where $r_{i}=0$ or $1$ according as $k_{i}$ is even or odd.]

Relevant Equations

None.

Proof:

Suppose $n>1$ is a positive integer.
Let $n=p_{1}^{k_{1}} p_{2}^{k_{2}}\dotsb p_{r}^{k_{r}}$ be the prime factorization of $n$
such that each $k_{i}$ is a positive integer and $p_{i}'s$ are prime for $i=1,2,3,...,r$ with
$p_{1}<p_{2}<p_{3}<\dotsb <p_{r}$.
Applying the Division Algorithm produces:
$k_{i}=2q_{i}+r_{i}$ for $0\leq r_{i}<2$,
where there exist unique integers $q_{i}$ and $r_{i}$.
Since $0\leq r_{i}<2$,
it follows that $k_{i}=2q_{i}$ or $k_{i}=2q_{i}+1$.
Now we consider two cases.
Case #1: Suppose $k_{i}=2q_{i}$.
Then we have $n=p_{1}^{k_{1}} p_{2}^{k_{2}}\dotsb p_{r}^{k_{r}}$
$=p_{1}^{2q_{1}} p_{2}^{2q_{2}}\dotsb p_{r}^{2q_{r}}$
$=(p_{1}^{q_{1}} p_{2}^{q_{2}}\dotsb p_{r}^{q_{r}})^2$
$=1\cdot (p_{1}^{q_{1}} p_{2}^{q_{2}}\dotsb p_{r}^{q_{r}})^2$.
Thus, $1$ is a square-free integer and $(p_{1}^{q_{1}} p_{2}^{q_{2}}\dotsb p_{r}^{q_{r}})^2$
is a perfect square.
Case #2: Suppose $k_{i}=2q_{i}+1$.
Then we have $n=p_{1}^{k_{1}} p_{2}^{k_{2}}\dotsb p_{r}^{k_{r}}$
$=p_{1}^{2q_{1}+1} p_{2}^{2q_{2}+1}\dotsb p_{r}^{2q_{r}+1}$
$=(p_{1} p_{2}\dotsb p_{r})(p_{1}^{2q_{1}} p_{2}^{2q_{2}}\dotsb p_{r}^{2q_{r}})$
$=(p_{1} p_{2}\dotsb p_{r})(p_{1}^{q_{1}} p_{2}^{q_{2}}\dotsb p_{r}^{q_{r}})^2$.
Thus, $p_{1} p_{2}\dotsb p_{r}$ is a square-free integer and $(p_{1}^{q_{1}} p_{2}^{q_{2}}\dotsb p_{r}^{q_{r}})^2$
is a perfect square.
Therefore, every integer $n>1$ is the product of a square-free integer and a perfect square.

 

[PeroK]

The primes factors will typically have a mixture of odd and even powers. You've assumed that either they are all even or they are all odd. That won't do at all.

 

[Math100]

The primes factors will typically have a mixture of odd and even powers. You've assumed that either they are all even or they are all odd. That won't do at all.

Then how should I revise this proof?

 

[nuuskur]

1. Suppose $k_i = 2q_i$ for every $i$.
2. Suppose $k_i=2q_i+1$ for at least one $i$.

What did I tell you about quantifying your statements? Convince yourself that these are two mutually exclusive cases and that if you show the result in both cases you will have completed the proof.

Is the statement true for perfect squares? Why is the statement true for something like $2\cdot 3$, say?

 

[Math100]

1. Suppose $k_i = 2q_i$ for every $i$.
2. Suppose $k_i=2q_i+1$ for at least one $i$.

What did I tell you about quantifying your statements? Convince yourself that these are two mutually exclusive cases and that if you show the result in both cases you will have completed the proof.

Is the statement true for perfect squares? Why is the statement true for something like $2\cdot 3$, say?

Can you please specify 'perfect squares'? Are you talking about $(k_i)^2=(2q_i)^2$ or something else?

 

[nuuskur]

"Perfect square" appears in your problem statement. My assumption is you know all relevant definitions. If not, then make sure you do before you attempt to prove the statement.

Spoiler

On a cynical note - do not try to trick your grader into giving you definitions. I don't take kindly to that with my students and I bet a full wheel of mince pizza that others wouldn't, either.